Infrared radiation or infrared rays, this is electromagnetic radiation occupying the spectral region between red light (with a wavelength of 0.74 microns) and short-wave radio radiation (1-2 mm).

The discovery of infrared radiation occurred in 1800.
The English scientist W. Herschel discovered that in the obtained spectrum of the Sun beyond the red light boundary (ie, in the invisible part of the spectrum), the temperature of the thermometer rises. A thermometer placed outside the red part of the solar spectrum showed an increased temperature compared to control thermometers located on the side.

The infrared region of the spectrum according to the international classification is divided by:
- to near IR-A (from 0.7 to 1.4 µm);
- average IR-B (1.4 - 3 microns);
- far IR-C (over 3 microns).

All heated solids emit a continuous infrared spectrum. This means that in the radiation there are waves with all frequencies without exception, and talking about radiation at a particular wave is a meaningless exercise. A heated solid body radiates in a very wide range of wavelengths.

At low temperatures (below 400°C), the radiation of a heated solid body is located almost entirely in the infrared region, and such a body appears dark. As the temperature rises, the fraction of radiation in the visible region increases, and the body at first appears:

Dark red...................470-650°C

Cherry red ...............700°С

Light red.............800°C

Dense orange .............. 900 ° C

Orange-yellow .............. 1000°С

Light yellow ................1100°С

Straw yellow...........1150°С

White of different brightness......1200-1400°C

In this case, both the total energy of radiation and the energy of infrared radiation increase. At temperatures above 1000°C, a heated body begins to emit ultraviolet radiation.

The laws of thermal radiation

A special place in the theory of thermal radiation is occupied by the Absolute Black Body (ABB). So G. Kirchhoff called the body, in which at all frequencies and at any temperatures the absorption capacity is equal to one. A real body always reflects part of the energy of the radiation incident on it. Even soot approaches the properties of a completely black body only in the optical range.

An absolutely black body is a reference body in the theory of thermal radiation. And, although there is no absolutely black body in nature, it is enough to simply implement a model for which the absorptivity at all frequencies will differ negligibly from unity. Below are the laws that are valid for blackbody.

Planck's basic law of thermal radiation establishes the dependence of the emissivity of the body R on the wavelength λ and body temperature T.

The dependence of R on the wavelength at a constant temperature is shown in the figure. The radiation power has a maximum at a certain value λ max.

Although the spectrum changes with temperature, it has general patterns that do not depend on T if the waves are expressed in a dimensionless unit λ /λ max. Then the share of the radiated energy in different areas does not depend on temperature (the share in % of the total energy is shown in the figure). It is useful to remember that approximately 90% of the energy falls on the spectral intervalλ /λ max = 0.5 ... 3.0, i.e. from l max /2 to 3 l max .

Wien's displacement law . Wavelength lmax , corresponding to the maximum spectral density of the emissivity of the blackbody, inversely proportional to the temperature: l max = 2.9/T,where C is a constant.

LawStefan-Boltzmann. Blackbody emissivity, i.e. total radiation power with unit area, proportional to the fourth power of temperature: R= σT 4, where σ is the Stefan-Boltzmann constant.

In the theory of thermal radiation, an idealized model of real bodies is often used - the concept of a "gray body". A body is called "gray" if its absorption coefficient is the same for all frequencies and depends only on the temperature of the material and the state of its surface. In fact, a real physical body in its characteristics approaches a gray body only in a narrow range of radiation frequencies.

Kirchhoff's law of thermal radiation. The ratio of the spectral density of the energy luminosity of a body to its monochromatic absorption coefficient does not depend on the material of the body (that is, it is the same for all bodies) and is equal to the spectral density of the energy luminosity of a completely black body. This quantity is a function of only the temperature and radiation frequency.

Consequences of Kirchhoff's law.

Since the absorption coefficient for any body is less than unity, then the emissivity of any body for a given radiation frequency is less than that of a black body. In other words, a black body at any temperature and frequency of radiation is the most intense source of radiation.

If a body does not absorb radiation in any region of the spectrum, then it does not radiate in that region of the spectrum.

For a given temperature, those gray bodies that have a large absorption coefficient radiate more strongly.

AND intensity of irradiation from a heated surface or through the hole in the furnace can be determined by the formula (at L ≥F 0.5)

E \u003d 0.91F ((T / 1000) 4 -A) / L 2

where E is the irradiation intensity, W/m2; F is the area of ​​the radiating surface, m2; l - distance from the center of the radiating surface to the irradiated object, m; A = 85 - for human skin and cotton fabric; A \u003d 100 is a constant coefficient for cloth.

18.1. Find the temperature T of the furnace if it is known that the radiation from a hole in it with an area S = 6.1 cm 2 has a power N = 34.6 W. The radiation is considered to be close to the radiation of an absolutely black body.

18.2. What is the N radiation power of the Sun? The radiation of the Sun is considered close to the radiation of a completely black body. The temperature of the surface of the Sun T = 5800 K.

18.3. What energy luminosity R" E has hardened lead? The ratio of the energy luminosities of lead and a black body for a given temperature k =0.6.

18.4. Radiation power of a black body N = 34 kW. Find temperature T this body, if it is known that its surface S\u003d 0.6 m 2.

18.5. The radiation power of a hot metal surface is N = 0.67 kW. Surface temperature T = 2500K, its area S = 10 cm2. What radiation power N would this surface have if it were absolutely black? Find the ratio k of the energy luminosities of this surface and a completely black body at a given temperature.

18.6. The diameter of the tungsten filament in a light bulb d= 0.3 mm, length of the spiral l = 5 cm. U 127 V current flows through the bulb I \u003d 0.31 A. Find the temperature T spirals. Assume that after equilibrium is established, all the heat released in the filament is lost as a result of radiation. The ratio of the energy luminosities of tungsten and a black body for a given temperature is k = 0.31.

18.7. The temperature of a tungsten filament in a 25-watt light bulb is T = 2450 K. The ratio of its luminosity to the luminosity of a black body at a given temperature k = 0.3 . Find the area S of the radiating surface of the spiral.

18.8. Find the solar constant K, i.e., the amount of radiant energy sent by the Sun per unit time through a unit area perpendicular to the sun's rays and located at the same distance from it as the Earth. The temperature of the Sun's surface is T = 5800K. The radiation of the Sun is considered close to the radiation of a completely black body.

18.9. Assuming that the atmosphere absorbs 10% of radiant energy,. sent by the Sun, find the radiation power N received from the Sun by a horizontal section of the Earth with an area S= 0.5 ha. The height of the Sun above the horizon is φ = 30°. The radiation of the Sun is considered close to the radiation of a completely black body.

18.10. Knowing the value of the solar constant for the Earth (see problem 18.8), find the value of the solar constant for Mars.

18.11. What energy luminosity R e does an absolutely black body have if the maximum spectral density of its energy luminosity falls at a wavelength λ = 484 nm?

18.12. Radiation power of a completely black body N = 10 kW Find the area S of the radiating surface of the body if the maximum spectral density of its energy luminosity falls on the wavelength λ = 700 nm.

18.13. In what regions of the spectrum are the wavelengths corresponding to the maximum spectral density of energy luminosity, if the light source is: a) a spiral of an electric light bulb (T = 3000 K); b) the surface of the Sun (T = 6000 K); c) an atomic bomb, in which at the moment of explosion the temperature develops T = 10 7 K? The radiation is considered to be close to the radiation of an absolutely black body.

18.14. The figure shows the dependence curve of the spectral density of the energy luminosity of a black body r λ on the wavelength λ at a certain temperature. To what temperature T does this curve apply? What percentage of the emitted energy is in the visible spectrum at this temperature?

18.15. When heating a completely black body, the wavelength λ, which accounts for the maximum spectral density of energy luminosity, changed from 690 to 500 nm. How many times did the energy permeability of the body increase in this case?

18.16. What wavelength λ accounts for the maximum spectral density of the energy luminosity of a completely black body with a temperature equal to the temperature t = 37° human body, i.e. T = 310K?

18.17. The temperature T of an absolutely black body changed when heated from 1000 to 3000 K. How many times did its energy luminosity R e increase? How much has the wavelength λ changed, which accounts for the maximum spectral density of energy luminosity? How many times has its maximum spectral density of energy luminosity r λ increased ?

18.18. An absolutely black body has a temperature T 1 = 2900 K. As a result of the cooling of the body, the wavelength, which accounts for the maximum spectral density of energy luminosity, has changed by Δλ = 9 μm. To what temperature T2 has the body cooled?

18.19. The surface of the body is heated to a temperature T = 1000K. Then one half of this surface is heated by ΔT = 100K, the other half is cooled by ΔT = 100K. How many times will the energy luminosity change R uh the surface of this body?

18.20. What power N must be supplied to a blackened metal ball of radius r = 2 cm to keep the temperature at ΔT = 27K above the ambient temperature? Ambient temperature T = 293 K. Assume that heat is lost only due to radiation.

18.21. The blackened ball cools down from temperature T 1 = 300 K to T 2 = 293 K. How much has the wavelength λ changed , corresponding to the maximum spectral density of its energy luminosity?

18.22. By how much will the mass of the Sun decrease in a year due to radiation? How long will it take for the mass of the sun to halve? Sun surface temperature T= 5800K. Radiation from the Sun is assumed to be constant.

Absolutely white and gray bodies, having the same surface area, are heated to the same temperature. Compare the fluxes of thermal radiation of these bodies Ф 0 (white) and Ф (gray). Answer: 3. F 0 <Ф. Absolutely black and gray bodies, having the same surface area, are heated to the same temperature. Compare the fluxes of thermal radiation of these bodies Ф 0 (black) and Ф (gray). Answer: 2. F 0 >F. The black body is... Answer: 1. a body that absorbs all the energy of electromagnetic waves incident on it, regardless of the wavelength (frequency). An absolutely black body has a temperature T 1 =2900 K. As a result of the cooling of the body, the wavelength, which accounts for the maximum spectral density of energy luminosity, has changed by Δλ=9 μm. To what temperature T2 has the body been cooled? Vina constant with 1=2.9×10 -3 mK. Answer: 2. T 2 =290K. It is known that the maximum radiation energy of the Sun corresponds to the wave l 0 =0.48 μm. Radius of the Sun R= m, mass of the Sun M= kg. How long does it take for the Sun to lose 1,000,000 kg of its mass? Answer: 4. 2 × 10 -4 With. There are two absolutely black sources of thermal radiation. The temperature of one of them is T 1 =2500 K. Find the temperature of the other source if the wavelength corresponding to the maximum of its emissivity is l=0.50 µm longer than the wavelength corresponding to the maximum of the emissivity of the first source (Wien's displacement law constant b=0.29 cm× TO). Answer: 3.T 2 =1750K. There are two absolutely black sources of thermal radiation. The temperature of one of them is T 1 =2500 K. Find the temperature of another source if the wavelength corresponding to the maximum of its emissivity is ∆λ=0.50 µm greater than the wavelength corresponding to the maximum of the emissivity of the first source. Answer: 1. 1.75 kK. A metal surface with an area of ​​S=15 cm 2 , heated to a temperature of T=3 kK, emits 100 kJ per minute. Determine the ratio of the energy luminosities of this surface and the black body at a given temperature. weight: 2. 0.2. Can the absorption capacity of a gray body depend on: a) radiation frequency. b) temperature. Answer: 3. a) no; b) yes. Radiation power of absolutely black body N=34 kW. Find the temperature T of this body if it is known that its surface is S=0.6 m 2 . Stefan-Boltzmann constant d=5.67×10 -8 W/(m2×K2). Answer: 4. T=1000 K. The radiation power of a hot metal surface is P'=0.67 kW. Surface temperature T=2500 K, its area S=10 cm 2 . Find the ratio k of the energy luminosities of this surface and a completely black body at a given temperature (the Stefan-Boltzmann constant σ=5.67×10 -8 W/(m 2 ×K 4)). Answer: 1. k=0.3. answer: 1. 2. Find the temperature T of the furnace, if it is known that the radiation from the hole in it with an area of ​​S=6.1 cm 2 has a power of N=34.6 W. The radiation is considered close to the radiation of a completely black body (S=5.67×10 -8 W/(m 2 ×K 4)). Answer: 2. T=1000K. 2. λm=0.97 µm. Answer: 2. λm≈0.5 µm. The figure shows the dependence of the spectral density of substances (1, 2) on the wavelength. What can be said about these substances and their temperatures? 1) the substances are the same, T 1 >T 2. 2) different substances T 1 3) the substances are the same, it is impossible to draw a conclusion about the temperature ratio. 4) the substances are the same, T 1 5) the substances are different, it is impossible to draw a conclusion about the ratio of temperatures. 6) the substances are the same, T 1 \u003d T 2. 7) it is impossible to draw a conclusion about substances, T 1 > T 2. 8) substances cannot be concluded, T 1 9) there are no correct answers. Answer: 9. There are no correct answers. Figure shows graphs of the dependence of the spectral density of the energy luminosity of a blackbody on the wavelength of radiation at different temperatures T 1 and T 2, with T 1 > T 2 (T 1 tops in Ox is greater than T 2). Which of the figures correctly takes into account the laws of thermal radiation? Answer: 1. Correct. The surface of the body is heated to a temperature of T=1000 K. Then one half of this surface is heated by ΔT=100 K, the other is cooled by ΔT=100 K. How many times will the average energy luminosity Re of the surface of this body change? Answer: 3. 1.06 times. An electric current passes through the plate, as a result of which it reaches an equilibrium temperature T 0 \u003d 1400 K. After that, the power of the electric current decreased by 2 times. Determine the new equilibrium temperature T. 2. T=1174 K.	Choose the correct statement. Answer: 2. The radiation of a completely black body at a given temperature exceeds the radiation of any other bodies at the same temperature. Choose the correct statement regarding the way electromagnetic waves are emitted. Answer: 4. Electromagnetic waves are not emitted continuously, but in separate quanta at any temperature above 0 K. The diameter of the tungsten filament in a light bulb is d=0.3 mm, the length of the filament is l=5 cm. When the bulb is connected to a network with voltage U=127V, current I=0.31 A flows through the bulb. Find the temperature T of the filament. Assume that after equilibrium is established, all the heat released in the filament is lost as a result of radiation. The ratio of the energy luminosities of tungsten and a black body for a given temperature is k=0.31. Stefan-Boltzmann constant d = 5.67 × 10-8 W / (m 2 × K 2). Answer: 3. T=2600 K. There are two cavities (see Fig.) with small holes of the same diameter d=l.0 cm and absolutely reflective outer surfaces. The distance between the holes is l=10 cm. A constant temperature is maintained in cavity 1 T 1 =1700 K. Calculate the steady temperature in cavity 2. 3.T 2 =400 K. There are two cavities (see Fig.) with small holes of the same diameter d cm and absolutely reflective outer surfaces. The distance between the holes l see In the cavity 1 is maintained at a constant temperature T 1 . Calculate the steady state temperature in cavity 2. Note: Keep in mind that a black body is a cosine emitter. 1. T 2 =T1sqrt(d/2l). The study of the solar emission spectrum shows that the maximum spectral density of the emissivity corresponds to the wavelength l=500nm. Taking the Sun as a completely black body, determine the emissivity (Re) of the Sun. 2. Re=64mW/m 2 . Radiation power of absolutely black body N=10 kW. Find the area S of the radiating surface of the body if the maximum spectral density of its energy luminosity falls on the wavelength λ=700 nm. Stefan-Boltzmann constant d=5.67×10 -8 W/(m2×K2). Answer: 3.S=6.0 cm². a) the wavelength corresponding to the maximum spectral density of radiation (λ max). b) the maximum energy emitted by a wave of a given length per unit time from a unit surface (rλ, t) with an increase in the temperature of a heated body. 3. a) decrease; b) will increase. A heated body produces thermal radiation over the entire wavelength range. How will it change: a) the wavelength corresponding to the maximum spectral radiation density (λmax). b) the maximum energy emitted by a wave of a given length per unit time from a unit surface (rλ, t) with a decrease in the temperature of a heated body. Answer: 2. a) increase; b) decrease. Find how many times it is necessary to reduce the thermodynamic temperature of a black body so that its energy luminosity Re decreases by 16 times? Answer: 1. 2. Find the temperature T of the furnace, if it is known that the radiation from the hole in it with an area of ​​S=6.1 cm 2 has a power of N=34.6 W. The radiation is considered close to the radiation of a completely black body (S=5.67×10 -8 W/(m 2 ×K 4)). Answer: 2. T=1000K. Find the wavelength λm, corresponding to the maximum spectral density of energy luminosity, if the light source is a spiral of a light bulb (T=3000 K). The radiation is considered to be close to the radiation of an absolutely black body. (Constant Vina С 1 =2.9∙10-3 m∙K). Answer: 2. λm=0.97 µm. Find the wavelength λm corresponding to the maximum spectral density of energy luminosity if the surface of the Sun serves as the light source (T=6000 K). The radiation is considered close to the radiation of an absolutely black body (Wien's constant C 1 =2.9∙10 -3 m×K). Answer: 2. λm≈0.5 µm. Below are the characteristics of thermal radiation. Which of them is called the spectral density of energy luminosity? Answer: 3. Energy radiated per unit time per unit area of ​​the body surface in a unit interval of wavelengths, depending on the wavelength (frequency) and temperature. Determine how many times it is necessary to reduce the thermodynamic temperature of a black body so that its energy luminosity Re is weakened by 39 times? 3.T 1 /T 2 =2.5. Determine how and how many times the radiation power of a black body will change if the wavelength corresponding to the maximum spectral density of its energy luminosity shifts from 720 nm to 400 nm. Answer: 3. 10.5. Determine the body temperature at which, at ambient temperature t = 27 0 C, it emitted 8 times more energy than it absorbed. Answer: 2.504 K. A cavity with a volume of 1 liter is filled with thermal radiation at a temperature Τ, the entropy of which is ς = 0.8 10-21 J/K. What is Τ equal to? Answer: 1. 2000K. What is the area under the radiation energy distribution curve? Answer: 3. Energy luminosity.	To enhance the energy luminosity of a black body by 16 times, it is necessary to increase its temperature by λ times. Determine λ. Answer: 1. 2. To increase the energy luminosity of an absolutely black body by 16 times, it is necessary to reduce its temperature by λ times. Define λ. Answer: 3. 1/2. Does the emitting and absorbing ability of the gray body depend on: a) radiation frequencies. b) temperature. c) Does the ratio of the emitting power of a body to its absorbing power depend on the nature of the body? Answer: 2.a) Yes; b) yes; c) no. The blackened ball cools down from temperature T 1 =300 K to T 2 =293 K. How much has the wavelength λ corresponding to the maximum spectral density of its energy luminosity changed (constant in Wien's first law С 1 =2.9×10-3 mK)? Answer: 2. Δλ=0.23 µm. What characteristic of thermal radiation in SI is measured in W / m 2? 1. Energy luminosity. What statements for absolutely black bodies are true? 1 - all absolutely black bodies at a given temperature have the same distribution of radiant energy over wavelengths. 3 - the luminosity of all absolutely black bodies changes equally with temperature. 5 - the emissivity of a black body increases with increasing temperature. Answer: 1. 1, 3, 5. Which law does not apply at infrared wavelengths? Answer: 3. Rayleigh-Jeans law. Which of the figures correctly takes into account the laws of thermal radiation (T 1 > T 2)? Answer:O:3. What is the radiation power of the Sun? The radiation of the Sun is considered close to the radiation of a completely black body. The temperature of the surface of the Sun is T=5800K (R=6.96*108m is the radius of the Sun). Answer: 1. 3.9 × 1026 W. What energy luminosity Re does a black body have if the maximum spectral density of its energy luminosity falls on the wavelength l=484 nm. (C 1 =2.9×10 -3 m×K). Answer: 4. 73 mW/m 2 . What energy luminosity Re does an absolutely black body have if the maximum spectral density of its energy luminosity falls at a wavelength of λ=484 nm (Stefan-Boltzmann constant σ=5.67×10 -8 W/(m 2 × K 4), Wien constant C 1 =2.9×10 -3 m×K)? Answer: 3. Re=73.5 mW/m 2 . A metal surface with an area of ​​S=15 cm 2 , heated to a temperature of T=3 kK, emits 100 kJ per minute. Determine the energy emitted by this surface, assuming it is black. Answer: 3.413 kJ. What wavelength λ accounts for the maximum spectral density of the energy luminosity of an absolutely black body having a temperature equal to the temperature t=37 °C of the human body, i.e. T=310 K? Wien's constant c1=2.9×10 –3 m×K. Answer: 5.λm=9.3 µm. What length l accounts for the maximum spectral density of the energy luminosity of a completely black body, which has a temperature equal to t 0 =37 ° C of the human body. Answer: 3. 9.35 microns. The figure shows the radiation energy distribution curve of a completely black body at a certain temperature. What is the area under the distribution curve? Answer: 1. Re=89 mW/m 2 . The figure shows the dependence (vertices are different in Ox) of the spectral density of substances (1, 2) on the wavelength. What can be said about these substances and their temperatures? Answer: 7. Substances cannot be inferred, T 1 > T 2. Determine the maximum speed of photoelectrons ejected from the metal surface if the photocurrent stops when a delay voltage U 0 =3.7 V is applied. Answer: 5. 1.14 mm/s. Determine how the energy luminosity will change if the thermodynamic temperature of a black body is increased by 3 times? Answer: Increase 81 times. Determine the temperature T of the Sun, taking it as an absolutely black body, if it is known that the maximum intensity of the Sun's spectrum lies in the green region λ=5×10 ‾5 cm. Answer: 1. T=6000K. Determine the wavelength corresponding to the maximum intensity in the spectrum of a completely black body, the temperature of which is 106 K. Answer: 1.λ max =29Å. Determine how many times the radiation power of a black body will increase if the wavelength corresponding to the maximum of its spectral density of energy luminosity shifted from 720 nm to 400 nm. Answer: 4. 10.5. According to what law does the ratio of the emissivity rλ,T of a given substance to the absorptivity aλ,T change? Answer: 2. const. The cavity with a volume of 1 liter is filled with thermal radiation at a temperature of 2000K. Find the heat capacity of the cavity C (J/K). Answer: 4.2.4×10 -8 .
When studying stars A and stars B, the ratio of the masses lost by them per unit time was established: DmA=2DmB, and their radii: RA=2.5RB. The maximum radiation energy of the star B corresponds to the wave lB=0.55 µm. Which wave corresponds to the maximum radiation energy of the star A? Answer: 1. lA=0.73 µm. When a blackbody is heated, the wavelength λ, which accounts for the maximum spectral density of energy luminosity, changed from 690 to 500 nm. How many times did the energy luminosity of the body change in this case? Answer: 4. 3.63 times. When passing through the plate, light of wavelength λ is attenuated due to absorption by N 1 times, and light of wavelength λ 2 by N 2 times. Determine the absorption coefficient for light of wavelength λ 2 if the absorption coefficient for λ 1 is equal to k 1 . 3. k 2 =k 1 ×lnN 2 /lnN 1 . The equilibrium temperature of the body is T. The area of ​​the radiating surface S, absorption capacity a. The power released in the body has increased by P. Determine the new equilibrium temperature T 1. T 1 = sqrt^4(T^4+ P/ aS× psi). Assuming that heat losses are due only to radiation, determine how much power must be supplied to a copper ball with a diameter of d=2 cm in order to maintain its temperature equal to t=17 ˚C at ambient temperature t 0 = -13 ˚C. Take the absorption capacity of copper equal to A=0.6. Answer: 2. 0.1 W. Considering nickel as a black body, determine the power required to maintain the temperature of molten nickel at 1453 0 С unchanged if its surface area is 0.5 cm 2 . Answer: 1. 25 W. The temperature of the inner surface of the muffle furnace with an open hole with a diameter of 6 cm is 650 0 C. Assuming that the furnace hole radiates as a black body, determine what fraction of the power is dissipated by the walls if the power consumed by the furnace is 600 W. Answer: 1. h=0.806. Energy luminosity of a completely black body Re=3 × 10 4 W/m 2 . Determine the wavelength λm corresponding to the maximum emissivity of this body Answer: 1. λm=3.4×10 -6 m. Energy luminosity of a black body ME=3.0 W/cm 2 . Determine the wavelength corresponding to the maximum emissivity of this body (S=5.67×10 -8 W/m 2 K 4 , b=2.9×10 -3 m×K). Answer: 1. lm=3.4 µm. Energy luminosity of an absolutely black body ME. Determine the wavelength corresponding to the maximum emissivity of this body. 1. Lam= b× sqrt^4(psi/ M). Energy luminosity of absolutely black body Re=3×104 W/m 2 . Determine the wavelength λm corresponding to the maximum emissivity of this body Answer: 1. λm=3.4×10 -6 m When studying stars A and stars B, the ratio of the masses lost by them per unit time was established: m A =2m B , and their radii: R A =2.5 R B . The maximum radiation energy of the star B corresponds to the wave  B =0.55 µm. Which wave corresponds to the maximum radiation energy of the star A? Answer: 1. A =0.73 µm. Taking the Sun (radius is 6.95 × 10 8 m) for a black body and given that its maximum spectral density of energy luminosity corresponds to a wavelength of 500 nm, determine: a) the energy emitted by the Sun in the form of electromagnetic waves for 10 minutes. b) the mass lost by the Sun during this time due to radiation. Answer: 2. a) 2.34 × 10 29 J; b) 2.6×10 12 kg. A silver ball (heat capacity - 230 J/gK, density - 10500 kg/m3) with a diameter of d=1 cm was placed in an evacuated vessel, the wall temperature of which was maintained close to absolute zero. The initial temperature is T 0 =300 K. Considering the surface of the ball to be absolutely black, find how long it will take for its temperature to decrease by n=2 times. Answer: 4. 1.7 hours. The temperature (T) of the inner wall of the furnace with an open hole with an area (S \u003d 50 cm 2) is 1000 K. If we assume that the furnace hole radiates as a black body, then find how much power is lost by the walls due to their thermal conductivity, if the power consumed by the furnace is 1.2 kW? Answer: 2. 283 watts. The temperature of the tungsten filament in a 25-watt light bulb is T=2450 K. The ratio of its energy luminosity to the energy luminosity of a completely black body at a given temperature is k=0.3. Find the area S of the radiating surface of the spiral. (Stefan-Boltzmann constant σ=5.67×10 -8 W/(m2×K4)). Answer: 2.S=4×10 -5 m 2 . The temperature of the "blue" star is 30,000 K. Determine the integral radiation intensity and the wavelength corresponding to the maximum emissivity. Answer: 4. J=4.6×1010 W/m 2 ; λ=9.6×10 -8 m. The temperature T of a completely black body changed when heated from 1000 to 3000 K. How much did the wavelength λ change, which accounts for the maximum spectral density of energy luminosity (constant in Wien's first law C 1 \u003d 2.9 × 10 -3 m × K)? Answer: 1. Δλ=1.93 µm. The temperature T of an absolutely black body changed when heated from 1000 to 3000 K. How many times did its maximum spectral density of energy luminosity rλ increase? Answer: 5. 243 times. The black body was heated from a temperature Τ=500K to some Τ 1 , while its energy luminosity increased 16 times. What is the temperature Τ 1 equal to? Answer: 3. 1000 K.	The black body was heated from temperature Τo=500K to Τ 1=700K. How did the wavelength corresponding to the maximum spectral density of energy luminosity change? Answer: 1. Decreased by 1.7 microns. Silver ball (heat capacity - 230 J/g × K, density - 10500 kg/m 3) of diameter d=1 cm was placed in an evacuated vessel, the temperature of the walls of which is maintained close to absolute zero. The initial temperature is T 0 =300 K. Considering the surface of the ball to be absolutely black, find how long it will take for its temperature to decrease by n=2 times. Answer: 5. 2 hours. The gray body is ... Answer: 2. a body whose absorption capacity is the same for all frequencies and depends only on temperature, material and surface condition. Assuming nickel to be a black body, determine the power required to maintain the temperature of molten nickel at 1453 0 C unchanged if its surface area is 0.5 cm 2 . Answer: 1. 25.2 W. The temperature of one of the two absolutely black sources T 1 =2900 K. Find the temperature of the second source T 2 if the wavelength corresponding to the maximum of its emissivity is ∆λ=0.40 µm greater than the wavelength corresponding to the maximum emissivity of the first source. Answer: 1. 1219 K. The temperature of the inner surface of the muffle furnace with an open hole with an area of ​​30 cm2 is 1.3 kK. Assuming that the furnace opening radiates as a black body, determine what part of the power is dissipated by the walls if the power consumed by the furnace is 1.5 kW. Answer: 3.0.676. The surface temperature of a completely black body is T=2500 K, its area is S=10 cm 2 . What radiation power P does this surface have (Stefan-Boltzmann constant σ=5.67 × 10 -8 W / (m 2 × To 4))? Answer: 2. P=2.22 kW. The temperature T of an absolutely black body changed when heated from 1000 to 3000 K. How many times did its energy luminosity Re increase? Answer: 4. 81 times. The black body is at a temperature Τ 0 =2900 K. When it cools down, the wavelength corresponding to the maximum spectral density of the energy luminosity changed by 10 microns. Determine the temperature Τ 1 to which the body has cooled. Answer: 1.264 K. The black body was heated from a temperature Τ to Τ 1 , while its energy luminosity increased 16 times. Find the ratio Τ 1 /Τ. Answer: 2. 2. The black body was heated from the temperature T 1 =600 K to T 2 =2400 K. Determine how many times its energy luminosity has changed. Answer: 4. Increased 256 times. What happens to the maximum emissivity of a blackbody as the temperature rises? Answer: 3. Increases in magnitude, shifts to shorter wavelengths. Gate photoelectric effect… Answer: 3. consists in the occurrence of photo-EMF due to an internal photoelectric effect near the contact surface of the metal - conductor or semiconductor with a p-n junction. The valve photoelectric effect is... Answer: 1. The occurrence of EMF (photo-EMF) when illuminating the contact of two different semiconductors or a semiconductor and a metal (in the absence of an external electric field). External photoelectric effect... Answer: 1. consists in pulling out electrons from the surface of solid and liquid substances under the action of light. Internal photoelectric effect... Answer: 2. consists in pulling out electrons from the surface of solid and liquid substances under the action of light. What is the maximum kinetic energy of photoelectrons when illuminating a metal with work function А=2 eV by light with a wavelength λ=6.2×10 -7 m? Answer: 10 eV. The efficiency of a 100-watt electric lamp in the visible light region is η=1%. Estimate the number of photons emitted in one second. Assume that the emitted wavelength is 500 nm. Answer: 2.2.5×10 18 ph/s The red border of the photoelectric effect for some metal λ 0 . What is the kinetic energy of photoelectrons when this metal is illuminated by light with a wavelength λ (λ<λ 0). Постоянная Планка h, скорость света C. Answer: 3.h× C×(λ 0 - λ )/ λλ 0 . The red border of the photoelectric effect for some metal  max =275 nm. What is the minimum value of the photon energy that causes the photoelectric effect? Answer: 1. 4.5 eV. The figure shows the current-voltage characteristics of two photocathodes illuminated by the same light source. Which photocathode has the highest work function? Answer: 2>1. The figure shows the current-voltage characteristic of the photocell. Determine the number N of photoelectrons leaving the cathode surface per unit time. Answer: 4.3.75×10 9 .	The internal photoelectric effect is... Answer: 2. caused by electromagnetic radiation, the transitions of electrons inside a semiconductor or dielectric from bound states to free ones without escaping to the outside. In what photoelectric effect does the concentration of free current carriers increase under the action of incident light? Answer: 2. Internal. In Stoletov's experiment, a charged negative zinc plate was irradiated with light from a volt arc. To what maximum potential will a zinc plate be charged when irradiated with monochromatic light with a wavelength of  = 324 nm, if the work function of electrons from the surface of zinc is Aout = 3.74 eV? Answer: 2.1.71 B. The electrons knocked out by light during the photoelectric effect, when the photocathode is irradiated with visible light, are completely delayed by the reverse voltage U=1.2 V. The wavelength of the incident light is λ=400 nm. Determine the red border of the photoelectric effect. 4. 652 nm. Choose the correct statements: 1. Electrons escape from the metal if the frequency of the light incident on the metal is less than a certain frequency ν gr. 2. Electrons escape from the metal if the frequency of the light incident on the metal is greater than a certain frequency ν gr. 3. Electrons escape from the metal if the wavelength of the light incident on the metal is greater than a certain wavelength λ gr. 4. λ gr - wavelength, which is constant for each metal. 5. ν gr - the frequency is different for each substance: 6. Electrons break out of the metal if the wavelength of the light incident on the metal is less than a certain wavelength λ gr. Answer: b) 2, 5. The retarding voltage for a platinum plate (work function 6.3 eV) is 3.7 V. Under the same conditions for another plate, the retarding voltage is 5.3 V. Determine the work function of electrons from this plate. Answer: 1. 4.7 eV. It is known that the wavelength of light incident on a metal can be determined by the formula. Determine the physical meaning of the coefficients a, b, c. Answer: 4.ais Planck's constant,bis the work function,cis the speed of light in vacuum. How will the dependence of the photocurrent on the voltage between the photocathode and the grid change if the number of photons falling on the photocathode per unit time decreases by half, and the wavelength increases by 2 times. Compare with the chart. Answer: 1. Potassium is illuminated with monochromatic light with a wavelength of 400 nm. Determine the smallest retarding voltage at which the photocurrent will stop. The work function of electrons from potassium is 2.2 eV. Answer: 3.0.91 V. What is the maximum kinetic energy of photoelectrons when a metal is illuminated with the work function A=2 eV by light with a wavelength of λ=550 nm? Answer: 1. 0.4 eV. The red border of the photoelectric effect for metal () is 577 nm. Find the minimum value of the photon energy (E min) that causes the photoelectric effect Answer: 1. 2.15 eV. The red border of the photoelectric effect for metal () is 550 nm. Find the minimum value of the photon energy (E min) that causes the photoelectric effect. Answer: 1. 2.24 eV. The maximum initial velocity (maximum initial kinetic energy) of photoelectrons ... Answer: 2. does not depend on the intensity of the incident light. There is a distance S between the photocathode and the anode, and such a potential difference is applied that the fastest photoelectrons can only travel half S. What distance will they fly if the distance between the electrons is halved for the same potential difference? Answer:S/4. The largest wavelength of light at which the photoelectric effect occurs for tungsten is 275 nm. Find the maximum speed of electrons ejected from tungsten by light with a wavelength of 250 nm. Answer: 2. 4 × 10 5 . Find the potential to which a solitary metal ball with the work function A = 4 eV will be charged when irradiated with light with a wavelength λ = 3 × 10 -7 m. Answer: 1. 0.14 V. Find to what potential a solitary metal ball with work function A=4 eV will be charged when irradiated with light with a wavelength λ=3×10 -7 . Answer: 2.8.5×10 15 . Find the wavelength of radiation whose photon mass is equal to the electron rest mass. Answer: 3.2.43 p.m. Find the voltage at which the X-ray tube would operate in such a way that the minimum radiation wave was equal to 0.5 nm. Answer: 2. 24.8 kV. Find the frequency ν of light pulling out electrons from the metal, which are completely delayed by the potential difference Δφ=3 V. The cutoff frequency of the photoelectric effect is ν 0 =6×10 14 Hz. Answer: 1. ν \u003d 13.2 × 10 14 Hz. Monochromatic light (λ=0.413 µm) falls on a metal plate. The flow of photoelectrons escaping from the metal surface is completely stopped when the potential difference of the retarding electric field reaches U=1 V. Determine the work function. Answer: 2.A=3.2×10 -19 J. Every second, 10 19 photons of monochromatic light with a power of 5 W fall on the metal surface. To stop the emission of electrons, you need to apply a delaying potential difference of 2 V. Determine the work function of the electrons (in eV). Answer: 1.1.125.
Every second, 1019 photons of monochromatic light with a power of 6.7 W fall on the metal surface. To stop the emission of electrons, you need to apply a restraining potential difference of 1.7 V. Determine: a) work function of electrons b) the maximum speed of photoelectrons. Answer: 1. a) 2.5 eV; b) 7.7×10 5 m/s. Monochromatic light with a wavelength of λ=310 nm falls on the surface of lithium. To stop the photocurrent, it is necessary to apply a delaying potential difference Uz of at least 1.7 V. Determine the work function of electrons from lithium. Answer: 2. 2.31 eV. Figure 1 shows the current-voltage characteristics of one photocell when it is illuminated with monochromatic light from two sources with frequencies V 1 (curve 1) and V 2 (curve 2). Compare the magnitudes of light fluxes, assuming that the probability of knocking out electrons does not depend on frequency. Answer: 2. F 1 <Ф 2 . Figure 1 shows the current-voltage characteristics of one photocell when it is illuminated with monochromatic light from two sources with frequencies V 1 (curve 1) and V 2 (curve 2). Compare the frequencies of V 1 and V 2 . Options: Answer: 1.V 1 > V 2 . The figure shows the current-voltage characteristics for a photocell. Which statements are true? ν is the frequency of the incident light, Ф is the intensity. Answer: 1. v 1 >ν 2 , F 1 =F 2 . The figure shows the dependence of the retarding potential difference Uz on the frequency of the incident light ν for some materials (1, 2). How do the work functions A out for these materials compare? Answer: 2. A 2 >A 1 . The figure shows the current-voltage characteristics of one photocell when it is illuminated with monochromatic light from two sources with frequencies v  and v 2 . Compare the frequencies v  and v 2 . Answer: 2.v  > v 2 . The figure shows the current-voltage characteristic of the photoelectric effect. Determine which curve corresponds to the high illumination (Ee) of the cathode, at the same frequency of light. Answer: 1. Curve 1. The figure shows the current-voltage characteristic of the photoelectric effect. Determine which curve corresponds to a higher frequency of light, with the same illumination of the cathode. Answer: 3. The frequencies are equal. The figure shows the current-voltage characteristics of one photocell when it is illuminated with monochromatic light from two sources with frequencies v  and v 2 . Answer: 2.v  > v 2. The work function of an electron from the surface of one metal A1=1 eV, and from another A2=2 eV. Will the photoelectric effect be observed in these metals if the photon energy of the radiation incident on them is 4.8×10 -19 J? Answer: 3. Will be for both metals. The work function of an electron from the surface of one metal A1=1 eV, and from another A2=2 eV. Will the photoelectric effect be observed in these metals if the photon energy of the radiation incident on them is 2.8×10 -19 J? Answer: 1. Only for metal with work function A1. The work function of an electron from the surface of cesium is A out \u003d 1.89 eV. With what maximum speed v do electrons fly out of cesium if the metal is illuminated with yellow light with a wavelength of =589 nm? Answer: 4. ν=2.72×10 5 m/s. The work function of an electron from the surface of one metal A1=1 eV, and from another A2=2 eV. Will the photoelectric effect be observed in these metals if the photon energy of the light incident on them is 4.8×10 -19 J? Answer: 4. No, for both metals. The dimension in the SI system of the expression h×k, where h is Planck's constant, k is the wave number, is: Answer: 5. kg×m/s. An X-ray tube operating under voltage U=50 kV and consuming current I, emits photons with an average wavelength λ in a time tN. Determine the efficiency η. Answer:Nhc/ IUtλ. How many photons fall for 1 sv of a human eye if the eye perceives light with a wavelength of 1 micron at a light flux power of 4 × 10 -17 W? Answer: 1.201. How many photons does E=10 7 J of radiation with a wavelength =1 µm contain? Answer: 5.04×10 11 . Figure 1 shows the current-voltage characteristics of one photocell when it is illuminated with monochromatic light from two sources with frequencies n 1 (curve 1) and n 2 (curve 2). Compare the frequencies n 1 and n 2 . Answer: 1.n 1 >n 2 . Determine the work function. Answer: 2. A=3.2×10 -19 J. Determine the work function A of electrons from sodium if the red border of the photoelectric effect lр=500 nm (h=6.62×10 -34 J×s, s=3×108m/s). Answer: 1. 2.49 eV. Determine the maximum speed V max of photoelectrons escaping from the silver surface by ultraviolet radiation with a wavelength l=0.155 µm. at the work function for silver A=4.7 eV. Answer: 1.1.08 mm/s. Determine the wavelength of the "red border" of the photoelectric effect for aluminum. Work function And vy =3.74 Ev. Answer: 2.3.32×10 -7 . Determine the red limit Lam of the photoelectric effect for cesium, if when its surface is irradiated with violet light of a long wave λ=400 nm, the maximum speed of photoelectrons is 0.65 im/s (h=6.626×10 -34 J×s). Answer: 640nm.	Determine the "red border" of the photoelectric effect for silver if the work function is 4.74 eV. Answer: 2.λ 0 =2.64×10 -7 m. Determine the maximum speed of photoelectrons if the photocurrent is converted at a retarding potential difference of 1 V (electron charge 1.6×10 -19 C, electron mass 9.1×10 -31 kg). Answer: 1.0.6×10 6 m/s. Determine dependency order a) saturation current b) the number of photoelectrons leaving the cathode per unit time with the photoelectric effect from the energy illumination of the cathode. Answer: 3. a) 1; b) 1. The photocathode is illuminated by various monochromatic light sources. The dependence of the photocurrent on the voltage between the cathode and the anode with one light source is displayed by curve 1, and with another by curve 2 (Fig. 1). How are light sources different from each other? Answer: 2. The first light source has a higher radiation frequency than the second. Photons with energy E=5 eV pull out photoelectrons from the metal with work function A=4.7 eV. Determine the maximum momentum transferred to the surface of this metal when an electron is emitted. Answer: 4.2.96×10 -25 kg×m/s. Photoelectrons ejected from the surface of the metal are completely stopped when a reverse voltage U=3 V is applied. The photoelectric effect for this metal begins at the frequency of the incident monochromatic light ν=6 × 10 14 s -1 . Determine the work function of electrons from this metal. Answer: 2. 2.48 eV. Photoelectrons ejected from the surface of the metal are completely stopped at Uo=3 V. The photoelectric effect for this metal begins at a frequency of n 0 =6×10 14 s -1. Determine the frequency of the incident light. Answer: 1. 1.32×10 15 With -1 . a) a=h/A out; c=m/2h. b) a=h/A out; c=2h/m. c) a=A out /h; c=2h/m. d) there is no correct answer. Answer: d) there is no correct answer. a) a=h/A out; c=m/2h. b) a=h/A out; c=2h/m. c) a=A out / h; c=m/2h. d) a=A out /h; c=2h/m. Answer: c)a= A exit / h; c= m/2 h. Determine how many photons fall in 1 minute per 1 cm 2 of the Earth's surface, perpendicular to the sun's rays, if the average wavelength of sunlight  cf \u003d 550 nm, the solar constant  \u003d 2 cal / (cm 2 min). Answer: 3.n=2.3×10 19 . Determine the speed of photoelectrons pulled out from the silver surface by ultraviolet rays (λ=0.15 µm, m e =9.1×10 -31 kg). Answer: 3.1.1×10 6 m/s. On what quantities does the "red border" n 0 of the photoelectric effect depend? Answer: 1. From the chemical nature of the substance and the state of its surface. A cesium plate is illuminated with light with a wavelength of =730 nm. The maximum electron escape velocity is v=2.5×10 5 m/s. A polarizer was installed in the path of the light beam. Degree of polarization P=0.16. What will be the maximum electron escape velocity if the work function for cesium is A out = 1.89 eV? Answer: 4.v 1 =2.5×10 5 m/s. Planck's constant h has a dimension. Answer: 5. J×s. It is generally accepted that in photosynthesis it takes about 9 photons to convert one molecule of carbon dioxide into hydrocarbon and oxygen. Let us assume that the wavelength incident on the plant is 670 nm. What is the efficiency of photosynthesis? Consider that 29% is required for the reverse chemical reaction. 2. 29%. When one metal is replaced by another, the wavelength corresponding to the "red border" decreases. What can be said about the work function of these two metals? Answer: 2. The second metal has more. It is generally accepted that in photosynthesis it takes about 9 photons to convert one molecule of carbon dioxide into hydrocarbon and oxygen. Let us assume that the wavelength of light falling on a plant is 670 nm. What is the efficiency of photosynthesis? Take into account that 4.9 eV is released during the reverse chemical reaction. Answer: 2. 29%. What is the wavelength of the red edge of the photoelectric effect for zinc? Work function for zinc A=3.74 eV (Planck constant h=6.6 × 10 -34 J × With; electron charge e=1.6 × 10 -19 C). 3.3.3x10 -7 m. What is the maximum speed of an electron knocked out from the surface of sodium (work function - 2.28 eV) by light with a wavelength of 550 nm? Answer: 5. There is no correct answer. What is the maximum speed of an electron knocked out from the surface of sodium (work function - 2.28 eV) by light with a wavelength of 480 nm? Answer: 3. 3 × 105 m / s. An electron accelerated by an electric field acquired a speed at which its mass became equal to twice the rest mass. Find the potential difference passed by the electron. Answer: 5. 0.51 mV. The energy of a photon of monochromatic light with a wavelength λ is equal to: Answer: 1.hc/λ.	Are the following statements true: a) scattering occurs when a photon interacts with a free electron, and the photoelectric effect occurs when interacting with bound electrons; b) the absorption of a photon by a free electron is impossible, since this process is in conflict with the laws of conservation of momentum and energy. 3. a) yes b) yes In what case is the inverse Compton effect observed, associated with a decrease in wavelength as a result of light scattering on a substance? 2. In the interaction of a photon with relativistic electrons As a result of the Compton effect, a photon colliding with an electron was scattered through an angle q = 900. The energy e' of a scattered photon is 0.4 MeV. Determine the photon energy (e) before scattering. 1.1.85 MeV As a result of Compton scattering, in one case, the photon flew at an angle to the original direction of the incident photon, and in the other, at an angle. In which case is the radiation wavelength after scattering longer and in which case did the electron participating in the interaction receive more energy? 4. 2 , 2 As a result of the Compton effect, a photon colliding with an electron was scattered through an angle =90 0 . Scattered photon energy E’=6.4*10^-14 J. Determine photon energy E before scattering. (s=3*10^8m/s, me=9.1*10^-31kg). 2. 1.8*10^-18J What is the difference between the nature of the interaction of a photon and an electron in the photoelectric effect (PE) and the Compton effect (EC)? 2. PE: a photon interacts with a bound electron and it is absorbed EC: a photon interacts with a free electron and it is scattered For what wavelengths is the Compton effect noticeable? 1. X-ray waves For what wavelengths is the Compton effect noticeable? The Compton effect is noticeable for the X-ray spectrum of waves ~ 10 -12 m. 1 - intense for substances with low atomic weight. 4 - weak for substances with a large atomic weight. 2) 1,4 Which of the following regularities is subject to Compton scattering? 1 - at the same scattering angles, the change in wavelength is the same for all scattering substances. 4. change in wavelength during scattering increases with increasing scattering angle 2) 1,4 What was the wavelength of X-ray radiation, if at the Compton scattering of this radiation by graphite at an angle of 60º the wavelength of the scattered radiation turned out to be equal to 2.54∙10-11m. 4. 2.48∙10-11 m What was the wavelength l0 of X-ray radiation, if during the Compton scattering of this radiation by graphite at an angle j = 600, the wavelength of the scattered radiation turned out to be equal to l = 25.4 pm 4. l0= 24.2*10-12m Which of the following expressions is the formula experimentally obtained by Compton (q is the scattering angle)? 1.∆l= 2h*(sinQ/2)^2/ m* c What was the wavelength of X-rays, if, when this radiation is scattered by some substance at an angle of 60 °, the wavelength of scattered X-rays is λ1 = 4 * 10-11 m 4.λ=2.76*10-11m What energy must a photon have in order for its mass to be equal to the rest mass of an electron? 4.8.19*10-14 J A Compton electron was ejected at an angle of 30°. Find the change in the wavelength of a photon with an energy of 0.2 MeV, when it is scattered by a free electron at rest. 4.3.0 pm Compton found that the optical difference between the wavelength of scattered and incident radiation depends on: 3. Scattering angle The Compton wavelength (when a photon is scattered into electrons) is equal to: 1. h/ m* c Can a free electron absorb a photon? 2. no Find the kinetic energy of a recoil electron if a photon with a wavelength λ=4pm is scattered at an angle of 90 0 by a free electron at rest. 5) 3.1*10 5 eV. Find the change in the frequency of a photon scattered by an electron at rest. h- constant bar; m 0 - rest mass of an electron; c is the speed of light; ν is the frequency of the photon; ν′ is the frequency of the scattered photon; φ - scattering angle; 2) ∆ν= h * ν * ν '*(1- cosφ ) / ( m 0 * c 2 ); Figure 3 shows the vector diagram of Compton scattering. Which of the vectors represents the momentum of the scattered photon? 1) 1 Figure 3 shows the vector diagram of Compton scattering. Which of the vectors represents the momentum of the recoil electron? 2) 2 2. 2.5*10^8m/s
The figure shows the dependences of the intensity of primary and secondary radiation on the wavelength of light when light is scattered on some substances. What can be said about the atomic weights (A 1 and A 2) of these substances (1,2)? λ is the wavelength of the primary radiation, λ / is the wavelength of the secondary radiation. 1) A 1 < A 2 Determine the maximum change in wavelength when light is scattered by protons. 2) ∆λ=2.64*10 -5 Ǻ; On what particles is it possible to observe the Compton effect? 1 - Free electrons 2 - Protons 3 - Heavy atoms 4 - Neutrons 5 - Positive metal ions 3) 1, 2, 3 A directed monochromatic light flux Ф falls at an angle a=30 o on absolutely black (A) and mirror (B) plates (Fig. 4) Compare the light pressure pa and pv on plates A and B, respectively, if the plates are fixed 3.pa Figure 2 shows a vector diagram of Compton scattering. Scattering angle φ=π/2. which of the vectors corresponds to the momentum of the scattered photon? 3. φ=180 O Figure 2 shows a vector diagram of Compton scattering. At what angle of scattering of photons is the change in their wavelength ∆λ maximum? 3 . φ=180 O Determine the maximum speed of electrons emitted from the metal under the action of γ-radiation with a wavelength of λ=0.030A. 2. 2.5*10^8m/s Determine the wavelength λ of X-ray radiation if, with Compton scattering of this radiation at an angle Θ \u003d 60 °, the wavelength of the scattered radiation λ 1 turned out to be 57 pm. 5) λ = 55.8 * 10 -11 m The discovery of the Compton effect proved that... b) a photon can behave both as a particle and as a wave e) when an electron and a photon interact, the energy of the photon decreases2) b, e The light rays scattered by the particles of matter passed through a converging lens and produced an interference pattern. What does it say? 5. The binding energy of electrons in the atoms of matter is greater than the energy of a photon X-rays (λ = 5 pm) are scattered by the wax. Find the length λ 1 of the X-ray wave scattered at an angle of 145° (Λ is the Compton wavelength). 3) λ 1 = 4,65 * 10 -11 m X-rays with a wavelength of 0.2Ǻ (2.0 * 10 -11 m) experience Compton scattering at an angle of 90º. Find the kinetic energy of the recoil electron. 2)6,6*10 3 eV; X-rays with a wavelength of  0 =70.8 pm experience Compton scattering on paraffin. Find the wavelength λ of X-rays scattered in the direction =/2( c =2.22pm).64.4pm 4. 73.22pm X-rays with a wavelength λ 0 = 7.08 * 10 -11 m experience Compton scattering on paraffin. Find the wavelength of X-rays scattered at an angle of 180º. 3)7,57*10 -11 m; X-rays with a wavelength l0=70.8pm experience Compton scattering on paraffin. Find the wavelength l of X-rays scattered in the direction j=p/2 (mel=9.1*10-31kg). 3.73.22*10-12m X-rays with a wavelength l0=70.8pm experience Compton scattering on paraffin. Find the wavelength l of X-rays scattered in the direction j=p(mel=9.1*10-31kg). 2.75.6 *10-12m X-ray radiation with a wavelength of l=55.8 pm is scattered by a graphite tile (Compton effect). Determine the wavelength l' of light scattered at an angle q = 600 to the direction of the incident light beam 1. 57pm A photon with an energy of 1.00 MeV was scattered by a free electron at rest. Find the kinetic energy of the recoil electron if the frequency of the scattered photon has changed by a factor of 1.25. 2) 0.2 MeV The energy of the incident photon hυ=0.1 MeV, the maximum kinetic energy of the recoil electron is 83 KeV. Determine the length of the primary wave. 3) λ=10 -12 m; A photon with energy e=0.12 MeV was scattered by an initially resting free electron. It is known that the wavelength of the scattered photon has changed by 10%. Determine the kinetic energy of the recoil electron (T). 1. 20 keV A photon with energy e = 0.75 MeV was scattered by a free electron at an angle q = 600. Assuming that the kinetic energy and momentum of the electron before the collision with the photon were negligibly small, determine the energy e of the scattered photon. 1. 0.43 MeV A photon with energy E=1.025 MeV was scattered by an initially resting free electron. Determine the scattering angle of a photon if the wavelength of the scattered photon turned out to be equal to the Compton wavelength λc=2.43 pm. 3. 60 ˚ A photon with energy j=1.025 MeV was scattered by a free electron at rest. The wavelength of the scattered photon turned out to be equal to the Compton wavelength lK=2.43 pm. Find the scattering angle q. 5. 600 A photon with energy j=0.25 MeV was scattered by a free electron at rest. Determine the kinetic energy of the recoil electron if the wavelength of the scattered photon has changed by 20%. 1. =41.7 keV	A narrow beam of monochromatic X-ray radiation is incident on a scattering substance. The wavelengths of radiation scattered at angles q1=600 and q2=1200 differ by 1.5 times. Determine the wavelength of the incident radiation if scattering occurs on free electrons. 3. 3.64 pm A narrow beam of monochromatic X-ray radiation is incident on a scattering substance. It turns out that the wavelengths of the radiation scattered at angles θ1=60˚ and θ2=120˚ differ by 1.5 times. Determine the wavelength of the incident radiation, assuming that scattering occurs on free electrons. 3.3.64 pm The photon was scattered at an angle θ=120˚ by an initially resting free electron. Determine the photon energy if the energy of the scattered photon is 0.144 MeV. 2) hν =250 keV; 2) W= hc TO /  ( +  TO ) A photon with a wavelength  experienced Compton perpendicular scattering by a free electron at rest. Compton wavelength  K. Find the energy of the recoil electron. 4) p= h* sqrt((1/  )2+(1/( +  TO ))2) A photon with a wavelength λ=6 pm was scattered at right angles by a free electron at rest. Find the wavelength of the scattered photon. 2) 8.4 pm A photon with a wavelength λ = 5 pm experienced Compton scattering at an angle υ = 90 0 on an initially resting free electron. Determine the change in wavelength upon scattering. 1) 2.43 pm A photon with a wavelength λ = 5 pm experienced Compton scattering at an angle Θ = 60°. Determine the change in wavelength upon scattering (Λ is the Compton wavelength). 2) Δλ=Λ/2 A photon with a wavelength λ = 5 pm experienced Compton scattering at an angle υ = 90 0 on an initially resting free electron. Determine the energy of the recoil electron. 3) 81 keV A photon with a wavelength λ = 5 pm experienced Compton scattering at an angle υ = 90 0 on an initially resting free electron. Determine the momentum of the recoil electron. 4) 1,6 *10 -22 kg*m/s A photon, having experienced a collision with a free electron, scattered at an angle of 180º. Find the Compton offset of the wavelength of the scattered photon (in pm): 3. 4.852 A photon with a wavelength of 100 pm is scattered at an angle of 180º by a free electron. Find the recoil kinetic energy (in eV): 4. 580 A photon with a wavelength of 8pm scattered at a right angle from a free electron at rest. Find the recoil kinetic energy (in keV): 2. 155 A photon with a wavelength λ = 5 pm experienced Compton scattering at an angle Θ = 60° Determine the change in wavelength during scattering. Λ - Compton wavelength 2. Δλ = ½*Λ A photon with momentum p=1.02 MeV/s, c is the speed of light, was scattered at an angle of 120º by a free electron at rest. How will the momentum of a photon change as a result of scattering. 4. decrease by 0.765 MeV/s A photon with energy hν=250 keV was scattered at an angle θ=120˚ by an initially resting free electron. Determine the energy of the scattered photon. 3) 0.144 MeV A photon with energy =1.025 MeV was scattered by a free electron at rest. The wavelength of the scattered photon turned out to be equal to the Compton wavelength  K = 2.43 pm. Find the scattering angle . 5) 60 0 A photon with energy =0.25 MeV is scattered by a free electron at rest. Determine the kinetic energy of the recoil electron T e if the wavelength of the scattered photon has changed by 20%. 1) T e =41.7 keV A photon with energy Е=6.4*10 -34 J scattered at an angle =90 0 on a free electron. Determine the energy E' of the scattered photon and the kinematic energy T of the recoil electron. 5. there is no correct answer A photon with energy E=4*10 -14 J was scattered by a free electron. Energy E=3.2*10 -14 J. Determine the scattering angle . (h=6.626*10 -34 J*s,  s=2.426 pm, s=3*10 8 m/s). 4. 3,2* 10 -14 The Compton effect is called... 1. elastic scattering of short-wavelength electromagnetic radiation on free electrons of a substance, accompanied by an increase in wavelength

Polarization

1) The magnetic rotation of the plane of polarization is determined by the following formula. 4

2) Determine the thickness of the quartz plate, for which the angle of rotation of the plane of polarization is 180. The specific rotation in quartz for a given wavelength is 0.52 rad/mm. 3

3) Plane polarized light, whose wavelength in vacuum is 600 nm, falls on a plate of Icelandic spar, perpendicular to its optical axis. The refractive indices for ordinary and extraordinary rays are 1.66 and 1.49, respectively. Determine the wavelength of an ordinary ray in a crystal. 3

4) Some substance was placed in the longitudinal magnetic field of a solenoid located between two polarizers. The length of the tube with the substance l. Find the Verdet constant if, at a field strength H, the angle of rotation of the polarization plane for one direction of the field and for the opposite direction of the field. 4

5) Monochromatic plane-polarized light with a circular frequency passes through a substance along a homogeneous magnetic field with a strength H. Find the difference in the refractive indices for the right and left circularly polarized components of the light beam if the Verdet constant is equal to V. 1

6) Find the angle between the main planes of the polarizer and analyzer if the intensity of natural light passing through the polarizer and analyzer is reduced by a factor of 4. 45

7) Linearly polarized light of intensity I0 falls on the analyzer, the vector E0 of which makes an angle of 30 with the transmission plane. What fraction of the incident light does the analyzer let through? 0.75

8) If we pass natural light through two polarizers, the main planes of which form an angle, then the intensity of this light is I \u003d 1/2 * Iest * cos ^ 2 (a). What is the intensity of the plane polarized light that comes out of the first polarizer? 1

9) Natural light passes through two polarizers, the main planes of which form an angle a between themselves. What is the intensity of the plane polarized light that comes out of the second polarizer? 4

10) The angle between the main planes of the polarizer and the analyzer is 60. Determine the change in the intensity of the light passing through them if the angle between the main planes becomes 45. 2

11) A beam of natural light falls on a system of 6 polarizers, the transmission plane of each of which is rotated by an angle of 30 relative to the transmission plane of the previous polarizer. What part of the light flux passes through this system? 12

12) A quartz plate 2 mm thick, cut perpendicular to the optical axis of the crystal, rotates the plane of polarization of monochromatic light of a certain wavelength by an angle of 30. Determine the thickness of the quartz plate placed between parallel nicols so that this monochromatic light is extinguished. 3

13) Natural light passes through a polarizer and an analyzer set so that the angle between their principal planes is phi. Both the polarizer and the analyzer absorb and reflect 8% of the light incident on them. It turned out that the intensity of the beam that emerged from the analyzer is equal to 9% of the intensity of natural light incident on the polarizer. 62

14) When adding two linearly polarized light waves oscillating in perpendicular directions with a phase shift ... 3

15) In what cases, when light passes through the analyzer, is the Malus law applicable? 2

16) What types of waves have the property of polarization? 3

17) What type of waves are electromagnetic? 2

18) Determine the intensity of the reflected light if the oscillations of the light vector of the incident light are perpendicular to the plane of incidence. 1

19) Light falls on the interface between two media with refractive indices n1 and n2, respectively. Denote the angle of incidence by a and let n1>n2. Total reflection of light occurs when ... 2

20) Determine the intensity of the reflected light, in which the oscillations of the light vector lie in the plane of incidence. 5

21) A crystal plate that creates a phase difference between ordinary and extraordinary rays is placed between two polarizers. The angle between the plane of transmission of the polarizers and the optical axis of the plate is 45. In this case, the intensity of the light transmitted through the polarizer will be maximum under the following conditions ... 1

22) Which statements about partially polarized light are true? 3

23) Which statements about plane polarized light are true? 3

24) Two polarizers are placed in the path of the natural light beam, the axes of the polarizers are oriented in parallel. How are the vectors E and B oriented in the beam of light coming out of the second polarizer? 1

25) Which of the following statements are true only for plane polarized electromagnetic waves? 3

26) Which of the following statements are true for both plane polarized electromagnetic waves and non-polarized ones? 4

27) Determine the path difference for a quarter-wave plate cut parallel to the optical axis? 1

28) What is the difference between the refractive indices of ordinary and extraordinary rays in the direction perpendicular to the optical axis in the case of deformation. 1

29) A parallel beam of light falls normally on a 50 mm thick plate of Icelandic spar cut parallel to the optical axis. Taking the refractive indices of Icelandic spar for ordinary and extraordinary rays, respectively, 1.66 and 1.49, determine the difference in the path of these rays passing through this plate. 1

30) A linearly polarized light beam is incident on a polarizer rotating around the beam axis at an angular velocity of 27 rad/s. The energy flux in the incident beam is 4 mW. Find the light energy passing through the polarizer in one revolution. 2

31) A beam of polarized light (lambda = 589nm) falls on a plate of Icelandic spar. Find the wavelength of an ordinary ray in a crystal if its refractive index is 1.66. 355

32) A linearly polarized light beam falls on a polarizer whose transmission plane rotates around the beam axis with an angular velocity w. Find the light energy W passing through the polarizer in one revolution if the energy flux in the incident beam is phi. 1

33) A beam of plane polarized light (lambla = 640nm) falls on a plate of Icelandic spar perpendicular to its optical axis. Find the wavelengths of the ordinary and extraordinary rays in a crystal if the refractive index of Icelandic spar for the ordinary and extraordinary rays is 1.66 and 1.49. 1

34) Plane polarized light falls on an analyzer rotating around the beam axis at an angular velocity of 21 rad/s. Find the light energy passing through the analyzer in one revolution. The intensity of polarized light is 4 W. 4

35) Determine the difference between the refractive index of the ordinary and extraordinary rays of a substance, if the smallest thickness of a half-wave crystalline plate made of this substance for lambda0 \u003d 560 nm is 28 microns. 0.01

36) Plane polarized light, with a wavelength of lambda \u003d 589 nm in vacuum, falls on a crystal plate perpendicular to its optical axis. Finding nm (in modulus) is the difference in wavelengths in a crystal, if the refractive index of the ordinary and extraordinary rays in it is 1.66 and 1.49, respectively. 40

37) Determine the smallest thickness of the crystal plate in half a wave for lambda = 589 nm, if the difference between the refractive indices of ordinary and extraordinary rays for a given wavelength is 0.17. 1.73

38) A parallel beam of light falls normally on a 50 mm thick Icelandic spar plate cut parallel to the optical axis. Taking the refractive indices of the ordinary and extraordinary rays to be 1.66 and 1.49, respectively, determine the difference in the path of the rays that have passed through the plate. 8.5

39) Determine the path difference for a half-wave plate cut parallel to the optical axis? 2

40) A linearly polarized light beam is incident on a polarizer, the transmission plane of which rotates around the beam axis with an angular velocity of 20. Find the light energy W passing through the polarizer in one revolution if the power of the incident beam is 3 W. 4

41) A beam of natural light falls on a glass prism with an angle at the base of 32 (see figure). Determine the refractive index of the glass if the reflected beam is plane polarized. 2

42) Determine at what angle to the horizon the Sun should be in order for the rays reflected from the surface of the lake (n=1.33) to be maximally polarized. 2

43) Natural light falls on glass with a refractive index of n=1.73. Determine the angle of refraction, to the nearest degree, at which the light reflected from the glass is completely polarized. thirty

44) Find the refractive index n of glass if, when light is reflected from it, the reflected beam is completely polarized at an angle of refraction of 35. 1.43

45) Find the angle of full polarization when light is reflected from glass, the refractive index of which is n \u003d 1.57 57.5

46) A beam of light reflected from a dielectric with a refractive index n is completely polarized when the reflected beam forms an angle of 90 with the refracted beam. At what angle of incidence is complete polarization of the reflected light achieved? 3

47) A beam of light falls on the surface of the water (n=1.33). Determine the angle of refraction to the nearest degree if the reflected beam is completely polarized. 37

48) In what case is Brewster's law inaccurate? 4

49) A natural beam of light falls on the surface of a glass plate with refractive indices n1 = 1.52, placed in a liquid. The reflected beam makes an angle of 100 with the incident beam and is completely polarized. Determine the refractive index of the liquid. 1.27

50) Determine the speed of propagation of light in glass if, when light falls from air onto glass, the angle of incidence corresponding to the full polarization of the reflected beam is 58. 1

51) Angle of total internal reflection at the "glass-air" interface 42. Find the angle of incidence of a light beam from the air onto the glass surface, at which the beam is completely polarized to within a degree. 56

52) Determine the refractive index of the medium to the second decimal place, upon reflection from which at an angle of 57 the light will be completely polarized. 1.54

53) Find the refractive index of glass if, when light is reflected from it, the reflected beam is completely polarized at an angle of refraction of 35. 1.43

54) A beam of natural light falls on a glass prism, as shown in the figure. Angle at the base of the prism 30. Determine the refractive index of the glass if the reflected beam is plane polarized. 1.73

55) Determine at what angle to the horizon the Sun should be so that the rays reflected from the surface of the lake (n = 1.33) are maximally polarized. 37

56) A beam of natural light falls on a glass prism with an angle at the base a (see figure). Refractive index of glass n=1.28. Find the angle a to the nearest degree if the reflected beam is plane polarized. 38

57) Determine the refractive index of glass if, when light is reflected from it, the reflected beam is completely polarized at the angle of refraction. 4

58) A beam of plane polarized light falls on the surface of water at the Brewster angle. Its plane of polarization makes an angle of 45 with the plane of incidence. Find the reflection coefficient. 3

59) Determine the refractive index of glass if, when light is reflected from it, the reflected beam is completely polarized at an angle of incidence of 55. 4

60) The degree of polarization of partially polarized light is 0.2. Determine the ratio of the maximum intensity of light transmitted by the analyzer to the minimum. 1.5

61) What are Imax, Imin, P for plane polarized light, where ... 1

62) Determine the degree of polarization of partially polarized light if the amplitude of the light vector corresponding to the maximum light intensity is twice the amplitude corresponding to the minimum intensity. 0.6

63) Determine the degree of polarization of partially polarized light if the amplitude of the light vector corresponding to the maximum light intensity is three times the amplitude corresponding to the maximum intensity. 1

64) The degree of polarization of partially polarized light is 0.75. Determine the ratio of the maximum intensity of light transmitted by the analyzer to the minimum. 1

65) Determine the degree of polarization P of light, which is a mixture of natural light with plane polarized light, if the intensity of polarized light is 3 times the intensity of natural light. 3

66) Determine the degree of polarization P of light, which is a mixture of natural light with plane polarized light, if the intensity of polarized light is 4 times the intensity of natural light. 2

67) Natural light falls at the Brewster angle on the surface of the water. In this case, part of the incident light is reflected. Find the degree of polarization of the refracted light. 1

68) Natural light falls at the Brewster angle on the glass surface (n=1.5). Determine the reflection coefficient in percent. 7

69) Natural light falls at the Brewster angle on the glass surface (n=1.6). Determine the reflection coefficient in percent using Fresnel formulas. 10

70) Determine, using the Fresnel formulas, the reflection coefficient of natural light at normal incidence on the glass surface (n=1.50). 3

71) The reflectance of natural light at normal incidence on the surface of a glass plate is 4%. What is the refractive index of the plate? 3

72) Degree of polarization of partially polarized light P=0.25. Find the ratio of the intensity of the polarized component of this light to the intensity of the natural component. 0.33

73) Determine the degree of polarization P of light, which is a mixture of natural light with plane polarized light, if the intensity of polarized light is equal to the intensity of natural light. 4

74) Degree of polarization of partially polarized light P=0.75. Find the ratio of the intensity of the polarized component of this light to the intensity of the natural component. 3

75) Determine the degree of polarization P of light, which is a mixture of natural light with plane polarized light, if the intensity of polarized light is half the intensity of natural light. 0.33

76) A narrow beam of natural light passes through a gas of optically isotropic molecules. Find the degree of polarization of light scattered at an angle a to the beam. 1

POLARIZATION A beam of natural light falls on the polished surface of a glass (n=1.5) plate immersed in a liquid. The light beam reflected p from the plate makes an angle φ=970 with the incident beam. Determine the refractive index n of the liquid if the reflected light is completely polarized. Answer: 1. n=1.33. A beam of natural light falls on a glass prism with an angle of refraction =30. Determine the refractive index of the glass if the reflected beam is plane polarized. Answer:1. n=1,73. A beam of polarized light (=589nm) falls on a plate of Icelandic spar perpendicular to its optical axis. Find the wavelength  o of an ordinary ray in a crystal if the refractive index of Icelandic spar for an ordinary ray is n o =1.66. Answer: 2. 355 nm.	A) Determine the angle of incidence of light on the water surface (n=1.33), at which the reflected light will be plane polarized. b) Determine the angle of the refracted light. Answer:2. a) 53 ; b) 37 . The analyzer attenuates by a factor of 4 the intensity of the polarized light incident on it from the polarizer. What is the angle between the principal planes of the polarizer and analyzer? Answer:3 . 60  . In which of the following cases will the phenomenon of polarization be observed: Answer: 1. When transverse waves pass through an anisotropic medium. The angle between the main planes of the polarizer and analyzer is  1 =30. Determine the change in the intensity of the light passing through them if the angle between the main planes is  2 \u003d 45. Answer: 3.I 1 / I 2 =1,5.	It is possible to observe interference in natural light, which is a mixture of differently oriented waves, because: a) in an interference experiment, we force the waves sent almost simultaneously by the same atom to meet. b) interference occurs between parts of the same polarized wave. Answer: 2. a) yes; b) yes. Choose the correct statement regarding the degree of polarization P and the type of refracted wave at an angle of incidence B equal to the Brewster angle. Answer: 3. Degree of polarizationP- maximum: refracted wave - partially polarized. Select the conditions necessary for the occurrence of birefringence when light passes through a polarizer. Answer: b) the light beam is partially polarized before refraction and the polarizer is anisotropic; c) the light beam is completely unpolarized before refraction and the polarizer is anisotropic.
Natural monochromatic light falls on a system of two crossed polarizers, between which there is a quartz plate cut perpendicular to the optical axis. Find the minimum thickness of the plate, at which this system will transmit h=0.30 of the light flux, if the quartz rotation constant is a=17 arcsec. deg/mm. Answer: 4. 3.0 mm. Natural light falls at the Brewster angle on the surface of the water. In this case, part of the incident light  is reflected. Find the degree of polarization of the refracted light. Answer: 1.r/(1- r) . Natural light is incident at the Brewster angle on the glass surface (n=1.5). Determine the reflection coefficient in this case. Answer: 2. 7%.	Which of the following statements are true for natural light received from a heat source: Answer: 1. The initial phases of electromagnetic waves emitted by a heat source are different. 2. The frequencies of electromagnetic waves emitted by a heat source are different. 4. Electromagnetic waves are emitted from different points on the surface of a heat source in different directions. Which statements about partially polarized light are true? Answer: a) Characterized by the fact that one of the directions of oscillation is predominant. c) Such light can be considered as a mixture of natural and polarized light. What are the degrees of polarization for plane polarized light P 1 and natural light P 2 ? Answer: 2. R 1 =1 ; R 2 =0.	A linearly polarized light beam is incident on a polarizer whose transmission plane rotates around the beam axis with an angular velocity ω. Find the light energy W passing through the polarizer in one revolution if the energy flux in the incident beam is . Answer: 1. W=pi×fi/w. The magnetic rotation of the polarizer plane is determined by the following formula: Answer: 4. = V× B× l. Linearly polarized light falls on the analyzer, the vector E of which makes an angle =30 0 with the transmission plane. Find the intensity of the transmitted light. Answer: 2. 0.75;I 1 . Two polarizers are placed on the path of the natural light beam, the axes of the polarizers are oriented mutually perpendicular. How are the vectors E and B oriented in the beam of light coming out of the second polarizer? Answer: 4. Modules of vectors E and B are equal to 0.
The figure shows the surface of radial velocities of a uniaxial crystal. Define: 1. Commensurability of the propagation speeds of the ordinary and the extraordinary. 2. Positive or negative uniaxial crystal. Answer: 3.v e > v o , negative. Find the refractive index n of glass if, when light is reflected from it, the reflected beam is completely polarized at an angle of refraction =30. Answer: 3.n=1,73. Find the angle φ between the main planes of the polarizer and analyzer if the intensity of natural light passing through the polarizer and analyzer decreases by a factor of 3. Answer: 3.35˚. Find the angle φ between the principal planes of the polarizer and analyzer if the intensity of natural light passing through the polarizer and analyzer is reduced by a factor of 4. Answer:3. 45  .	Find the angle i B of full polarization when light is reflected from glass, the refractive index of which is n=1.57. Answer: 1.57.5 . Unpolarized light passes through two polaroids. The axis of one of them is vertical, and the axis of the other forms an angle of 60° with the vertical. What is the intensity of the transmitted light? Answer:2. I=1/8 I 0 . An ordinary beam of light falls on a polaroid, and birefringence occurs in it. Which of the following laws is valid for double refraction for an extraordinary ray? O is an ordinary ray. E - extraordinary ray. Answer: 1. sinA/sinB=n 2 /n 1 = const. An ordinary beam of light falls on a polaroid, and birefringence occurs in it. Which of the following laws is valid for birefringence for an ordinary beam? O is an ordinary ray. E - extraordinary ray. Answer: 3. sinA/sinB=f(A)#const.	Determine the smallest thickness of a crystalline plate in half a wave for λ=640 nm, if the difference between the refractive indices of ordinary and extraordinary rays for a given wavelength is n0-ne=0.17? Answer:3. d=1.88 µm. Determine the refractive index of glass if, when light is reflected from it, the reflected beam is completely polarized at an angle of refraction . Answer: 4.n= sin(90  -  )/ sin . Determine the refractive index of glass if, when light is reflected from it, the reflected beam is completely polarized at an angle of =35. Answer:4. 1,43. Determine at what angle to the horizon the Sun should be in order for the rays reflected from the surface of the lake (n=1.33) to be maximally polarized. Answer: 2.36° .
Determine at what angle to the horizon the sun must be in order for its rays reflected from the surface of the water to be completely polarized (n=1.33). Answer: 4. 37°. Determine the degree of polarization P of light, which is a mixture of natural light with plane polarized light, if the intensity of polarized light is equal to the intensity of natural light. Answer: 4.0.5 Determine the degree of polarization P of light, which is a mixture of natural light with plane polarized light, if the intensity of polarized light is 5 times greater than the intensity of natural light. Answer: 2.0.833. The degree of polarization of partially polarized light is 0.75. Determine the ratio of the maximum intensity of light transmitted by the analyzer to the minimum. Answer: 1. 7. The limiting angle of total internal reflection for some substance i=45 0 . Find the Brewster angle of total polarization for this substance. Answer: 3.55 0 . The degree of polarization of partially polarized light P = 0.1. Find the ratio of the intense polarized component to the intense natural component. Answer: 1. 1/9.	Estimate the ratio of the maximum intensity of the light wave transmitted by the analyzer to the minimum, provided that the degree of polarization of partially polarized light is 0.5. Answer:2. 3. A parallel beam of light is incident normally on a 50 mm thick Icelandic spar plate cut parallel to the optical axis. Taking the refractive indices of Icelandic spar for ordinary and extraordinary rays, respectively, N o =1.66 and N e =1.49, determine the path difference of these rays that have passed through this plate. Answer:1. 8.5 µm. A quartz plate with a thickness of d 1 =2 mm, cut perpendicular to the optical axis of the crystal, rotates the plane of polarization of monochromatic light of a certain wavelength by an angle  1 =30 0 . Determine the thickness d 2 of a quartz plate placed between parallel nickels so that this monochromatic light is completely extinguished. Answer: 3.6 mm. The degree of polarization of partially polarized light is P = 0.25. Find the ratio of the intensity of the polarized component of this light to the intensity of the natural component. Answer: 4.0.3. The degree of polarization of partially polarized light is 0.5. Determine the ratio of the maximum intensity of light transmitted by the analyzer to the minimum. Answer: 1. 3.	A flat beam of natural light with intensity I 0 is incident at the Brewster angle on the water surface. Refractive index n=4/3 . What is the degree of reflection of the light flux, if the intensity of the refracted light is reduced by 1.4 times compared to I 0 . Answer:1. ρ=0.047. The polarizer and analyzer absorb 2% of the light incident on them. The intensity of the beam leaving the analyzer is equal to 24% of the intensity of natural light incident on the polarizer. Find the angle φ between the principal planes of the polarizer and analyzer. Answer: 1.45 . The degree of polarization of partially polarized light P=0.1. Find the ratio of the intense natural component to the intense polarized component. Answer: 1. 9. The degree of polarization of partially polarized light is P=0.25. Find the ratio of the intensity of the polarized component of this light to the intensity of the natural component. Answer: 3.I floor / I eating = p/(1- p). Determine the degree of polarization of partially polarized light if the amplitude of the light vector corresponding to the maximum light intensity is three times the amplitude corresponding to the minimum intensity. Answer: 1. 0.8.

3) The gray body is... 2

5) In fig. graphs of the dependence of the spectral density of the energy luminosity of a black body on the wavelength of radiation at different temperatures T1 and T2 are given, and T1>

Quantum mechanics

quantum mechanics

8) A particle with charge Q and rest mass m0 accelerates in an electric field, having passed a potential difference U. Can the de Broglie wavelength of a particle be less than its Compton wavelength. (Maybe if QU>0.41m0*c^2)

10) Determine at what numerical value of the speed the de Broglie wavelength for an electron is equal to its Compton wavelength. (2.12e8. lambda(c)=2pi*h/m0*c; lambda=2pi*h*sqrt(1-v^2/c^2)/m0*v; lambda(c)=lambda; 1/ c=sqrt(1-v^2/c^2)/v;v^2=c^2*(1-v^2/c^2);v^2=c^2-v^2;v =c/sqrt(2);v=2.12e8 m/s)

<=x<=1. Используя условие нормировки, определите нормировочный множитель. (A=sqrt(2/l))

>Dpr)

32) The uncertainty relation for energy and time means that (the lifetime of the state of the system (particle) and the uncertainty of the energy of this state of relations >=h)

35) Which of the following ratios is not the Heisenberg ratio. (VEV(x)>=h)

quantum mechanics

1) The kinetic energy of a moving electron is 0.6 MeV. Determine the de Broglie wavelength of an electron. (1.44 pm; 0.6 MeV = 9.613*10^-14 J; lambda=2pi*h/(sqrt(2mT))=1.44 pm)

2) Find the de Broglie wavelength for a proton with a kinetic energy of 100 eV. (2.86 rm. phi=h/sqrt(2m*E(k))=2.86 pm)

3) The kinetic energy of the neutron is 1 keV. Determine the de Broglie wavelength. (0.91 pm. 1keV=1600*10^-19 J. lambda=2pi*h/sqrt(2m*T))=0.91pm)

4) a) Is it possible to represent the De Broglie wave as a wave packet? b) How will the group velocity of the wave packet U and the particle velocity V be related in this case? (no, u=v)

5) Find the ratio of the Compton wavelength of the proton to the De Broglie wavelength for a proton moving at a speed of 3*10^6 m/s. (0.01. lambda(c)=2pi*h/mc=h/mc; lambda=2pi*h/sqrt(2m*T); lambda(c)/phi=0.01)

6) The kinetic energies of two electrons are 3 keV and 4 keV, respectively. Determine the ratio of the corresponding De Broglie lengths. (1.15. lambda=2pi*h/sqrt(2mT); phi1/phi2=1.15)

7) Calculate the de Broglie wavelength of a 0.2 kg ball flying at a speed of 15 m/s. (2.2*10^-34; lambda=h/mv=2.2*10^-34)

8) A particle with charge Q and rest mass m0 accelerates in an electric field, having passed a potential difference U. Can the de Broglie wavelength of a particle be less than its Compton wavelength. (Maybe if QU>0.41m0*c^2)

9) Determine what accelerating potential difference a proton must pass in order for the de Broglie wavelength for it to be 1 nm. (0.822 mV. lambda=2pi*h/sqrt(2m0*T); lambda^2*2m0*T=4*pi^2*h^2; T=2*pi^2*h^2/lambda^2 *m0=2.39e-19; T=eU; U=T/e=2pi^2*h^2/lambda^2*m0*e=0.822 mV)

10) Determine at what numerical value of the speed the de Broglie wavelength for an electron is equal to its Compton wavelength. (2.12e8. lambda(c)=2pi*h/m0*c; lambda=2pi*h*sqrt(1-v^2/c^2)/m0*v; lambda(c)=lambda; 1/ c=sqrt(1-v^2/c^2)/v;v^2=c^2*(1-v^2/c^2);v^2=c^2-v^2;v =c/sqrt(2);v=2.12e8 m/s)

11) Determine the minimum probable energy for a quantum particle located in an infinitely deep potential well of width a. (E=h^2/8ma^2)

12) A particle of mass m is in a one-dimensional potential rectangular well with infinitely high walls. Find the number dN of energy levels in the energy interval (E, E+dE) if the levels are very dense. (dN=l/pi*n*sqrt(m/2E)dE)

13) A quantum particle is located in an infinitely deep potential well of width L. At what points of the electron location on the first (n=1) energy level is the function maximum. (x=L/2)

14) A quantum particle is in an infinitely deep potential well of width a. At what points of the third energy level the particle cannot be. (a, b, d, e)

15) The particle is in an infinitely deep hole. At what energy level is its energy defined as 2h^2/ml^2? (4)

16) The wave function psi(x)=Asin(2pi*x/l) is defined only in the region 0<=x<=1. Используя условие нормировки, определите норировочный множитель. (A=sqrt(2/l))

17) The particle is in the ground state (n=1) in a one-dimensional infinite deep potential well of lambda width with absolutely impenetrable walls (0

18) The particle is in a one-dimensional rectangular potential well with infinitely high walls. Find the quantum number of the energy level of the particle if the energy intervals to the neighboring levels (upper and lower) are related as n:1, where n=1.4. (2.)

19) Determine the wavelength of the photon emitted during the transition of an electron in a one-dimensional rectangular potential well with infinitely high walls of width 1 from state 2 to the state with the lowest energy. (lambda=8cml^2/3h.)

20) The electron hits a potential barrier of finite height. At what energy value of an electron will it not pass through a potential barrier of height U0. (no correct answers)

21) Complete the definition: The tunnel effect is a phenomenon in which a quantum particle passes through a potential barrier at (E

22) The coefficient of transparency of the potential barrier - (the ratio of the flux density of passing particles to the flux density of incident particles)

23) What will be the transparency coefficient of the potential barrier if its width is doubled? (D^2)

24) A particle of mass m falls on a rectangular potential barrier, and its energy E >Dpr)

25) A proton and an electron, having the same energy, move in the positive direction of the X axis and meet a rectangular potential barrier on their way. Determine how many times it is necessary to narrow the potential barrier so that the probability of its passage by a proton is the same as for an electron. (42.8)

26) The rectangular potential barrier has a width of 0.3 nm. Determine the energy difference at which the probability of an electron passing through the barrier is 0.8. (5.13)

27) An electron with an energy of 25 eV meets on its way a low potential step with a height of 9 eV. Determine the refractive index of de Broglie waves at the step boundary. (0.8)

28) A proton with an energy of 100 eV changes the de Broglie wavelength by 1% when passing through a potential step. Determine the height of the potential barrier. (2)

29) The uncertainty relation for position and momentum means that (it is possible to simultaneously measure the coordinates and momentum of a particle only with a certain accuracy, and the product of the uncertainties of position and momentum must be at least h / 2)

30) Estimate the uncertainty of the speed of an electron in a hydrogen atom, assuming the size of the hydrogen atom is 0.10 nm. (1.16*10^6)

31) The uncertainty relation for position and momentum means that (it is possible to simultaneously measure the coordinates and momentum of a particle only with a certain accuracy, and the product of the uncertainties of the position and momentum must be at least h/2)

32) The uncertainty relation for energy and time means that (the lifetime of the state of the system (particle) and the uncertainty of the energy of this state of relations >=h)

33) The uncertainty relation follows from (wave properties of microparticles)

34) The average kinetic energy of an electron in an atom is 10 eV. What is the order of the smallest error with which the coordinate of an electron in an atom can be calculated. (10^-10)

35) Which of the following ratios is not the Heisenberg ratio. (VEV(x)>=h)

36) The uncertainty relation for the position and momentum of a particle means that (it is possible to simultaneously measure the coordinates and momentum of a particle only with a certain accuracy, and the uncertainties of the position and momentum must be at least h/2)

37) Choose the INCORRECT statement (with n=1 an atom can only be in the first energy level for a very small amount of time n=1)

38) Determine the ratio of the uncertainties of the speed of an electron and a dust particle with a mass of 10 ^ -12 kg, if their coordinates are set with an accuracy of 10 ^ -5 m. (1.1 * 10 ^ 18)

39) Determine the speed of an electron in the third orbit of a hydrogen atom. (v=e^2/(12*pi*E0*h))

40) Derive the relationship between the radius of a circular electron orbit and the de Broglie wavelength, where n is the number of the stationary orbit. (2pi*r=n*lambda)

41) Determine the energy of a photon emitted during the transition of an electron in a hydrogen atom from the third energy level to the second. (1.89 eV)

42) Determine the speed of the electron in the third Bohr orbit of the hydrogen atom. (0.731 mm/s)

43) Using Bohr's theory for hydrogen, determine the speed of an electron in an excited state at n=2. (1.14 mm/s)

44) Determine the period of revolution of an electron located in a hydrogen atom in a stationary state (0.15 * 10^-15)

45) An electron is knocked out of a hydrogen atom in a stationary state by a photon whose energy is 17.7. Determine the speed of an electron outside the atom. (1.2 mm/s)

46) Determine the maximum and minimum photon energies in the visible series of the hydrogen spectrum (Bolmer series). (5/36hR, 1/4hR)

47) Calculate for the hydrogen atom the radius of the second Bohr orbit and the speed of the electron on it. (2.12*10^-10, 1.09*10^6)

48) Using Bohr's theory, determine the orbital magnetic moment of an electron moving along the third orbit of a hydrogen atom. (2.8*10^-23)

49) Determine the binding energy of an electron in the ground state for the He+ ion. (54.5)

50) Based on the fact that the ionization energy of the hydrogen atom is 13.6 eV, determine the first excitation potential of this atom. (10.2)

51) An electron is knocked out of a hydrogen atom, which is in the ground state, by a photon of energy e. Determine the speed of an electron outside the atom. (sqrt(2(E-Ei)/m))

52) What is the maximum speed that electrons must have in order to transfer a hydrogen atom from the first state to the third by impact. (2.06)

53) Determine the energy of a photon emitted during the transition of an electron in a hydrogen atom from the third energy level to the second. (1.89)

54) What orbit from the main one will an electron in a hydrogen atom go to when a photon with an energy of 1.93 * 10^-18 J is absorbed. (3)

55) As a result of the absorption of a photon, the electron in the hydrogen atom moved from the first Bohr orbit to the second. What is the frequency of this photon? (2.5*10^15)

56) An electron in a hydrogen atom passes from one energy level to another. What transitions correspond to the absorption of energy. (1,2,5)

57) Determine the minimum electron velocity required for the ionization of a hydrogen atom if the ionization potential of a hydrogen atom is 13.6. (2.2*10^6)

58) At what temperature do mercury atoms have translational kinetic energy sufficient for ionization? The ionization potential of the mercury atom is 10.4 V. The molar mass of mercury is 200.5 g/mol, the universal gas constant is 8.31. (8*10^4)

59) The binding energy of an electron in the ground state of the He atom is 24.6 eV. Find the energy required to remove both electrons from this atom. (79)

60) With what minimum kinetic energy must a hydrogen atom move in order for one of them to be able to emit a photon in an inelastic head-on collision with another hydrogen atom at rest. It is assumed that both atoms are in the ground state before the collision. (20.4)

61) Determine the first excitation potential of the hydrogen atom, where R is the Rydberg constant. (3Rhc/4e)

62) Find the difference between the wavelengths of the head lines of the Lyman series for atoms of light and heavy hydrogen. (33 pm)

1) Choose the correct statement regarding the way electromagnetic waves are emitted. 4

2) Absolutely black and gray bodies, having the same surface area, are heated to the same temperature. Compare the fluxes of thermal radiation of these bodies Ф0 (black) and Ф (gray). 2

3) The gray body is... 2

4) The characteristics of thermal radiation are given below. Which of them is called the spectral density of energy luminosity? 3

5) In fig. graphs of the dependence of the spectral density of the energy luminosity of a blackbody on the wavelength of radiation at different temperatures T1 and T2, and T1>T2 are shown. Which of the figures correctly takes into account the laws of thermal radiation? 1

6) Determine how many times it is necessary to reduce the thermodynamic temperature of a black body so that its energy luminosity R is weakened by 39 times? 3

7) A completely black body is ... 1

8) Can the absorption capacity of a gray body depend on a) radiation frequency b) temperature? 3

9) When studying stars A and stars B, the ratio of the masses lost by them per unit time (delta)mA=2(delta)mB and their radii Ra=2.5Rb was established. The maximum radiation energy of the star B corresponds to the wave lambda B = 0.55 μm. Which wave corresponds to the maximum radiation energy of the star A? 1

10) Choose the correct statement. (absolutely white body) 2

11) Find the wavelength of lambda0 light corresponding to the red border of the photoelectric effect for lithium. (Work function A=2.4 eV). Planck's constant h=6.62*10^-34 J*s. 1

12) Find the wavelength of lambda0 light corresponding to the red border of the photoelectric effect for sodium. (Work function A=2.3 eV). Planck's constant h=6.62*10^-34 J*s. 1

13) Find the wavelength of lambda0 light corresponding to the red border of the photoelectric effect for potassium. (Work function A=2.0 eV). Planck's constant h=6.62*10^-34 J*s. 3

14) Find the wavelength of lambda0 light corresponding to the red border of the photoelectric effect for cesium. (Work function A=1.9 eV). Planck's constant h=6.62*10^-34 J*s. 653

15) The wavelength of light corresponding to the red border of the photoelectric effect for some metal lambda0. Find the minimum energy of a photon that causes the photoelectric effect. 1

16) The wavelength of light corresponding to the red border of the photoelectric effect for some metal lambda0. Find the work function A of an electron from a metal. 1

17) The wavelength of light corresponding to the red border of the photoelectric effect for some metal is equal to lambda0. Find the maximum kinetic energy W of electrons ejected from the metal by light with a wavelength of lambda. 1

18) Find the retarding potential difference U for electrons ejected when a certain substance is illuminated by light with a wavelength of lambda, where A is the work function for this substance. 1

19) Photons with energy e eject electrons from the metal with work function A. Find the maximum momentum p transferred to the surface of the metal when each electron is emitted. 3

20) A vacuum photocell consists of a central cathode (tungsten ball) and an anode (the inner surface of the bulb silvered from the inside). The contact potential difference between the electrodes U0 accelerates the outgoing electrons. The photocell is illuminated by light with a wavelength of lambda. What speed v will the electrons get when they reach the anode if no potential difference is applied between the cathode and the anode? 4

21) In fig. Graphs of the dependence of the maximum energy of photoelectrons on the energy of photons incident on the photocathode are shown. In which case does the photocell cathode material have the lowest work function? 1

22) Einstein's equation for the many photon photoelectric effect has the form. 1

23) Determine the maximum speed of electrons emitted from the cathode if U=3V. 1

24) External photoelectric effect - ... 1

25) Internal photoelectric effect - ... 2

26) Valve photoelectric effect - ... 1) consists ... 3

27) Determine the speed of photoelectrons pulled out from the surface of silver by ultraviolet rays (lambda \u003d 0.15 microns, m \u003d 9.1 * 10^-31 kg), if the work function is 4.74 eV. 3

28) Determine the "red border" of the photoelectric effect for silver if the work function is 4.74 eV. 2

29) The red border of the photoelectric effect for metal (lambda0) is 550 nm. Find the minimum value of the photon energy (Emin) that causes the photoelectric effect. 1

30) The work function of an electron from the surface of one metal is A1=1 eV, and from the other - A2=2 eV. Will the photoelectric effect be observed in these metals if the photon energy of the radiation incident on them is 4.8 * 10^-19 J? 3

31) Valve photoelectric effect is ... 1) occurrence ... 1

32) The figure shows the current-voltage characteristic of the photoelectric effect. Determine which curve corresponds to the high illumination of the cathode, at the same frequency of light. 1

33) Determine the maximum speed Vmax of photoelectrons ejected from the surface of silver by ultraviolet radiation with a wavelength of 0.155 μm at a work function for silver of 4.7 eV. 1

34) Compton found that the optical difference between the wavelength of the scattered and incident radiation depends on ... 3

35) The Compton wavelength (when a photon is scattered into electrons) is equal to. 1

36) Determine the wavelength of X-ray radiation if, with Compton scattering of this radiation at an angle of 60, the wavelength of the scattered radiation turned out to be 57 pm. 5

37) A photon with a wavelength of 5 pm experienced Compton scattering at an angle of 60. Determine the change in wavelength during scattering. 2

38) What was the wavelength of X-rays, if, when this radiation is scattered by some substance at an angle of 60, the wavelength of the scattered X-rays is 4 * 10^-11 m.

39) Are the following statements true: a) scattering occurs when a photon interacts with a free electron, and the photoelectric effect occurs when interacting with bound electrons b) absorption of a photon by a free electron is impossible, since this process is in conflict with the laws of conservation of momentum and energy. 3

40) Figure 3 shows a vector diagram of Compton scattering. Which of the vectors represents the momentum of the scattered photon? 2

41) Directed monochromatic light flux Ф falls at an angle of 30 on absolutely black (A) and mirror (B) plates (Fig. 4). Compare the light pressure on plates A and B, respectively, if the plates are fixed. 3

42) Which of the following expressions is the formula experimentally obtained by Compton? 1

43) Can a free electron absorb a photon? 2

44) A photon with an energy of 0.12 MeV was scattered by an initially resting free electron. It is known that the wavelength of the scattered photon has changed by 10%. Determine the kinetic energy of the recoil electron (T). 1

45) X-ray radiation with a wavelength of 55.8 pm is scattered by a graphite tile (Compton effect). Determine the wavelength of light scattered at an angle of 60 to the direction of the incident light beam. 1

85) In Young's experiment, the hole is illuminated with monochrome light (lambda=600 nm). The distance between the holes is d=1 nm, the distance from the holes to the screen is L=3 m. Find the position of the first three light bands. 4

86) The installation for obtaining Newton's rings is illuminated by monochromatic light falling normally. Light wavelength lambda=400 nm. What is the thickness of the air wedge between the lens and the glass plate for the third bright ring in reflected light? 3

87) In Young's experiment (interference of light from two narrow slits), a thin glass plate was placed in the path of one of the interfering rays, as a result of which the central bright stripe shifted to the position originally occupied by the fifth bright stripe (not counting the central one). The beam is incident perpendicular to the surface of the plate. The refractive index of the plate n=1.5. Wavelength lambda=600 nm. What is the thickness h of the plate? 2

88) Installation for the observation of Newton's rings is illuminated by monochromatic light with a wavelength of lambda = 0.6 microns, falling normally. Observation is carried out in reflected light. The radius of curvature of the lens is R=4 m. Determine the refractive index of the liquid that fills the space between the lens and the glass plate if the radius of the third light ring is r=2.1 mm. It is known that the refractive index of a liquid is less than that of glass. 3

89) Determine the length of the segment l1, on which the same number of wavelengths of monochromatic light in vacuum fit as they fit on the cutter l2=5 mm in glass. Refractive index of glass n2=1.5. 3 http://ivandriver.blogspot.ru/2015/01/l1-l25-n15.html

90) A normally parallel beam of monochromatic light (lambda = 0.6 μm) falls on a thick glass plate covered with a very thin film, the refractive index of which is n = 1.4. At what minimum film thickness will reflected light be maximally attenuated? 3

91) What should be the allowable width of the slits d0 in Young's experiment so that the interference pattern is visible on a screen located at a distance L from the slits. Distance between slots d, wavelength lambda0. 1

92) A point source of radiation contains wavelengths in the range from lambda1=480 nm to lambda2=500 nm. Estimate the coherence length for this radiation. 1

93) Determine how many times the width of the interference fringes on the screen will change in the experiment with Fresnel mirrors if the violet light filter (0.4 microns) is replaced by red (0.7 microns). max: delta=+-m*lambda, delta=xd/l, xd/l=+-m*lambda, x=+-(ml/d)*lambda, delta x=(ml*lambda/d)-( (m-1)l*lambda/d)=l*lambda/d, delta x1/delta x2=lambda2/lambda1 = 1.75 (1)

94) In the Young installation, the distance between the slots is 1.5 mm, and the screen is located at a distance of 2 m from the slots. Determine the distance between the interference fringes on the screen if the wavelength of monochromatic light is 670 nm. 3

95) Two coherent beams (lambda = 589 nm) maximize each other at a certain point. A normal soap film was placed on the path of one of them (n=1.33). At what minimum thickness d of the soap film will these coherent rays maximally attenuate each other at some point. 3

96) Installation for obtaining Newton's rings is illuminated by monochromatic light incident along the normal to the surface of the plate. The radius of curvature of the lens R=15 m. Observation is carried out in reflected light. The distance between the fifth and twenty-fifth light Newton's rings l=9 mm. Find the lambda wavelength of monochromatic light. r=sqrt((2m-1)lambda*R/2), delta d=r2-r1=sqrt((2*m2-1)lambda*R/2)-sqrt((2*m1-1)lambda* R/2)=7sqrt(lambda*R/2)-3sqrt(lambda*R/2)=4sqrt(lambda*R/2), lambda=sqr(delta d)/8R = 675 nm.

97) The two slots are 0.1 mm apart and 1.20 m from the screen. From a remote source, light with a wavelength of lambda=500 nm falls on the slit. How far apart are the light stripes on the screen? 2

98) Monochromatic light with a wavelength of lambda = 0.66 microns falls on the installation for obtaining Newton's rings. The radius of the fifth light ring in reflected light is 3 mm. Determine the radius of curvature of the lens. 3m or 2.5m

100) An interference pattern from two coherent light sources with a wavelength of lambda \u003d 760 nm is observed on the screen. By how many fringes will the interference pattern on the screen shift if a plastic of fused quartz d = 1 mm thick with a refractive index n = 1.46 is placed in the path of one of the rays? The beam falls on the plate normally. 2

101) An interference pattern from two coherent light sources with a wavelength of 589 nm is observed on the screen. By how many fringes will the interference pattern on the screen shift if a fused quartz plastic 0.41 mm thick with a refractive index n = 1.46 is placed in the path of one of the rays? The beam falls on the plate normally. 3

103) If you look, squinting your eyes, at the filament of an incandescent lamp, then the filament seems to be bordered by light highlights in two perpendicular directions. If the lamp filament is parallel to the observer's nose, then a series of iridescent images of the filament can be observed. Explain the reason for this phenomenon. 4

104) Light falls normally on a transparent diffraction grating of width l=7 cm. Determine the smallest wave difference that this grating can resolve in the lambda=600 nm region. Type your answer in pm to the nearest tenth. 7.98*10^-12=8.0*10^-12

105) Let the intensity of a monochromatic wave be I0. The diffraction pattern is observed using an opaque screen with a round hole on which the given wave is incident perpendicularly. Considering the hole to be equal to the first Fresnel zone, compare the intensities I1 and I2, where I1 is the light intensity behind the screen with the hole fully open, and I2 is the light intensity behind the screen with the hole half closed (in diameter). 2

106) Monochromatic light with a wavelength of 0.6 microns normally falls on a diffraction grating. The diffraction angle for the fifth maximum is 30, and the minimum wavelength difference resolvable by the grating is 0.2 nm for this maximum. Determine: 1) diffraction grating constant; 2) the length of the diffraction grating. 4

107) A parallel beam of light falls on a diaphragm with a round hole. Determine the maximum distance from the center of the hole to the screen, at which a dark spot will still be observed in the center of the diffraction pattern, if the radius of the hole is r=1 mm, the wavelength of the incident light is 0.5 μm. 2

108) Normally monochromatic light falls on a narrow slit. Its direction to the fourth dark diffraction band is 30. Determine the total number of diffraction peaks. 4

109) A normally monochromatic wave of lambda length falls on a diffraction grating with a period of d \u003d 2.8 * lambda. What is the highest order of the diffraction maximum given by the grating? Determine the total number of maxima? 1

110) Light with a wavelength of 750 nm passes through a slit with a width of D \u003d 20 microns. What is the width of the central maximum on the screen located at a distance L=20 cm from the slit? 4

111) A beam of light from a discharge tube normally falls on a diffraction grating. What should be the constant d of the diffraction grating so that the maxima of the lines lambda1=656.3 nm and lambda2=410.2 nm coincide in the direction phi=41. 1

112) Using a diffraction grating with a period of 0.01 mm, the first diffraction maximum was obtained at a distance of 2.8 cm from the central maximum and at a distance of 1.4 m from the grating. Find the length of the light wave. 4

113) A point source of light with a wavelength of 0.6 microns is located at a distance a = 110 cm in front of a diaphragm with a round hole with a radius of 0.8 mm. Find the distance b from the diaphragm to the observation point for which the number of Fresnel zones in the hole is k=2. 3

114) A point light source (lambda = 0.5 μm) is located at a distance a = 1 m in front of a diaphragm with a round hole of diameter d = 2 mm. Determine the distance b (m) from the diaphragm to the observation point if the hole opens three Fresnel zones. 2 http://studyport.ru/images/stories/tasks/Physics/difraktsija-sveta/1.gif

116) Normally monochromatic light with a wavelength of 550 nm falls on a diffraction grating with a length l \u003d 15 mm, containing N \u003d 3000 strokes. Find: 1) the number of maxima observed in the spectrum of the diffraction grating 2) the angle corresponding to the last maximum. 2

117) How does the pattern of the diffraction spectrum change as the screen moves away from the grating? 2

118) A parallel beam of monochromatic light with a wavelength of 0.5 microns normally falls on a screen with a round hole of radius r \u003d 1.5 mm. The observation point is located on the axis of the hole at a distance of 1.5 m from it. Determine: 1) the number of Fresnel zones that fit in the hole 2) a dark or light ring is observed in the center of the diffraction pattern if a screen is placed at the observation point. r=sqrt(bm*lambda), m=r^2/b*lambda=3 - odd, light ring. 2

119) A plane wave falls normally on a diaphragm with a round hole. Determine the radius of the fourth Fresnel zone if the radius of the second Fresnel zone = 2 mm. 4

120) The angular dispersion of the diffraction grating in the spectrum of the first order dphi / dlambda \u003d 2.02 * 10 ^ 5 rad / m. Find the linear dispersion D of the diffraction grating if the focal length of the lens projecting the spectrum onto the screen is F=40 cm. 3

Finally, there is another way to characterize electromagnetic radiation - by specifying its temperature. Strictly speaking, this method is suitable only for the so-called black-body or thermal radiation. In physics, an absolutely black body is an object that absorbs all radiation falling on it. However, ideal absorbing properties do not prevent the body from emitting radiation itself. On the contrary, for such an idealized body it is possible to accurately calculate the shape of the radiation spectrum. This is the so-called Planck curve, the shape of which is determined by a single parameter - temperature. The famous hump of this curve shows that a heated body radiates little at both very long and very short wavelengths. The maximum radiation falls on a well-defined wavelength, the value of which is directly proportional to the temperature.

When specifying this temperature, one must keep in mind that this is not a property of the radiation itself, but only the temperature of an idealized absolutely black body, which has a maximum radiation at a given wavelength. If there is reason to believe that the radiation is emitted by a heated body, then by finding the maximum in its spectrum, one can approximately determine the temperature of the source. For example, the surface temperature of the Sun is 6,000 degrees. This just corresponds to the middle of the visible radiation range. This is hardly accidental - most likely, the eye during evolution has adapted to make the most efficient use of sunlight.

Temperature ambiguity

The point of the spectrum, which accounts for the maximum of black-body radiation, depends on which axis we are plotting on. If the wavelength in meters is evenly plotted along the abscissa axis, then the maximum will fall on

λ max = b/T= (2.9 10 -3 m· TO)/T ,

Where b= 2.9 10 -3 m· TO. This is the so-called Wien's displacement law. If we plot the same spectrum, plotting the radiation frequency uniformly on the y-axis, the location of the maximum is calculated by the formula:

ν max = (α k/h) · T= (5.9 10 10 Hz/TO) · T ,

where α = 2.8, k= 1.4 10 –23 J/TO is the Boltzmann constant, h is Planck's constant.

Everything would be fine, but as it turns out λ max and v max correspond to different points of the spectrum. This becomes obvious if we calculate the wavelength corresponding to ν max, then you get:

λ" max = With/ν max = (ch/α k)/T= (5.1 10 -3 m K) / T .

Thus, the maximum of the spectrum, determined by frequency, in λ" max/ν max = 1,8 times different in wavelength (and hence in frequency) from the maximum of the same spectrum determined from wavelengths. In other words, the frequency and wavelength of the maximum black-body radiation do not correspond to each other: λ max ≠ With/ν max .

In the visible range, it is customary to indicate the maximum of the thermal radiation spectrum along the wavelength. In the spectrum of the Sun, as already mentioned, it falls in the visible range. However, in terms of frequency, the maximum solar radiation lies in the near infrared range.

And here is the maximum of cosmic microwave radiation with a temperature of 2.7 TO it is customary to indicate by frequency - 160 MHz, which corresponds to a wavelength of 1.9 mm. Meanwhile, in the graph by wavelengths, the maximum of the cosmic microwave background falls on 1.1 mm.

All this shows that temperature must be used with great care to describe electromagnetic radiation. It can be used only in the case of radiation close in spectrum to thermal radiation, or for a very rough (up to an order of magnitude) characteristic of the range. For example, visible radiation corresponds to a temperature of thousands of degrees, X-ray - millions, microwave - about 1 kelvin.

thermal radiation- electromagnetic radiation emitted by a substance and arising due to its internal energy.

It is caused by the excitation of particles of matter during collisions in the process of thermal motion of oscillating ions.

The intensity of radiation and its spectral composition depend on body temperature, so thermal radiation is not always perceived by the eye.

Body. Heated to a high temperature, a significant part of the energy emits in the visible range, and at room temperature, energy is emitted in the infrared part of the spectrum.

According to international standards, there are 3 areas of infrared radiation:

1. Infrared A

λ from 780 to 1400 nm

2. Infrared B

λ from 1400 to 3000 nm

3. Infrared C

λ from 3000 to 1000000 nm.

Features of thermal radiation.

1. Thermal radiation - this is a universal phenomenon inherent in all bodies and occurring at temperatures other than absolute zero (-273 K).

2. The intensity of thermal radiation and the spectral composition depend on the nature and temperature of the bodies.

3. Thermal radiation is in equilibrium, i.e. in an isolated system at a constant temperature, bodies radiate as much energy per unit time per unit area as they receive from outside.

4. Along with thermal radiation, all bodies have the ability to absorb thermal energy from the outside.

2 . Main characteristics of absorption.

1. Radiant energy W (J)

2. Radiant flux P \u003d W / t (W)

(Radiation Flux)

3. Emissivity (energy luminosity) is the energy of electromagnetic radiation radiated in all possible directions per unit of time per unit area at a given temperature

RT= W/St (W/m2)

4. Absorption capacity (absorption coefficient) is equal to the ratio of the radiant flux absorbed by a given body to the radiant flux incident on the body at a given temperature.

αt = Rdeep / Rfall.

3. Thermal radiators and their characteristics.

The concept of a completely black body.

Thermal emitters - these are technical devices for obtaining a thermal radiant flux. Each heat source is characterized by emissivity, absorption capacity, temperature of the radiating body, spectral composition of the radiation.

The concept of an absolutely black body (A.Ch.T.) has been introduced as a standard.

When light passes through a substance, the radiant flux is partly reflected, partly absorbed, scattered, and partly passes through the substance.

If the body completely absorbs the light flux falling on it, then it is called absolutely black body.

For all wavelengths and at any temperature, the absorption coefficient α=1. There is no absolutely black body in nature, but you can point to a body that is close to it in its properties.

Model A.Ch.T. is a cavity with a very small opening whose walls are blackened. The beam that enters the hole after multiple reflections from the walls will be absorbed almost completely.

If you heat such a model to a high temperature, the hole will glow, such radiation is called black radiation. To a.h.t. the absorption properties of black velvet are close.

α for soot = 0.952

α for black velvet = 0.96

An example is the pupil of the eye, a deep well, etc.

If α=0, then it is an absolutely mirror surface. More often α is in the range from 0 to 1, such bodies are called gray.

For gray bodies, the absorption coefficient depends on the wavelength, the incident radiation, and to a large extent on temperature.

4. The laws of thermal radiation and their characteristics

1. Kirkhoff's law:

the ratio of the emissivity of a body to the absorptivity of a body at the same temperature and at the same wavelength is a constant value.

2. Stefan-Boltzmann law:

emissivity of A.Ch.T. proportional to the fourth power of its absolute temperature.

δ is the Stefan-Boltzmann constant.

δ=5.669*10-8 (W/m2*K4)

W=Pt=RTSt= δStT4

T-temperature

As the temperature (T) increases, the radiation power increases very rapidly.

With an increase in time (t) to 800, the radiation power will increase by 81 times.

The laws of thermal radiation. Radiant warmth.

Maybe for someone it will be news, but the transfer of temperature occurs not only by heat conduction through the touch of one body to another. Each body (solid, liquid and gaseous) emits thermal rays of a certain wave. These rays, leaving one body, are absorbed by another body, and take on heat. And I will try to explain to you how this happens, and how much heat we lose by this radiation at home. (I think many will be interested to see these figures). At the end of the article, we will solve a problem from a real example.

The article will contain three-level formulas and integral expressions for mathematicians, but you should not be afraid of them, you may not even delve into these formulas. In the problem, I will give you formulas that can be solved one or two and you don’t even need to know higher mathematics, it’s enough to know elementary arithmetic.

I was convinced of this more than once that sitting by the fire (usually large) my face was burned by these rays. And if I covered the fire with my palms and at the same time my arms were outstretched, it turned out that my face stopped burning. It is not difficult to guess that these rays are straight as light. It is not the air circulating around the fire that burns me, and not even the air, but the direct, invisible heat rays coming from the fire.

In space, there is usually a vacuum between the planets and therefore the transfer of temperatures is carried out exclusively by thermal rays (All rays are electromagnetic waves).

Thermal radiation has a nature such as light and electromagnetic rays (waves). Simply, these waves (rays) have different wavelengths.

For example, wavelengths in the range of 0.76 - 50 microns are called infrared. All bodies having a room temperature of + 20 °C emit mainly infrared waves with wavelengths close to 10 microns.

Any body, if only its temperature is different from absolute zero (-273.15 ° C), is capable of sending radiation into the surrounding space. Therefore, any body radiates rays to the surrounding bodies and, in turn, is influenced by the radiation of these bodies.

Any furniture in the house (chair, table, walls and even a sofa) emits heat rays.

Thermal radiation can be absorbed or passed through the body, or it can simply be reflected off the body. The reflection of heat rays is similar to the reflection of a light beam from a mirror. The absorption of heat radiation is similar to how a black roof gets very hot from the sun's rays. And the penetration or passage of rays is similar to how rays pass through glass or air. The most common type of electromagnetic radiation in nature is thermal radiation.

Very close in its properties to a black body is the so-called relic radiation, or the cosmic microwave background - radiation filling the Universe with a temperature of about 3 K.

In general, in the science of heat engineering, in order to explain the processes of thermal radiation, it is convenient to use the concept of a black body in order to qualitatively explain the processes of thermal radiation. Only a black body is capable of facilitating calculations in some way.

As described above, any body is capable of:

black body- this is a body that completely absorbs thermal energy, that is, it does not reflect rays and thermal radiation does not pass through it. But do not forget that the black body radiates thermal energy.

Therefore, it is so easy to apply calculations to this body.

What difficulties arise in the calculation if the body is not a black body?

A body that is not a black body has these factors:

These two factors complicate the calculation so much that "mom don't cry." It's very difficult to count. And scientists on this occasion did not really explain how to calculate the gray body. By the way, a gray body is a body that is not a black body.

There is also a concept: White body and transparent body, but more on that below.

thermal radiation has different frequencies (different waves), and each individual body can have a different wave of radiation. In addition, when the temperature changes, this wavelength can change, and its intensity (radiation strength) can also change.

All these factors will complicate the process so much that it is difficult to find a universal formula for calculating energy losses due to radiance. And therefore, in textbooks and in any literature, a black body is used for calculation, and other gray bodies are used as part of a black body. The emissivity factor is used to calculate the gray body. These coefficients are given in reference books for some materials.

Consider an image that confirms the complexity of emissivity calculation.

The figure shows two balls that have particles of this ball in themselves. The red arrows are the rays emitted by the particles.

Consider a black body.

Inside the black body, deep inside, there are some particles that are indicated in orange. They emit rays that absorb nearby other particles, which are indicated in yellow. The rays of the orange particles of the black body are not able to pass through other particles. And therefore, only the outer particles of this ball emit rays over the entire area of ​​the ball. Therefore, the calculation of the black body is easy to calculate. It is also commonly believed that a black body emits the entire spectrum of waves. That is, it emits all available waves of various lengths. A gray body can emit part of the wave spectrum, only of a certain wavelength.

Consider a gray body.

Inside the gray body, the particles inside emit some part of the rays that pass through other particles. And that's why the calculation becomes much more complicated.

thermal radiation- this is electromagnetic radiation arising from the conversion of the energy of the thermal motion of body particles into radiation energy. It is the thermal nature of the excitation of elementary emitters (atoms, molecules, etc.) that opposes thermal radiation to all other types of luminescence and determines its specific property to depend only on the temperature and optical characteristics of the radiating body.

Experience shows that thermal radiation is observed in all bodies at any temperature other than 0 K. Of course, the intensity and nature of the radiation depend on the temperature of the radiating body. For example, all bodies with a room temperature of + 20 ° C emit mainly infrared waves with wavelengths close to 10 microns, and the Sun emits energy, the maximum of which falls on 0.5 microns, which corresponds to the visible range. At T → 0 K, bodies practically do not radiate.

Thermal radiation leads to a decrease in the internal energy of the body and, consequently, to a decrease in body temperature, to cooling. A heated body, due to thermal radiation, gives off internal energy and cools down to the temperature of the surrounding bodies. In turn, by absorbing radiation, cold bodies can heat up. Such processes, which can also occur in a vacuum, are called radiation.

Completely black body- a physical abstraction used in thermodynamics, a body that absorbs all electromagnetic radiation falling on it in all ranges and reflects nothing. Despite the name, a black body itself can emit electromagnetic radiation of any frequency and visually have a color. The radiation spectrum of a black body is determined only by its temperature.

Table:

(Temperature range in Kelvin and their Color)

up to 1000 Red

1000-1500 Orange

1500-2000 Yellow

2000-4000 Pale yellow

4000-5500 Yellowish white

5500-7000 Pure white

7000-9000 Bluish white

9000-15000 White-blue

15000-∞ Blue

By the way, according to the wavelength (color), the temperature of the sun was determined, it is about 6000 Kelvin. Embers usually glow red. Doesn't this remind you of anything? Temperature can be determined by color. That is, there are such devices that measure the wavelength, thereby determining the temperature of the material.

The blackest real substances, for example, soot, absorb up to 99% of the incident radiation (i.e., have an albedo equal to 0.01) in the visible wavelength range, but they absorb infrared radiation much worse. The deep black color of some materials (charcoal, black velvet) and the pupil of the human eye is explained by the same mechanism. Among the bodies of the solar system, the Sun has the properties of an absolutely black body to the greatest extent. By definition, the Sun reflects practically no radiation at all. The term was introduced by Gustav Kirchhoff in 1862.

According to the spectral classification, the Sun belongs to the type G2V (“yellow dwarf”). The surface temperature of the Sun reaches 6000 K, so the Sun shines with almost white light, but due to the absorption of part of the spectrum by the Earth's atmosphere near the surface of our planet, this light acquires a yellow tint.

Absolutely black body - absorbs 100% and at the same time heats up, and vice versa! a heated body - 100% radiates, which means that there is a strict pattern (the radiation formula of an absolutely black body) between the temperature of the Sun - and its spectrum - since both the spectrum and temperature have already been determined - yes, the Sun has no deviations from these parameters!

In astronomy, there is such a diagram - "Spectrum-Luminosity", and so our Sun belongs to the "main sequence" of stars, to which most other stars belong, that is, almost all stars are "absolutely black bodies", strange as it may seem ... Exceptions - white dwarfs, red giants and New, Super-New...

This is someone who did not study physics at school.

A black body absorbs ALL radiation and emits more than all other bodies (the more the body absorbs, the more it heats up; the more it heats up, the more it radiates).

Suppose we have two surfaces - gray (with a blackness factor of 0.5) and absolutely black (coefficient of 1).

The emissivity is the absorption coefficient.

Now on these surfaces by directing the same flux of photons, say, 100 pieces.

A gray surface will absorb 50 of them, a black surface will absorb all 100.

Which surface emits more light - in which "sits" 50 photons or 100?

The radiation of a completely black body was first correctly calculated by Planck.

The radiation of the Sun approximately obeys Planck's formula.

And so we begin to study the theory ...

Under the radiation (radiation) understand the emission and propagation of electromagnetic waves of any kind. Depending on the wavelength, there are: X-ray, ultraviolet, infrared, light (visible) radiation and radio waves.

x-ray radiation- electromagnetic waves, the photon energy of which lies on the scale of electromagnetic waves between ultraviolet radiation and gamma radiation, which corresponds to wavelengths from 10−2 to 103 Angstroms. 10 Angstroms = 1 nm. (0.001-100 nm)

Ultraviolet radiation(ultraviolet, UV, UV) - electromagnetic radiation, occupying the range between the violet border of visible radiation and X-ray radiation (10 - 380 nm).

Infrared radiation- electromagnetic radiation occupying the spectral region between the red end of visible light (with a wavelength of λ = 0.74 μm) and microwave radiation (λ ~ 1-2 mm).

Now the entire range of infrared radiation is divided into three components:

Shortwave region: λ = 0.74-2.5 µm;

Medium wave region: λ = 2.5-50 µm;

Long-wavelength region: λ = 50-2000 µm;

Visible radiation- electromagnetic waves perceived by the human eye. The sensitivity of the human eye to electromagnetic radiation depends on the wavelength (frequency) of the radiation, with the maximum sensitivity at 555 nm (540 terahertz), in the green part of the spectrum. Since the sensitivity drops to zero gradually with distance from the maximum point, it is impossible to indicate the exact boundaries of the spectral range of visible radiation. Usually, a section of 380-400 nm (750-790 THz) is taken as a short-wave boundary, and 760-780 nm (385-395 THz) as a long-wave boundary. Electromagnetic radiation with such wavelengths is also called visible light, or simply light (in the narrow sense of the word).

radio emission(radio waves, radio frequencies) - electromagnetic radiation with wavelengths of 5 10−5-1010 meters and frequencies, respectively, from 6 1012 Hz and up to several Hz. Radio waves are used in the transmission of data in radio networks.

thermal radiation is a process of propagation in space of the internal energy of a radiating body by means of electromagnetic waves. The causative agents of these waves are the material particles that make up the substance. The propagation of electromagnetic waves does not require a material medium; in vacuum they propagate at the speed of light and are characterized by a wavelength λ or an oscillation frequency ν. At temperatures up to 1500 °C, the main part of the energy corresponds to infrared and partly to light radiation (λ=0.7÷50 µm).

It should be noted that the radiation energy is not emitted continuously, but in the form of certain portions - quanta. The carriers of these portions of energy are the elementary particles of radiation - photons, which have energy, number of motions and electromagnetic mass. When it hits other bodies, the radiation energy is partially absorbed by them, partially reflected, and partially passes through the body. The process of conversion of radiation energy into the internal energy of an absorbing body is called absorption. Most solid and liquid bodies emit energy of all wavelengths in the range from 0 to ∞, that is, they have a continuous emission spectrum. Gases emit energy only in certain wavelength ranges (selective emission spectrum). Solids radiate and absorb energy by the surface, and gases by volume.

The energy radiated per unit time in a narrow range of wavelengths (from λ to λ+dλ) is called the flux of monochromatic radiation Qλ. The radiation flux corresponding to the entire spectrum in the range from 0 to ∞ is called the integral, or total, radiant flux Q(W). The integral radiant flux emitted from a unit surface of the body in all directions of the hemispherical space is called the integral radiation density (W / m 2).

To understand this formula, consider an image.

It was not by chance that I depicted two versions of the body. The formula is valid only for a square-shaped body. Since the radiating area must be flat. Provided that it radiates only the surface of the body. Internal particles do not radiate.

Knowing the radiation density of the material, it is possible to calculate how much energy is spent on radiation:

It must be understood that the rays emanating from the plane have different radiation intensities with respect to the normal of the plane.

Lambert's Law. The radiant energy emitted by the body propagates in space in different directions with different intensities. The law that establishes the dependence of the intensity of radiation on the direction is called Lambert's law.

Lambert's law establishes that the amount of radiant energy emitted by a surface element in the direction of another element is proportional to the product of the amount of energy emitted along the normal by the value of the spatial angle made by the direction of radiation with the normal

See image.

The intensity of each ray can be found using the trigonometric function:

That is, it is a kind of angle coefficient and it strictly obeys the trigonometry of the angle. The coefficient only works for a black body. Since nearby particles will absorb side rays. For a gray body, it is necessary to take into account the number of rays passing through the particles. The reflection of the rays must also be taken into account.

Consequently, the largest amount of radiant energy is emitted in a direction perpendicular to the radiation surface. Lambert's law is completely valid for a completely black body and for bodies that have diffuse radiation at a temperature of 0 - 60 ° C. For polished surfaces, Lambert's law does not apply. For them, radiation at an angle will be greater than in the direction normal to the surface.

Below we will definitely consider more voluminous formulas for calculating the amount of heat lost by the body. But for now, there is something more to be learned about the theory.

A little about definitions. Definitions come in handy to express yourself correctly.

Note that most solid and liquid bodies have a continuous (continuous) emission spectrum. This means that they have the ability to emit rays of all wavelengths.

Even an ordinary table in a room, as a solid body, can emit X-ray or ultraviolet radiation, but its intensity is so small that we don’t notice it, its value in relation to other waves can approach zero.

The radiant flux (or radiation flux) is the ratio of radiant energy to the radiation time, W:

where Q is the radiation energy, J; t - time, s.

If a radiant flux emitted by an arbitrary surface in all directions (i.e. within a hemisphere of arbitrary radius) is carried out in a narrow wavelength range from λ to λ + Δλ, then it is called a monochromatic radiation flux

The total radiation from the surface of the body over all wavelengths of the spectrum is called the integral or total radiation flux Ф

The integral flux emitted from a unit surface is called the surface flux density of the integral radiation or emissivity, W / m 2,

The formula can also be applied to monochromatic radiation. If thermal monochromatic radiation falls on the surface of a body, then in the general case, a part equal to B λ of this radiation will be absorbed by the body, i.e. will turn into another form of energy as a result of interaction with matter, part of F λ will be reflected, and part of D λ will pass through the body. If we assume that the radiation incident on the body is equal to unity, then

In λ +F λ +D λ =1

where B λ , F λ , D λ are the absorption and reflection coefficients, respectively

and transmission of the body.

When B, F, D remain constant within the spectrum, i.e. do not depend on the wavelength, then there is no need for indices. In this case

If B \u003d 1 (F \u003d D \u003d 0), then a body that completely absorbs all radiation incident on it, regardless of the wavelength, direction of incidence and state of radiation polarization, is called a black body or a full radiator.

If F=1 (B=D=0), then the radiation incident on the body is completely reflected. In the case when the surface of the body is rough, then the rays are reflected diffusely (diffuse reflection), and the body is called white, and when the surface of the body is smooth and the reflection follows the laws of geometric optics, then the body (surface) is called mirror. In the case when D \u003d 1 (B \u003d F \u003d 0), the body is permeable to thermal rays (diathermic).

Solids and liquids are practically opaque to thermal rays (D = 0), i.e. athermic. For such bodies

Absolutely black, as well as transparent or white bodies, do not exist in nature. Such bodies must be regarded as scientific abstractions. But still, some real bodies can come close enough in their properties to such idealized bodies.

It should be noted that some bodies have certain properties with respect to rays of a certain wavelength, and others with respect to rays of a different wavelength. For example, a body may be transparent to infrared rays and opaque to visible (light) rays. The surface of a body can be smooth for rays of one wavelength and rough for rays of another wavelength.

Gases, especially those under low pressure, in contrast to solids and liquids, emit a line spectrum. Thus, gases absorb and emit rays of only a certain wavelength, while they can neither emit nor absorb other rays. In this case, one speaks of selective (selective) absorption and emission.

In the theory of thermal radiation, an important role is played by a quantity called the spectral density of the radiation flux, or spectral emissivity, which is the ratio of the density of the radiant flux emitted in an infinitely small interval of wavelengths from λ to λ + Δλ, to the size of this interval of wavelengths Δλ, W / m 2,

where E is the surface density of the radiant flux, W/m 2 .

Now I hope you understand that the calculation process becomes extremely difficult. We still have to work and work in this direction. It is each material that needs to be tested at different temperatures. But for some reason, there is practically no data on materials. Rather, I did not find an experimental guide to materials.

Why is there no such material guide? Because the thermal radiation is very small, and I think it is unlikely to exceed 10% in our living conditions. Therefore, they are not included in the calculation. That's when we often fly into space, then all the calculations will appear. Rather, in our astronautics, data on materials have accumulated, but they are not yet freely available.

Law of absorption of radiant energy

Each body is capable of absorbing some part of the radiant energy, more on that below.

If a radiant flux falls on any body of thickness l (see figure), then in the general case, when passing through the body, it decreases. It is assumed that the relative change in the radiant flux along the path Δl is directly proportional to the path of the stream:

The coefficient of proportionality b is called the absorption index, which generally depends on the physical properties of the body and the wavelength.

Integrating from l to 0 and keeping b constant, we get

Let us establish the relationship between the spectral absorption coefficient of the body B λ and the spectral absorption index of the substance b λ .

From the definition of the spectral absorption coefficient B λ we have

After substituting the values ​​into this equation, we obtain the relationship between the spectral absorption coefficient B λ and the spectral absorption index B λ .

The absorption coefficient B λ is zero for l 1 = 0 and b λ = 0. For a large value of bλ, a very small value of l is sufficient, but still not equal to zero, so that the value of B λ is arbitrarily close to unity. In this case, we can say that absorption occurs in a thin surface layer of the substance. Only in this understanding is it possible to speak of surface absorption. For most solids, due to the large value of the absorption index b λ, “surface absorption” takes place in the indicated sense, and therefore the state of its surface has a great influence on the absorption coefficient.

Bodies, although with a small value of the absorption index, such as gases, can, with their sufficient thickness, have a large absorption coefficient, i.e. are made opaque to rays of a given wavelength.

If b λ \u003d 0 for the interval Δλ, and for other wavelengths b λ is not equal to zero, then the body will absorb the incident radiation only of certain wavelengths. In this case, as mentioned above, one speaks of a selective (selective) absorption coefficient.

Let us emphasize the fundamental difference between the absorption index of a substance b λ and the absorption coefficient B λ of a body. The first characterizes the physical properties of matter in relation to rays of a certain wavelength. The value of В λ depends not only on the physical properties of the substance of which the body consists, but also on the shape, size and condition of the surface of the body.

Laws of radiation of radiant energy

Max Planck theoretically, on the basis of electromagnetic theory, established a law (called Planck's law), expressing the dependence of the spectral emissivity of a black body E 0λ on wavelength λ and temperature T.

where E 0λ (λ, T) is the emissivity of the black body, W / m 2; T - thermodynamic temperature, K; C 1 and C 2 are constant; C 1 \u003d 2πhc 2 \u003d (3.74150 ± 0.0003) 10-16 W m 2; C 2 =hc/k=(1.438790±0.00019) 10 -2; m K (here h=(6.626176±0.000036) 10 -34 J s - Planck's constant; c=(299792458±1.2) m/s - propagation velocity of electromagnetic waves in free space: k - Boltzmann's constant. )

It follows from Planck's law that the spectral emissivity can be zero at a thermodynamic temperature equal to zero (T=0), or at a wavelength λ = 0 and λ→∞ (at T≠0).

Consequently, a black body radiates at any temperature greater than 0 K. (T> 0) rays of all wavelengths, i.e. has a continuous (continuous) emission spectrum.

From the above formula, you can get the calculated expression for the black body emissivity:

Integrating within the range of λ from 0 to ∞, we obtain

As a result of expanding the integrand into a series and integrating it, a calculated expression for the blackbody radiance is obtained, called the Stefan-Boltzmann law:

where E 0 is the emissivity of the black body, W / m 2;

σ - Stefan Boltzmann's constant, W / (m 2 K 4);

σ = (5.67032 ± 0.00071) 10 -8;

T is thermodynamic temperature, K.

The formula is often written in a more convenient form for calculation:

We will use this formula for calculations. But this is not the final formula. It is valid only for black bodies. How to use for gray bodies will be described below.

where E 0 is the emissivity of a black body; C 0 \u003d 5.67 W / (m 2 K 4).

The Stefan-Boltzmann law is formulated as follows: the emissivity of a black body is directly proportional to its thermodynamic temperature to the fourth power.

Spectral distribution of black body radiation at different temperatures

λ - wavelength from 0 to 10 microns (0-10000 nm)

E 0λ - should be understood as follows: As if in the volume (m 3) of a black body there is a certain amount of energy (W). This does not mean that it radiates such energy only from the outer particles. Simply, if we collect all the particles of a black body in a volume and measure the emissivity of each particle in all directions and add them all, then we will get the total energy on the volume, which is indicated on the graph.

As can be seen from the location of the isotherms, each of them has a maximum, and the higher the thermodynamic temperature, the greater the value of E0λ corresponding to the maximum, and the maximum point itself moves to the region of shorter waves. The shift of the maximum spectral emissivity E0λmax to shorter wavelengths is known as

Wien's displacement law, according to which

T λ max \u003d 2.88 10 -3 m K \u003d const and λ max \u003d 2.88 10 -3 / T,

where λ max is the wavelength corresponding to the maximum value of the spectral emissivity E 0λmax .

So, for example, at T = 6000 K (the approximate temperature of the surface of the Sun), the maximum E 0λ is located in the region of visible radiation, on which about 50% of the solar radiation falls.

The elementary area under the isotherm, shaded on the graph, is equal to E 0λ Δλ. It is clear that the sum of these areas, i.e. the integral is the black body emissivity E 0 . Therefore, the area between the isotherm and the x-axis depicts the black body emissivity on a conventional scale of the diagram. At low values ​​of the thermodynamic temperature, the isotherms pass in close proximity to the abscissa axis, and the indicated area becomes so small that it can practically be considered equal to zero.

The concepts of so-called gray bodies and gray radiation play an important role in technology. Gray is a non-selective thermal emitter capable of emitting a continuous spectrum, with a spectral emissivity E λ for waves of all wavelengths and at all temperatures, which is a constant fraction of the spectral emissivity of a black body E 0λ i.e.

The constant ε is called the emissivity of the heat emitter. For gray bodies, emissivity ε

The graph schematically shows the wavelength distribution curves of the spectral emissivity of an absolutely black body E λ (ε = 1) and the spectral emissivity of a gray body Eλ of the same temperature as the black body (at ε = 0.5 and ε = 0.25). Gray body emissivity

Work

is called the gray body emissivity.

The values ​​of the emissivity obtained from experience are given in the reference literature.

Most of the bodies used in technology can be mistaken for gray bodies, and their radiation - for gray radiation. More precise studies show that this is possible only as a first approximation, but sufficient for practical purposes. The deviation from the Stefan-Boltzmann law for gray bodies is usually taken into account by the fact that the emissivity C is assumed to be temperature dependent. In this regard, the tables indicate the temperature range for which the value of the emissivity coefficient C is experimentally determined.

In what follows, to simplify the conclusions, we will assume that the emissivity of a gray body does not depend on temperature.

Emissivity of some materials

(Material / Temperature in °C / E value)

Oxidized aluminum / 200-600 / 0.11 -0.19

Aluminum polished / 225-575 / 0.039-0.057

Brick red / 20 / 0.93

Refractory brick / - / 0.8-0.9

Oxidized copper / 200-600 / 0.57-0.87

Lead oxidized / 200 / 0.63

Polished steel / 940-1100 / 0.55-0.61

Turned cast iron / 830-910 / 0.6-0.7

Oxidized cast iron / 200-600 / 0.64-0.78

Aluminum polished / 50-500 / 0.04-0.06

Bronze / 50 / 0.1

Iron sheet galvanized, bright / 30 / 0.23

Tinplate white, old / 20 / 0.28

Polished gold / 200 - 600 / 0.02-0.03

Matt brass / 20-350 / 0.22

Copper polished / 50-100 / 0.02

Nickel polished / 200-400 / 0.07-0.09

Tin shiny / 20-50 / 0.04-0.06

Silver polished / 200-600 / 0.02-0.03

Steel sheet / 50 / 0.56

Oxidized steel / 200-600 / 0.8

Steel heavily oxidized / 500 / 0.98

Cast iron / 50 / 0.81

Asbestos cardboard / 20 / 0.96

Planed wood / 20 / 0.8-0.9

Refractory brick / 500-1000 / 0.8-0.9

Fireclay brick / 1000 / 0.75

Brick red, rough / 20 / 0.88-0.93

Lacquer black, matte / 40-100 / 0.96-0.98

Lacquer white / 40-100 / 0.8-0.95

Oil paints of various colors / 100 / 0.92-0.96

Lamp black / 20-400 / 0.95

Glass / 20-100 / 0.91-0.94

Enamel white / 20 / 0.9

Kirchhoff's law

Kirchhoff's law establishes the relationship between the emissivity and the absorption coefficient of a gray body.

Consider two parallel gray bodies of infinite length with flat surfaces of area A each.

An infinitely extended plane makes it possible to approximate calculations for finding real radiation in practical and theoretical experiments. In theoretical experiments, the real value is found through integral expressions, and in experiments, a large plane brings the calculations closer to real values. Thus, we, as it were, with a large infinite plane, extinguish the influence of unnecessary lateral and angular radiations that fly away and are not absorbed by the experimental plates.

That is, if the coefficient is multiplied by the emissivity, then we get the resulting radiation value (W).

It can be assumed that all the rays sent by one body completely fall on the other. Let us assume that the transmission coefficients of these bodies are D 1 = D 2 = 0 and there is a heat-transparent (diathermic) medium between the surfaces of two planes. Denote by E 1 , B 1 , F 1 , T 1 , and E 2 , B 2 , F 2 , T 2 respectively the emissivities, absorption coefficients, reflections and temperatures of the surfaces of the first and second bodies.

The flux of radiant energy from surface 1 to surface 2 is equal to the product of the radiance of surface 1 and its area A, i.e. E 1 A, from which part of E 1 B 2 A is absorbed by surface 2, and part of E 1 F 2 A is reflected back to surface 1. From this reflected flow E 1 F 2 A, surface 1 absorbs E 1 F 2 B 1 A and reflects E 1 F 1 F 2 A. FROM the reflected energy flow E 1 F 1 F 2 A, surface 2 will again absorb E 1 F 1 F 2 B 2 A and reflect E 1 F 1 F 2 A, etc.

Similarly, there is a transfer of radiant energy by the flow E 2 from surface 2 to surface 1. As a result, the radiant energy flux absorbed by surface 2 (or given away by surface 1)

The flux of radiant energy absorbed by surface 1 (or given off by surface 2),

Ultimately, the flux of radiant energy transmitted by surface 1 to surface 2 will be equal to the difference between the radiant fluxes Ф 1→2 and Ф 2→1 i.e.

The resulting expression is valid for all temperatures T 1 and T 2 and, in particular, for T 1 = T 2 . In the latter case, the system under consideration is in dynamic thermal equilibrium, and on the basis of the second law of thermodynamics, it is necessary to put Ф 1→2 = Ф 2→1, from which it follows

E 1 B 2 \u003d E 2 B 1 or

The resulting equality is called Kirchhoff's law: the ratio of the emissivity of a body to its absorption coefficient for all gray bodies at the same temperature is the same and equal to the emissivity of a black body at the same temperature.

If a body has a small absorption coefficient, such as a well-polished metal, then this body also has a low emissivity. On this basis, in order to reduce heat loss by radiation to the external environment, heat-releasing surfaces are covered with sheets of polished metal for thermal insulation.

When deriving Kirchhoff's law, gray radiation was considered. The conclusion remains valid even if the thermal radiation of both bodies is considered only in a certain part of the spectrum, but nevertheless has the same character, i.e. both bodies emit rays whose wavelengths lie in the same arbitrary spectral region. In the limiting case, we arrive at the case of monochromatic radiation. Then

those. for monochromatic radiation, Kirchhoff's law should be formulated as follows: the ratio of the spectral emissivity of a body at a certain wavelength to its absorption coefficient at the same wavelength is the same for all bodies at the same temperatures, and is equal to the spectral emissivity of a black body at the same wavelength waves at the same temperature.

We conclude that for a gray body B = ε, i.e. the concepts of "absorption coefficient" B and "blackness coefficient" ε for a gray body coincide. By definition, the blackness coefficient does not depend on either temperature or wavelength, and, consequently, the absorption coefficient of a gray body also does not depend on either wavelength or temperature.

Gas emission

The radiation of gases differs significantly from the radiation of solid bodies. Absorption and emission of gases - selective (selective). Gases absorb and emit radiant energy only in certain, rather narrow intervals Δλ of wavelengths - the so-called bands. In the rest of the spectrum, gases do not emit or absorb radiant energy.

Diatomic gases have a negligible ability to absorb radiant energy, and consequently, a small ability to radiate it. Therefore, these gases are usually considered diathermic. Unlike diatomic gases, polyatomic gases, including triatomic gases, have a significant ability to emit and absorb radiant energy. Of the triatomic gases in the field of heat engineering calculations, carbon dioxide (CO 2) and water vapor (H 2 O), which each have three emission bands, are of the greatest practical interest.

In contrast to solids, the absorption coefficient for gases (of course, in the region of absorption bands) is small. Therefore, for gaseous bodies it is no longer possible to speak of "surface" absorption, since the absorption of radiant energy occurs in a finite volume of gas. In this sense, the absorption and emission of gases are called volumetric. In addition, the absorption coefficient b λ for gases depends on temperature.

According to the absorption law, the spectral absorption coefficient of a body can be determined from:

For gaseous bodies, this dependence is somewhat complicated by the fact that the gas absorption coefficient is affected by its pressure. The latter is explained by the fact that absorption (radiation) proceeds the more intensely, the more molecules the beam meets on its way, and the volumetric number of molecules (the ratio of the number of molecules to volume) is directly proportional to pressure (at t = const).

In technical calculations of gas radiation, usually absorbing gases (CO 2 and H 2 O) are included as components in the mixture of gases. If the pressure of the mixture is p, and the partial pressure of the absorbing (or emitting) gas is p i, then it is necessary to substitute the value p i 1 instead of l. The value p i 1, which is the product of the gas pressure and its thickness, is called the effective layer thickness. Thus, for gases, the spectral absorption coefficient

The spectral absorption coefficient of a gas (in space) depends on the physical properties of the gas, the shape of the space, its dimensions, and the temperature of the gas. Then, in accordance with the Kirchhoff law, the spectral radiance

Emissivity within one band of the spectrum

This formula determines the radiance of a gas into free space (emptiness). (Free space can be considered as a black space at 0 K.) But the gas space is always limited by the surface of a solid, in the general case having a temperature T st ≠ T g and emissivity ε st

The emissivity of a gas in a closed space is equal to the sum of the emissivities taken over all spectral bands:

Experimental studies have shown that the radiance of gases does not follow the Stefan-Boltzmann law, i.e. depending on the fourth power of absolute temperature.

However, for practical calculations of gas radiation, the law of fourth powers is used, introducing an appropriate correction to the value of the gas blackness coefficient ε g:

Here ε g = f(T, p l)

Average beam path length

where V is the gas volume; A is the surface area of ​​the shell.

Emissivity of a gas whose components are CO 2 and H 2 O (gaseous products of combustion) to the shell of a gray body

in which the last term takes into account the self-radiation of the shell.

The so-called effective coefficient of emissivity of the shell ε "st, greater than ε st, due to the presence of radiating gas.

Gas blackness coefficient at gas temperature t g

The values ​​of the emissivity ε CO2 and ε H2O depending on the temperature at various values ​​of the parameter p i l are shown in the figure.

The correction factor β is determined according to the schedule.

The emission and absorption bands for CO 2 and H 2 0 somewhat overlap each other, and therefore part of the energy emitted by one gas is absorbed by another. Therefore, the emissivity of a mixture of carbon dioxide and water vapor at a wall temperature t st

where Δε g is the correction taking into account the indicated absorption. For gaseous combustion products of the usual composition, Δε g \u003d 2 - 4% and it can be neglected.

It can be assumed that at ε st \u003d 0.8 + 1.0, the effective emissivity coefficient of the shell is ε "st = 0.5 (ε st + 1).

These features of the emission and absorption of gases make it possible to establish the mechanism of the so-called "greenhouse effect", which has a significant impact on the formation and change of the Earth's climate.

Most of the solar radiation passes through the atmosphere and warms the Earth's surface. In turn, the Earth emits infrared radiation, as a result of which it cools. However, part of this radiation is absorbed by the polyatomic (“greenhouse”) gases of the atmosphere, which, as a result, plays the role of a “blanket” that retains heat. At the same time, such "greenhouse" gases as carbon dioxide (55%), freons and related gases (25%), methane (15%), etc. have the greatest impact on global warming.

On the next page, some more laws will be touched upon. There will also be a detailed explanation of how thermal radiation occurs through the window. Some factors influencing heat transfer by radiation will be described, and there will also be real problems for radiation.