When a parallel beam of monochromatic light is incident perpendicularly (normally) onto a diffraction grating on a screen in the focal plane of a collecting lens located parallel to the diffraction grating, a non-uniform pattern of illumination distribution in different areas of the screen (diffraction pattern) is observed.
Main the maxima of this diffraction pattern satisfy the following conditions:
Where n- order of the main diffraction maximum, d - constant (period) diffraction grating, λ - wavelength of monochromatic light,φn- angle between the normal to the diffraction grating and the direction to the main diffraction maximum n th order.
Constant (period) of the diffraction grating length l
where N - the number of slits (lines) per section of the diffraction grating with length I.
Along with the wavelengthfrequently used frequency v waves.
For electromagnetic waves(light) in a vacuum
where c = 3 * 10 8 m/s - speed propagation of light in a vacuum.
Let us select from formula (1) the most difficult mathematically determined formulas for the order of the main diffraction maxima:
where denotes the whole part numbers d*sin(φ/λ).
Underdetermined analogues of formulas (4, a, b) without the symbol [...] on the right-hand sides contain the potential danger of substituting a physically based selection operation integer part of a number operation rounding a number d*sin(φ/λ) to an integer value according to formal mathematical rules.
Subconscious tendency (false trail) to substitute the operation of isolating an integer part of a number d*sin(φ/λ) rounding operation
this number to an integer value according to mathematical rules is even more intensified when it comes to test tasks type B to determine the order of the main diffraction maxima.
In any type B test tasks, the numerical values of the required physical quantities by agreementrounded to integer values. However, in the mathematical literature there are no uniform rules for rounding numbers.
IN reference book V. A. Gusev, A. G. Mordkovich in mathematics for students and Belarusian textbook L. A. Latotina, V. Ya. Chebotarevsky in mathematics for fourth grade give essentially the same two rules for rounding numbers. They are formulated as follows: “When rounding decimal Before any digit, all digits following this digit are replaced by zeros, and if they are after the decimal point, they are discarded. If the first digit following this digit is greater than or equal to five, then the last remaining digit is increased by 1. If the first digit following this digit is less than 5, then the last remaining digit is not changed."
In M. Ya. Vygodsky’s reference book on elementary mathematics, which has gone through twenty-seven (!) editions, it is written (p. 74): “Rule 3. If the number 5 is discarded, and there are no significant figures behind it, then rounding is done to the nearest even number, i.e. The last digit stored remains unchanged if it is even, and is enhanced (increased by 1) if it is odd."
In view of the existence of various rules for rounding numbers, the rules for rounding decimal numbers should be explicitly formulated in the “Instructions for Students” attached to the tasks of centralized testing in physics. This proposal acquires additional relevance, since not only citizens of Belarus and Russia, but also other countries, enter Belarusian universities and undergo mandatory testing, and it is certainly unknown what rules for rounding numbers they used when studying in their countries.
In all cases, we will round decimal numbers according to rules, given in , .
After a forced retreat, let us return to the discussion of the physical issues under consideration.
Taking into account zero ( n= 0) of the main maximum and the symmetrical arrangement of the remaining main maxima relative to it, the total number of observed main maxima from the diffraction grating is calculated using the formulas:
If the distance from the diffraction grating to the screen on which the diffraction pattern is observed is denoted by H, then the coordinate of the main diffraction maximum n th order when counting from the zero maximum is equal to
If then (radians) and
Problems on the topic under consideration are often offered during physics tests.
Let's start the review by looking at Russian tests used by Belarusian universities at initial stage, when testing in Belarus was optional and carried out by separate educational institutions at your own peril and risk as an alternative to the usual individual written and oral form of entrance examinations.
Test No. 7
A32. The highest spectral order that can be observed by diffraction of light with a wavelength λ on a diffraction grating with a period d=3.5λ equals
1) 4; 2) 7; 3) 2; 4) 8; 5) 3.
Solution
Monochromaticno light spectra out of the question. In the problem statement, we should talk about the main diffraction maximum of the highest order when monochromatic light is perpendicularly incident on the diffraction grating.
According to formula (4, b)
From an underdetermined condition
on the set of integers, after rounding we getn max=4.
Only due to the mismatch of the integer part of the number d/λ with its rounded integer value the correct solution is ( n max=3) differs from incorrect (nmax=4) at the test level.
An amazing miniature, despite the flaws in the wording, with a delicately verified false trail across all three versions of rounding numbers!
A18. If the diffraction grating constant d= 2 µm, then for white light normally incident on the grating 400 nm<λ < 700 нм наибольший полностью наблюдаемый порядок спектра равен
1)1; 2)2; 3)3; 4)4; 5)5.
Solution
It's obvious that n sp =min(n 1max, n 2max)
According to formula (4, b)
Rounding numbers d/λ to integer values according to the rules - , we get:
Due to the fact that the integer part of the number d/λ 2 differs from its rounded integer value, this task allows you to objectively distinguish the correct solution(n sp = 2) from incorrect ( n sp =3). A great problem with one false lead!
CT 2002 Test No. 3
B5. Find the highest spectral order for the yellow Na line (λ = 589 nm), if the diffraction grating constant is d = 2 µm.
Solution
The task is formulated scientifically incorrectly. Firstly, when illuminating the diffraction gratingmonochromaticWith light, as noted above, there can be no talk of a spectrum (spectra). The problem statement should deal with the highest order of the main diffraction maximum.
Secondly, the task conditions should indicate that light falls normally (perpendicularly) onto a diffraction grating, since only this particular case is considered in the physics course of secondary educational institutions. This limitation cannot be considered as implied by default: all restrictions must be specified in tests obviously! Test tasks must be self-sufficient, scientifically correct tasks.
The number 3.4, rounded to an integer value according to the rules of arithmetic - , also gives 3. Exactly therefore, this task should be considered simple and, by and large, unsuccessful, since at the test level it does not allow one to objectively distinguish the correct solution, determined by the integer part of the number 3.4, from the incorrect solution, determined by the rounded integer value of the number 3.4. The difference is revealed only with a detailed description of the solution process, which is what is done in this article.
Addendum 1. Solve the above problem by replacing in its condition d=2 µm by d= 1.6 µm. Answer: nmax = 2.
CT 2002 Test 4
B5. Light from a gas-discharge lamp is directed onto the diffraction grating. The diffraction spectra of the lamp radiation are obtained on the screen. Line with wavelength λ 1 = 510 nm in the fourth order spectrum coincides with the wavelength line λ 2 in the third order spectrum. What is it equal to λ 2(in [nm])?
Solution
In this problem, the main interest is not the solution of the problem, but the formulation of its conditions.
When illuminated by a diffraction gratingnon-monochromatic light( λ 1 , λ 2) quite it is natural to talk (write) about diffraction spectra, which in principle do not exist when illuminating a diffraction gratingmonochromatic light.
The task conditions should indicate that the light from the gas-discharge lamp falls normally on the diffraction grating.
In addition, the philological style of the third sentence in the task condition should be changed. The turnover of the "line with wavelength" hurts the ear λ "" , it could be replaced by “a line corresponding to radiation with a wavelength λ "" or in shorter form - “a line corresponding to the wavelength λ "" .
Test formulations must be scientifically correct and literary impeccable. Tests are formulated completely differently from research and Olympiad problems! In tests, everything should be precise, specific, unambiguous.
Taking into account the above clarification of the task conditions, we have:
Since according to the conditions of the task That
CT 2002 Test No. 5
B5. Find the highest order of the diffraction maximum for the yellow sodium line with a wavelength of 5.89·10 -7 m if the diffraction grating period is 5 µm.
Solution
Compared to task B5 from test No. 3 TsT 2002, this task is formulated more precisely, however, in the task conditions, we should talk not about the “diffraction maximum”, but about “ main diffraction maximum".
Along with main diffraction maxima there are always also secondary diffraction maxima. Without explaining this nuance in a school physics course, it is all the more necessary to strictly adhere to the established scientific terminology and talk only about the main diffraction maxima.
In addition, it should be noted that the light falls normally on the diffraction grating.
Taking into account the above clarifications
From an undefined condition
according to the rules of mathematical rounding of the number 8.49 to an integer value, we again obtain 8. Therefore, this task, like the previous one, should be considered unsuccessful.
Addendum 2. Solve the above problem by replacing in its condition d =5 µm per (1=A µm. Answer:nmax=6.)
RIKZ manual 2003 Test No. 6
B5. If the second diffraction maximum is located at a distance of 5 cm from the center of the screen, then when the distance from the diffraction grating to the screen increases by 20%, this diffraction maximum will be located at a distance... cm.
Solution
The condition of the task is formulated unsatisfactorily: instead of “diffraction maximum” you need “main diffraction maximum”, instead of “from the center of the screen” - “from the zero main diffraction maximum”.
As can be seen from the above figure,
From here
RIKZ manual 2003 Test No. 7
B5. Determine the highest spectral order in a diffraction grating having 500 lines per 1 mm when illuminated with light with a wavelength of 720 nm.
Solution
The conditions of the task are formulated extremely unsuccessfully from a scientific point of view (see clarifications of tasks No. 3 and 5 from the CT 2002).
There are also complaints about the philological style of wording the assignment. Instead of the phrase “in a diffraction grating” one would have to use the phrase “from a diffraction grating”, and instead of “light with a wavelength” - “light whose wavelength”. The wavelength is not the load on the wave, but its main characteristic.
Taking into account clarifications
Using all three rules for rounding numbers above, rounding 2.78 to a whole number results in 3.
The last fact, even with all the shortcomings in the formulation of the task conditions, makes it interesting, since it allows us to distinguish the correct (nmax=2) and incorrect (nmax=3) solutions.
Many tasks on the topic under consideration are contained in the CT 2005.
In the conditions of all these tasks (B1), you need to add the keyword “main” before the phrase “diffraction maximum” (see comments to task B5 CT 2002 Test No. 5).
Unfortunately, in all versions of the V1 TsT 2005 tests, the numerical values d(l,N) And λ poorly chosen and always given in fractions
the number of “tenths” is less than 5, which does not allow at the test level to distinguish the operation of separating an integer part of a fraction (correct decision) from the operation of rounding a fraction to an integer value (false trace). This circumstance calls into question the advisability of using these tasks to objectively test applicants’ knowledge on the topic under consideration.
It seems that the test compilers were carried away, figuratively speaking, by preparing various “side dishes for the dish”, without thinking about improving the quality of the main component of the “dish” - the selection of numerical values d(l,N) And λ in order to increase the number of "tenths" in fractions d/ λ=l/(N* λ).
CT 2005 Option 4
B1. On a diffraction grating whose periodd 1=1.2 µm, a normally parallel beam of monochromatic light with a wavelength of λ =500 nm. If we replace it with a lattice whose periodd 2=2.2 µm, then the number of maxima will increase by... .
Solution
Instead of "light with wavelength λ"" you need "light wavelength λ "" . Style, style and more style!
Because
then, taking into account the fact that X is const, and d 2 >di,
According to formula (4, b)
Hence, ΔN total max =2(4-2)=4
When rounding the numbers 2.4 and 4.4 to integer values, we also get 2 and 4, respectively. For this reason, this task should be considered simple and even unsuccessful.
Addendum 3. Solve the above problem by replacing in its condition λ =500 nm at λ =433 nm (blue line in the hydrogen spectrum).
Answer: ΔN total. max=6
CT 2005 Option 6
B1. On a diffraction grating with a period d= A normally parallel beam of monochromatic light with a wavelength of λ =750 nm. Number of maxima that can be observed within an angle A=60°, the bisector of which is perpendicular to the plane of the lattice, is equal to... .
Solution
The phrase "light with a wavelength λ " has already been discussed above in CT 2005, option 4.
The second sentence in the conditions of this task could be simplified and written as follows: “The number of observed main maxima within the angle a = 60°” and further according to the text of the original task.
It's obvious that
According to formula (4, a)
According to formula (5, a)
This task, like the previous one, does not allow objectively determine the level of understanding of the topic being discussed by applicants.
Appendix 4. Complete the above task, replacing in its condition λ =750 nm at λ = 589 nm (yellow line in the sodium spectrum). Answer: N o6ш =3.
CT 2005 Option 7
B1. On a diffraction grating havingN 1- 400 strokes per l=1 mm in length, a parallel beam of monochromatic light with a wavelength of λ =400 nm. If it is replaced with a lattice havingN 2=800 strokes per l=1 mm in length, then the number of diffraction maxima will decrease by... .
Solution
We will omit the discussion of inaccuracies in the wording of the task, since they are the same as in previous tasks.
From formulas (4, b), (5, b) it follows that
(α) to the diffraction grating, its wavelength (λ), grating (d), diffraction angle (φ) and spectral order (k). In this formula, the product of the grating period by the difference between the angles of diffraction and incidence is equated to the product of the order of the spectrum by monochromatic light: d*(sin(φ)-sin(α)) = k*λ.
Express the order of the spectrum from the formula given in the first step. As a result, you should get an equality, on the left side of which the desired value will remain, and on the right side there will be the ratio of the product of the lattice period by the difference between the sines of two known angles to the wavelength of light: k = d*(sin(φ)-sin(α)) /λ.
Since the grating period, wavelength and angle of incidence in the resulting formula are constant values, the order of the spectrum depends only on the diffraction angle. In the formula it is expressed through the sine and appears in the numerator of the formula. It follows from this that the larger the sine of this angle, the higher the order of the spectrum. The maximum value that the sine can take is one, so simply replace sin(φ) with one in the formula: k = d*(1-sin(α))/λ. This is the final formula for calculating the maximum order value of the diffraction spectrum.
Substitute numerical values from the conditions of the problem and calculate the specific value of the desired characteristic of the diffraction spectrum. In the initial conditions, it can be said that the light incident on the diffraction grating is composed of several shades with different wavelengths. In this case, use the one that has the least value in your calculations. This value is in the numerator of the formula, so the largest value of the spectrum period will be obtained at the smallest wavelength.
Light waves are deflected from their straight path when passing through small holes or past equally small obstacles. This phenomenon occurs when the size of obstacles or holes is comparable to the wavelength, and is called diffraction. Problems of determining the angle of deflection of light have to be solved most often in relation to diffraction gratings - surfaces in which transparent and opaque areas of the same size alternate.
Instructions
Find out the period (d) of the diffraction grating - this is the name given to the total width of one transparent (a) and one opaque (b) stripe: d = a+b. This pair is usually called one lattice stroke, and in the number of strokes per . For example, diffraction may contain 500 lines per 1 mm, and then d = 1/500.
For calculations, what matters is the angle (α) at which the light hits the diffraction grating. It is measured from the normal to the grating surface, and the sine of this angle is included in the formula. If the initial conditions of the problem say that light falls along the normal (α=0), this value can be neglected, since sin(0°)=0.
Find out the wavelength (λ) of the diffraction grating light. This is one of the most important characteristics that determine the diffraction angle. Normal sunlight contains a whole spectrum of wavelengths, but in theoretical problems and laboratory work, as a rule, we are talking about a point portion of the spectrum - “monochromatic” light. The visible region corresponds to lengths from approximately 380 to 740 nanometers. For example, one of the shades of green has a wavelength of 550 nm (λ = 550).
sinφ ≈ tanφ.
sinφ ≈ tanφ.
5 ≈ tanφ.
sinφ ≈ tanφ.
ν = 8.10 14 sinφ ≈ tanφ.
R=2 mm; a=2.5 m; b=1.5 m
a) λ=0.4 µm.
b) λ=0.76 µm
20) The screen is located at a distance of 50 cm from the diaphragm, which is illuminated by yellow light with a wavelength of 589 nm from a sodium lamp. At what aperture diameter will the geometric optics approximation be valid?
Solving problems on the topic “Diffraction grating”
1) A diffraction grating, the constant of which is 0.004 mm, is illuminated with light with a wavelength of 687 nm. At what angle to the grating must the observation be made in order to see the image of the second order spectrum.
2) Monochromatic light with a wavelength of 500 nm is incident on a diffraction grating having 500 lines per 1 mm. The light hits the grating perpendicularly. What is the highest order of the spectrum that can be observed?
3) The diffraction grating is located parallel to the screen at a distance of 0.7 m from it. Determine the number of lines per 1 mm for this diffraction grating if, under normal incidence of a light beam with a wavelength of 430 nm, the first diffraction maximum on the screen is located at a distance of 3 cm from the central light stripe. Consider that sinφ ≈ tanφ.
Diffraction grating formula
for small angles
tangent of angle = distance from maximum / distance to screen
lattice period
number of strokes per unit length (per mm)
4) A diffraction grating, the period of which is 0.005 mm, is located parallel to the screen at a distance of 1.6 m from it and is illuminated by a light beam of wavelength 0.6 μm incident normal to the grating. Determine the distance between the center of the diffraction pattern and the second maximum. Consider that sinφ ≈ tanφ.
5) Diffraction grating with a period of 10-5 m is located parallel to the screen at a distance of 1.8 m from it. The grating is illuminated by a normally incident beam of light with a wavelength of 580 nm. On the screen at a distance of 20.88 cm from the center of the diffraction pattern, maximum illumination is observed. Determine the order of this maximum. Assume that sinφ≈ tanφ.
6) Using a diffraction grating with a period of 0.02 mm, the first diffraction image was obtained at a distance of 3.6 cm from the central one and at a distance of 1.8 m from the grating. Find the wavelength of the light.
7) The spectra of the second and third orders in the visible region of the diffraction grating partially overlap with each other. What wavelength in the third-order spectrum corresponds to the wavelength of 700 nm in the second-order spectrum?
8)Plane monochromatic wave with frequency 8.10 14 Hz falls normal to the diffraction grating with a period of 5 μm. A collecting lens with a focal length of 20 cm is placed parallel to the grating behind it. The diffraction pattern is observed on the screen in the focal plane of the lens. Find the distance between its main maxima of 1st and 2nd orders. Consider that sinφ ≈ tanφ.
9) What is the width of the entire first-order spectrum (wavelengths ranging from 380 nm to 760 nm) obtained on a screen located 3 m from a diffraction grating with a period of 0.01 mm?
10) A normally parallel beam of white light falls on a diffraction grating. Between the grille and the screen, close to the grille, there is a lens that focuses the light passing through the grille onto the screen. What is the number of lines per 1 cm if the distance to the screen is 2 m and the width of the first order spectrum is 4 cm. The lengths of the red and violet waves are 800 nm and 400 nm, respectively. Consider that sinφ ≈ tanφ.
11) Plane monochromatic light wave with frequencyν = 8.10 14 Hz falls normal to the diffraction grating with a period of 6 μm. A collecting lens is placed behind it parallel to the grating. The diffraction pattern is observed in the rear focal plane of the lens. The distance between its main maxima of the 1st and 2nd orders is 16 mm. Find the focal length of the lens. Consider that sinφ ≈ tanφ.
12) What should be the total length of a diffraction grating having 500 lines per 1 mm in order to resolve two spectral lines with wavelengths of 600.0 nm and 600.05 nm?
13) Diffraction grating with a period of 10-5 m has 1000 strokes. Is it possible to resolve two lines of the sodium spectrum with wavelengths of 589.0 nm and 589.6 nm in the first-order spectrum using this grating?
14) Determine the resolution of a diffraction grating, the period of which is 1.5 μm and the total length is 12 mm, if light with a wavelength of 530 nm is incident on it.
15) Determine the resolution of a diffraction grating containing 200 lines per 1 mm if its total length is 10 mm. Radiation with a wavelength of 720 nm is incident on the grating.
16) What is the minimum number of lines the grating must contain so that two yellow sodium lines with wavelengths of 589 nm and 589.6 nm can be resolved in the first-order spectrum. What is the length of such a lattice if the lattice constant is 10 microns.
17) Determine the number of open zones with the following parameters:
R=2 mm; a=2.5 m; b=1.5 m
a) λ=0.4 µm.
b) λ=0.76 µm
18) A diaphragm with a diameter of 1 cm is illuminated with green light with a wavelength of 0.5 μm. At what distance from the diaphragm will the geometric optics approximation be valid?
19) A 1.2 mm slit is illuminated with green light with a wavelength of 0.5 µm. The observer is located at a distance of 3 m from the slit. Will he see the diffraction pattern?
20) The screen is located at a distance of 50 cm from the diaphragm, which is illuminated by yellow light with a wavelength of 589 nm from a sodium lamp. At what diaphragm diameter will the approximation ge be valid?metric optics.
21) A 0.5 mm slit is illuminated with green light from a laser with a wavelength of 500 nm. At what distance from the slit can the diffraction pattern be clearly observed?
3. Using a lens, a real image with a height of 18 cm was obtained from an object 3 cm high. When the object was moved 6 cm, a virtual image with a height of 9 cm was obtained. Determine the focal length of the lens (in centimeters).
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We solve the system of equations for d 1 or d 2. Define F= 12 cm.
Answer:F= 12 cm
4. A red light beam with a wavelength of 720 nm falls on a plate made of a material with a refractive index of 1.8 perpendicular to its surface. What is the smallest thickness of the plate that must be taken so that the light passing through the plate has maximum intensity?
minimal, then 0 " style="margin-left:7.8pt;border-collapse:collapse;border:none">
Given:
λ = 590 nm = 5.9×10–7 m
l= 10-3 m
Solution:
Condition max on the diffraction grating: d sinφ = kλ, Where k will be max if max is sinφ. And sinmaxφ = 1, then , where ; 
.
k max – ?
k can only take integer values, therefore k max = 3.
Answer: k max = 3.
6. The diffraction grating period is 4 µm. The diffraction pattern is observed using a lens with a focal length F= 40 cm. Determine the wavelength of light incident normally on the grating (in nm), if the first maximum is obtained at a distance of 5 cm from the central one.
Answer:λ = 500 nm
7. The height of the Sun above the horizon is 46°. In order for the rays reflected from a flat mirror to go vertically upward, the angle of incidence of the sun's rays on the mirror must be equal to:
1) 68° 2) 44° 3) 23° 4) 46° 5) 22°
Given: |
Angle of incidence equal to angle reflections α = α¢. From the figure it can be seen that α + α¢ + φ = 90° or 2α + φ = 90°, then |
Solution:
.Answer:
8. A point mirror is placed in the middle between two flat mirrors parallel to each other. If the source begins to move in a direction perpendicular to the planes of the mirrors at a speed of 2 m/s, then the first virtual images of the source in the mirrors will move relative to each other at a speed:
1) 0 m/s 2) 1 m/s 3) 2 m/s 4) 4 m/s 5) 8 m/s
Solution:
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Answer:
9. The limiting angle of total internal reflection at the interface between diamond and liquid nitrogen is 30°. The absolute refractive index of diamond is 2.4. How many times is the speed of light in a vacuum greater than the speed of light in liquid nitrogen?
1) 1.2 times 2) 2 times 3) 2.1 times 4) 2.4 times 5) 4.8 times
Given: | Solution: |
| Law of refraction: or for total internal reflection: ; n 1 = 2,4; |
With/υ2 – ? |
n 2 = n 1sinαpr = 1.2..gif" width="100" height="49 src=">.
Answer:
10. Two lenses - a diverging lens with a focal length of 4 cm and a converging lens with a focal length of 9 cm - are located so that their main optical axes coincide. At what distance from each other should the lenses be placed so that a beam of rays parallel to the main optical axis, passing through both lenses, remains parallel?
1) 4 cm 2) 5 cm 3) 9 cm 5) At any distance the rays will not be parallel.
Solution:
d = F 2 – F 1 = 5 (cm).
Given:
A= 10 cm
n st = 1.51
Solution:
; 
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Answer:b= 0.16 m
2. (7.8.3). At the bottom of a glass bath there is a mirror, on top of which a layer of water 20 cm high is poured. A lamp hangs in the air at a height of 30 cm above the surface of the water. At what distance from the surface of the water will an observer looking into the water see the image of a lamp in a mirror? The refractive index of water is 1.33. Present the result in SI units and round to the nearest tenth.
Given: h 1 = 20 cm h 2 = 30 cm n = 1,33 | Solution: |
| S` – virtual image; (1); (2); (3) a, b – small |
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Given:
O.C.= 4 m
S 1S 2 = 1 mm
L 1 = L 2 = OS
Solution:
D= k l – maximum condition
D= L 2 – L 1;
at 1 – ?
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2(OS)D = 2 ukd, from here 
; ; l = OS;
Given:
F= 0.15 m
f= 4.65 m
S= 4.32 cm2
Solution:
; ; S` = G 2 S
S– slide platform
; ; 
S` – ?
S` = 302 × 4.32 = 3888 (cm2) » 0.39 (m2)
Answer: S` = 0.39 m2
5. (7.8.28). Find the magnification factor of the object image AB given by a thin diverging lens with a focal length F. Round the result to hundredths.
Given: | Solution: |
|
|
G – ? |
; d 1 = 2F;https://pandia.ru/text/78/506/images/image708.gif" width="111" height="52 src=">; d 2 = F;
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l = d 1 – d 2 = F; https://pandia.ru/text/78/506/images/image712.gif" width="131" height="48 src=">
Answer: G = 0,17
OPTION No. 10
structure of the atom and nucleus. elements of the theory of relativity
Part A
1. Determine the retardation voltage required to stop the emission of electrons from the photocathode if radiation with a wavelength of 0.4 μm is incident on its surface and the red limit of the photoelectric effect is 0.67 μm. Planck's constant is 6.63×10-34 J×s, the speed of light in vacuum is 3×108 m/s. Provide your answer in SI units and round to the nearest hundredth.
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Answer: U h = 1.25 V
2. What is the mass of an X-ray photon with a wavelength of 2.5×10–10 m?
1) 0 kg 2) 3.8×10-33 kg 3) 6.6×10-32 kg 4) 8.8×10-31 kg 5) 1.6×10-19 kg
Given: l = 2.5×10-10 m | Solution: Photon energy: ; energy and mass are related by the relation: |
ε = mc 2. Then ; from here 
(kg).
Answer:
3. A beam of ultraviolet rays with a wavelength of 1×10-7 m imparts an energy of 10-6 J to a metal surface in 1 second. Determine the strength of the resulting photocurrent if the photoelectric effect is caused by 1% of incident photons.
1) 5×10-10 A 2) 6×10-14 A 3) 7×10-10 A 4) 8×10-10 A 5) 5×10-9 A
Given: D t= 1 s W= 10-6 J N 2 = 0,01N 1 | Solution: W = ε N 1, , where W– energy of all photons in the beam, N 1 – number of photons in the beam, – energy of one photon; ; N 2 = 0,01N 1; |
(A).
