sinφ ≈ tgφ.

sinφ ≈ tgφ.

5 ≈ tgφ.

sinφ ≈ tgφ.

ν = 8.10 14 sinφ ≈ tgφ.

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 approximation of geometrical optics be valid.

Solving problems on the topic "Diffraction grating"

1) A diffraction grating whose constant is 0.004 mm is illuminated by light with a wavelength of 687 nm. At what angle to the grating should 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. Light is incident on 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 at a distance of 3 cm from the central bright band. Think that sinφ ≈ tgφ.

Grating formula

for small angles
tangent of the angle = p-tion from u maximum / p-tion to the screen
grating period
number of strokes unit length (per mm)

4) A diffraction grating with a period of 0.005 mm is located parallel to the screen at a distance of 1.6 m from it and is illuminated by a beam of light with a wavelength of 0.6 μm incident along the normal to the grating. Determine the distance between the center of the diffraction pattern and the second maximum. Think that sinφ ≈ tgφ.

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. The maximum illumination is observed on the screen at a distance of 20.88 cm from the center of the diffraction pattern. Determine the order of this maximum. Assume that sinφ≈tgφ.

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) 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 a wavelength of 700 nm in the second order spectrum?

8) Plane monochromatic wave with a frequency of 8.10 14 Hz falls along the normal to the diffraction grating with a period of 5 μm. A converging 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. Think that sinφ ≈ tgφ.

9) What is the width of the entire first-order spectrum (wavelengths range from 380 nm to 760 nm) obtained on a screen 3 m away 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 grating and the screen, close to the grating, there is a lens that focuses the light passing through the grating onto the screen. What is the number of strokes per 1 cm, if the distance to the screen is 2 m, and the width of the spectrum of the first order is 4 cm. The lengths of the red and violet waves are respectively 800 nm and 400 nm. Think that sinφ ≈ tgφ.

11) Plane monochromatic light wave with frequency v = 8.10 14 Hz falls along the normal to the diffraction grating with a period of 6 μm. Parallel to the grating, a converging lens is placed behind it. 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. Think that sinφ ≈ tgφ.

12) What should be the total length of a diffraction grating with 500 lines per 1 mm in order to resolve two spectral lines with wavelengths of 600.0 nm and 600.05 nm with its help?

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 using this grating in the first-order spectrum?

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 falls 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 falls on the grating.

16) What is the smallest number of lines the grating should 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 grating if the grating constant is 10 µm.

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 microns. At what distance from the diaphragm will the geometrical optics approximation be valid?

19) A 1.2 mm slit is illuminated with green light at 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 aperture diameter will the approximationmetric 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?

With a perpendicular (normal) incidence of a parallel beam of monochromatic light on a diffraction grating on the screen in the focal plane of the converging lens, located parallel to the diffraction grating, an inhomogeneous pattern of illumination distribution of different parts of the screen (diffraction pattern) is observed.

Main the maxima of this diffraction pattern satisfy the following conditions:

where n is the order of the main diffraction maximum, d - constant (period) of the diffraction grating, λ is the wavelength of monochromatic light,φ n- the angle between the normal to the diffraction grating and the direction to the main diffraction maximum n th order.

The constant (period) of a diffraction grating with a length l

where N - the number of slots (strokes) per section of the diffraction grating with length I.

Along with the wavelengthfrequently used frequency v waves.

For electromagnetic waves (light) in vacuum

where c \u003d 3 * 10 8 m / s - speed propagation of light in a vacuum.

Let us single out from formula (1) the most difficult mathematically determined formulas for the order of the main diffraction maxima:

where denotes the integer part numbers d*sin(φ/λ).

Underdetermined analogues of formulas (4, a,b) without symbol [...] in the right parts contain the potential danger of substituting a physically based allocation operation the integer part of the number by the operation rounding number d*sin(φ/λ) to an integer value according to formal mathematical rules.

Subconscious tendency (false trace) to replace the operation of extracting the integer part of the number d*sin(φ/λ) rounding operation

this number to an integer value according to mathematical rules is even more enhanced when it comes to test tasks type B to determine the order of the main diffraction maxima.

In any test tasks of type B, the numerical values ​​of the required physical quantitiesby agreementrounded to integer values. However, in the mathematical literature there are no uniform rules(s) for rounding numbers.

In the reference book of V. A. Gusev, A. G. Mordkovich on mathematics for students and the Belarusian textbook L. A. Latotin, V. Ya. Chebotarevskii on mathematics for grade IV, essentially the same two rules for rounding numbers are given. In they are formulated as follows: "When rounding a decimal fraction to some digit, all the digits following this digit are replaced by zeros, and if they are after the decimal point, then they are discarded. If the first digit following this digit is greater than or equal to five, then the last remaining digit increase 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 performed to the nearest even number, i.e. the last stored digit remains unchanged if it is even, and amplifies (increases 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 not known what rounding rules they used when studying in their countries.

In all cases, decimal numbers will be rounded according to rules, given in , .

After a forced digression, 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 by 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 (radian) and

Problems on the topic under consideration are often offered at tests in physics.

Let's start the review with a review of Russian tests used by Belarusian universities at the initial stage, when testing in Belarus was optional and was conducted by individual educational institutions at their own peril and risk as an alternative to the usual individual written-oral form of entrance exams.

Test #7

A32. The highest order of the spectrum that can be observed in the 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.

Decision

Monochromaticno light spectra out of the question. In the condition of the problem, we should talk about the main diffraction maximum of the highest order for a perpendicular incidence of monochromatic light on a diffraction grating.

According to the 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 false trace finely adjusted for all three versions of rounding numbers!

A18. If the diffraction grating constant d= 2 μm, then for white light normally incident on the grating is 400 nm<λ < 700 нм наибольший полностью наблюдаемый порядок спектра равен

1)1; 2)2; 3)3; 4)4; 5)5.

Decision

It's obvious that n cn \u003d min (n 1max, n 2max)

According to the formula (4, b)

Rounding the numbers d/λ to integer values ​​according to the rules - , we get:

Due to the fact that the integer part of the number d/λ2 differs from its rounded integer value, this task allows you to objectively identify the correct solution(n cn = 2) from wrong ( n cn =3). Great problem with one false trail!

CT 2002 Test No. 3

AT 5. Find the highest order of the spectrum for the yellow line Na (λ = 589 nm) if the constant of the diffraction grating is d = 2 µm.

Decision

The task is formulated scientifically incorrectly. First, when illuminating the diffraction gratingmonochromaticlight, as noted above, there can be no question of the spectrum (spectra). In the condition of the problem, we should talk about the highest order of the main diffraction maximum.

Secondly, in the condition of the task it should be indicated that the light falls normally (perpendicularly) on the diffraction grating, because only this special case is considered in the physics course of secondary educational institutions. It is impossible to consider this restriction implied by default: in tests, all restrictions must be specified clearly! Test tasks should 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 recognized as 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 wrong solution, determined by the rounded integer value of the number 3.4. The difference is revealed only with a detailed description of the course of the solution, which is done in this article.

Addition 1. Solve the above problem by replacing in its condition d=2 µm to d= 1.6 µm. Answer: nmax = 2.

CT 2002 Test 4

AT 5. Light from a gas-discharge lamp is directed onto a diffraction grating. The diffraction spectra of the lamp radiation are obtained on the screen. Line with wavelength λ 1 = 510 nm in the spectrum of the fourth order coincides with the wavelength line λ2 in the spectrum of the third order. What is equal to λ2(in [nm])?

Decision

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( λ1 , λ2) quite it is natural to speak (write) about diffraction spectra, which, in principle, do not exist when a diffraction grating is illuminatedmonochromatic light.

The condition of the task 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 assignment should have been changed. Cuts hearing turnover 'line with a wavelength λ "" , it could be replaced by "a line corresponding to radiation of a wavelength λ "" or, more concisely, "a line corresponding to the wavelength λ "" .

Test formulations must be scientifically correct and literary impeccable. Tests are formulated in a completely different way than research and Olympiad tasks! In tests, everything should be accurate, specific, unambiguous.

Taking into account the above clarification of the task conditions, we have:

Since according to the condition of the assignment then

CT 2002 Test No. 5

AT 5. Find the highest order of the diffraction maximum for the yellow sodium line with a wavelength of 5.89·10 -7 m, if the period of the diffraction grating is 5 µm.

Decision

Compared to task AT 5 from test No. 3 of the TsT 2002, this task is formulated more precisely, however, in the condition of the task, we should talk not about the "diffraction maximum", but about " main diffraction maximum".

As well as main diffraction maxima, there are always also secondary diffraction peaks. Without explaining this nuance in a school physics course, all the more so it is necessary to strictly observe the established scientific terminology and talk only about the main diffraction maxima.

In addition, it should be pointed out that the light falls normally on the diffraction grating.

With 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 get 8. Therefore, this task, like the previous one, should be considered unsuccessful.

Supplement 2. Solve the above problem, replacing in its condition d \u003d 5 microns per (1 \u003d A microns. Answer:nmax=6.)

Benefit RIKZ 2003 Test No. 6

AT 5. If the second diffraction maximum is at a distance of 5 cm from the center of the screen, then with an increase in the distance from the diffraction grating to the screen by 20%, this diffraction maximum will be at a distance of ... cm.

Decision

The task condition is formulated unsatisfactorily: instead of "diffraction maximum" one should "main diffraction maximum", instead of "from the center of the screen" - "from the zero main diffraction maximum".

As can be seen from the given figure,

From here

Benefit RIKZ 2003 Test No. 7

AT 5. Determine the highest order of the spectrum in a diffraction grating having 500 lines per 1 mm when it is illuminated with light with a wavelength of 720 nm.

Decision

The condition of the task is formulated extremely unsuccessfully in scientific terms (see clarifications of tasks No. 3 and 5 from the 2002 CT).

There are also complaints about the philological style of the task formulation. Instead of the phrase "in a diffraction grating" one should have used the phrase "from a diffraction grating", and instead of "light with a wavelength" - "light whose wavelength". The wavelength is not the load to the wave, but its main characteristic.

Subject to clarifications

By all three of the above rules for rounding numbers, rounding the number 2.78 to an integer value gives 3.

The last fact, even with all the shortcomings in the formulation of the task condition, makes it interesting, since it allows you to distinguish the correct one at the test level (nmax=2) and incorrect (nmax=3) solutions.

Many tasks on the topic under consideration are contained in the 2005 CT.

In the conditions of all these tasks (B1), it is necessary to add the keyword "main" before the phrase "diffraction maximum" (see comments to task B5 of the CT 2002, Test No. 5).

Unfortunately, in all variants of tests B1 of the 2005 CT, the numerical values d(l,N) and λ chosen poorly and always given in fractions

the number of "tenths" is less than 5, which does not allow distinguishing the operation of extracting the integer part of a fraction (correct solution) from the operation of rounding the fraction to an integer value (false trace) at the test level. This circumstance casts doubt on the expediency of using these tasks for an objective test of applicants' knowledge on the topic under consideration.

It seems that the compilers of the tests were carried away, figuratively speaking, by preparing various "garnishes 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).

TT 2005 Option 4

IN 1. On a diffraction grating, the period of whichd1\u003d 1.2 μm, a normally parallel beam of monochromatic light falls with a wavelength λ =500 nm. If it is replaced by a lattice whose periodd2\u003d 2.2 μm, then the number of maxima will increase by ... .

Decision

Instead of "light with a wavelength λ"" need "light wavelength λ "" . Style, style and more style!

As

then, taking into account the fact that X is const, a d 2 >di,

According to the formula (4, b)

Hence, ∆Ntot. 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 recognized as simple and even unsuccessful.

Supplement 3. Solve the above problem by replacing in its condition λ =500 nm on λ =433 nm (blue line in the hydrogen spectrum).

Answer: ΔN total. max=6

TT 2005 Option 6

IN 1. On a diffraction grating with a period d= 2 µm incident normally parallel beam of monochromatic light with wavelength λ =750 nm. The number of maxima that can be observed within an angle a\u003d 60 °, the bisector of which is perpendicular to the plane of the lattice, is ... .

Decision

The phrase "light with a wavelength λ " has already been discussed above in TT 2005 Option 4.

The second sentence in the condition of this task could be simplified and written as follows: "The number of observed main maxima within the angle a = 60 °" and further in the text of the original task.

It's obvious that

According to the formula (4, a)

According to the formula (5, a)

This task, like the previous one, does not allow objectively determine the level of understanding of the topic under discussion by applicants.

Addendum 4. Complete the above task, replacing in its condition λ =750 nm on λ = 589 nm (yellow line in the spectrum of sodium). Answer: N o6sh \u003d 3.

TT 2005 Option 7

IN 1. on a diffraction grating withN 1- 400 strokes per l\u003d 1 mm in length, a parallel beam of monochromatic light falls with a wavelength λ =400 nm. If it is replaced by a lattice havingN 2=800 strokes per l\u003d 1 mm in length, then the number of diffraction maxima will decrease by ... .

Decision

We omit the discussion of inaccuracies in the formulation of the task, since they are the same as in the previous tasks.

From formulas (4, b), (5, b) it follows that

3. From an object 3 cm high, a real image 18 cm high was obtained using a lens. When the object was moved 6 cm, an imaginary image 9 cm high was obtained. Determine the focal length of the lens (in centimeters).

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https://pandia.ru/text/78/506/images/image653.gif" width="109" height="57 src=">.gif" width="122" height="54 src="> ( 3).

We solve the system of equations with respect to d 1 or d 2. Define F= 12 cm.

Answer:F= 12 cm

4. A red light beam with a wavelength of 720 nm is incident on a plate made of a material with a refractive index of 1.8 perpendicular to its surface. What is the minimum thickness of the plate that must be taken so that the light passing through the plate has the maximum intensity?

minimum, 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

Decision:

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, so k max = 3.

Answer: k max = 3.

6. The period of the diffraction grating is 4 μm. The diffraction pattern is observed using a lens with a focal length F\u003d 40 cm. Determine the wavelength of light incident normally on the grating light (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 upwards, 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:	Decision: The angle of incidence is equal to the angle of reflection α = α¢. The figure shows that α + α¢ + φ = 90° or 2α + φ = 90°, then .

Answer:

8. In the middle between two flat mirrors parallel to each other, a point is placed. If the source starts moving in the direction perpendicular to the planes of the mirrors at a speed of 2 m/s, then the first imaginary images of the source in the mirrors will move relative to each other at a speed of:

1) 0 m/s 2) 1 m/s 3) 2 m/s 4) 4 m/s 5) 8 m/s

Decision:

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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 faster is the speed of light in vacuum 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:	Decision:
	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 collecting 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, would remain parallel?

1) 4 cm 2) 5 cm 3) 9 cm cm 5) At any distance, the rays will not be parallel.

Decision:

d = F 2 – F 1 = 5 (cm).

Given:

a= 10 cm

n st = 1.51

Decision:

;

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Answer:b= 0.16 m

2. (7.8.3). At the bottom of the 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 the lamp in the mirror? The refractive index of water is 1.33. Express the result in SI units and round to tenths.

Given: h 1=20cm h 2 = 30 cm n = 1,33	Decision:
	S` – virtual image; (1); (2); (3) a, b are small

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Given:

OC= 4 m

S 1S 2 = 1 mm

L 1 = L 2 = OS

Decision:

D= k l - maximum condition

D= L 2 – L 1;

at 1 – ?

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2(OS)D = 2 ukd, hence ; ; l = OS;

Given:

F= 0.15 m

f= 4.65 m

S= 4.32 cm2

Decision:

; ; S` = G 2 S

S- transparencies platform

; ;

S` – ?

S` \u003d 302 × 4.32 \u003d 3888 (cm2) » 0.39 (m2)

Answer: S` = 0.39 m2

5. (7.8.28). Find the magnification factor of the image of the subject AB given by a thin diverging lens with focal length F. Round the result to the nearest hundredth.

Given:	Decision:
; d 1 = 2F;
G – ?

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Answer: G = 0,17

OPTION #10

structure of the atom and nucleus. elements of the theory of relativity

Part A

1. Determine the delay voltage required to stop the emission of electrons from the photocathode if radiation with a wavelength of 0.4 µm falls on its surface, and the red border of the photoelectric effect is 0.67 µm. Planck's constant 6.63×10-34 J×s, speed of light in vacuum 3×108 m/s. Give 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	Decision: Photon energy: ; energy and mass are related by:

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

Decision: W = ε N 1, , where W is the energy of all photons in the beam, N 1 is the number of photons in the beam, is the energy of one photon; ; N 2 = 0,01N 1; (BUT).

(α) on a diffraction grating, its wavelength (λ), gratings (d), diffraction angle (φ) and spectrum order (k). In this formula, the product of the grating period and the difference between the diffraction and incidence angles 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 and 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 are constant in the resulting formula, the order of the spectrum depends only on the diffraction angle. In the formula, it is expressed through a sine and is 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 a sine can take is one, so just replace sin(φ) with one in the formula: k = d*(1-sin(α))/λ. This is the final formula for calculating the maximum value of the order of the diffraction spectrum.

Substitute the 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 lowest value in the calculations. This value is in the numerator of the formula, so the largest value of the spectrum period will be obtained at the smallest value of the wavelength.

Light waves deviate from their rectilinear path when passing through small openings or passing similarly small obstacles. This phenomenon occurs when the size of obstacles or holes is comparable to the wavelength and is called diffraction. The tasks 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.

Instruction

Find out the period (d) of the diffraction grating - this is the name of the total width of one transparent (a) and one opaque (b) of its bands: d \u003d a + b. This pair is usually called one lattice stroke, and in the number of strokes on . For example, diffraction may contain 500 strokes per 1 mm, and then d = 1/500.

For calculations, the angle (α) under which the light enters the diffraction grating matters. It is measured from the normal to the surface of the lattice, and the sine of this angle is involved in the formula. If in the initial conditions of the problem it is said that the light falls along the normal (α=0), this value can be neglected, since sin(0°)=0.

Find out the wavelength (λ) on the diffraction grating of 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 section of the spectrum - about "monochromatic" light. The visible region corresponds to lengths from about 380 to 740 nanometers. For example, one of the shades of green has a wavelength of 550nm (λ=550).