DEFINITION
Diffraction grating- This is the simplest spectral device, consisting of a system of slits (transparent to light areas), and opaque gaps that are comparable to the wavelength.
A one-dimensional diffraction grating consists of parallel slits of the same width, which lie in the same plane, separated by gaps of the same width that are opaque to light. Reflective diffraction gratings are considered the best. They consist of a combination of areas that reflect light and areas that scatter light. These gratings are polished metal plates, on which light-scattering strokes are applied with a cutter.
The grating diffraction pattern is the result of mutual interference of waves coming from all slits. With the help of a diffraction grating, multipath interference of coherent light beams that have undergone diffraction and which come from all slits is realized.
A characteristic of a diffraction grating is its period. The period of the diffraction grating (d) (its constant) is called the value equal to:
where a is the slot width; b is the width of the opaque area.
Diffraction by a one-dimensional diffraction grating
Let us assume that a light wave with length is incident perpendicular to the plane of the diffraction grating. Since the grating slots are located at equal distances from each other, the path differences () coming from two adjacent slots for the direction will be the same for the entire diffraction grating under consideration:
The main intensity minima are observed in the directions determined by the condition:
In addition to the main minima, as a result of mutual interference of light rays that come from two slits, the rays cancel each other in some directions. As a result, additional intensity minima appear. They appear in those directions where the difference in the path of the rays is odd number half wave The condition for additional minima is the formula:
where N is the number of slits of the diffraction grating; - integer values except 0. In the event that the lattice has N slots, then between the two main maxima there is an additional minimum that separates the secondary maxima.
The main maxima condition for a diffraction grating is:
The value of the sine cannot be greater than one, then the number of main maxima:
Examples of solving problems on the topic "Diffraction grating"
EXAMPLE 1
| Exercise | A monochromatic beam of light with a wavelength is incident on a diffraction grating perpendicular to its surface. The diffraction pattern is projected onto a flat screen using a lens. The distance between two first-order intensity maxima is l. What is the constant of the diffraction grating if the lens is placed in close proximity to the grating and the distance from it to the screen is L. Consider that |
| Solution | As a basis for solving the problem, we use a formula that relates the constant of the diffraction grating, the wavelength of light and the deflection angle of the rays, which corresponds to the diffraction maximum number m: According to the condition of the problem Since the angle of deflection of the rays can be considered small (), we assume that: From Fig. 1 it follows that:
We substitute expression (1.3) into formula (1.1) and take into account that , we get:
From (1.4) we express the lattice period: |
| Answer |
EXAMPLE 2
| Exercise | Using the conditions of example 1, and the result of the solution, find the number of maxima that the lattice in question will give. |
| Solution | In order to determine the maximum angle of deflection of light rays in our problem, we find the number of maxima that our diffraction grating can give. For this we use the formula: where we assume that for . Then we get: |
So far we have considered spherical wave diffraction, studying the diffraction pattern at the observation point, which lies at a finite distance from the obstacle ( Fresnel diffraction ).
Type of diffraction in which a diffraction pattern is formed parallel beams, is called Fraunhofer diffraction . Parallel beams will appear if the source and screen are at infinity. In practice, two lenses are used: in the focus of one - the light source, and in the focus of the other - the screen.
Although Fraunhofer diffraction does not fundamentally differ from Fresnel diffraction, this case is practically important, since it is this type of diffraction that is used in many diffraction devices (diffraction grating, for example). In addition, here the mathematical calculation is simpler and allows us to solve the quantitative problem to the end (we considered Fresnel diffraction qualitatively).
Diffraction of light by a single slit
Let there be a slot in a continuous screen: slot width, slot length (perpendicular to the plane of the sheet) (Fig. 9.5). Parallel beams of light fall on the slit. To simplify the calculation, we assume that in the slot plane AB the amplitudes and phases of the incident waves are the same.
Let us divide the slit into Fresnel zones so that the optical path difference between the rays coming from neighboring zones is equal to .
If an even number of such zones fit within the slot width, then at the point ( side focus lenses) there will be a minimum of intensity, and if an odd number of zones, then a maximum of intensity:
The picture will be symmetrical about main focus points . The plus and minus sign corresponds to the angles measured in one direction or another.
Light intensity. As can be seen from fig. 9.5, the central maximum surpasses all others in intensity.
Let us consider the effect of the slot width.
Because the minimum condition has the form , hence
| . | (9.4.3) |
It can be seen from this formula that with increasing slot width b the positions of the minima shift towards the center, the central maximum becomes sharper.
With a decrease in the gap width b the whole picture expands, blurs, the central strip also expands, capturing an increasing part of the screen, and its intensity decreases.
Diffraction of light on a diffraction grating
A one-dimensional diffraction grating is a system of a large number N slots of the same width and parallel to each other in the screen, also separated by opaque gaps of the same width (Fig. 9.6).
The diffraction pattern on the grating is defined as the result of mutual interference of waves coming from all slits, i.e. in grating carried out multipath interference coherent diffracted beams of light coming from all slits.
Denote: b – slot width gratings; a - distance between slots; – grating constant.
The lens collects all the rays that fall on it at the same angle and does not introduce any additional path difference.
 |
 |
| Rice. 9.6 | Rice. 9.7 |
Let beam 1 fall on the lens at an angle φ ( diffraction angle ). light wave, going at this angle from the slit, creates a maximum intensity at the point. The second beam coming from the neighboring slot at the same angle φ will come to the same point. Both of these beams will come in phase and will amplify each other if the optical path difference is equal to mλ:
Condition maximum for a diffraction grating will look like:
| , | (9.4.4) |
where m= ± 1, ± 2, ± 3, … .
The maxima corresponding to this condition are called major highs . The value of the quantity m corresponding to one or another maximum is called order of the diffraction maximum.
At the point F 0 will always be observed null or central diffraction peak .
Since the light incident on the screen passes only through the slits in the diffraction grating, the condition minimum for gap and will be condition principal diffraction minimum for lattice:
| . | (9.4.5) |
Of course, when large numbers slits, at the points of the screen corresponding to the main diffraction minima, light will fall from some slits and there will form side effects diffraction maxima and minima(Fig. 9.7). But their intensity, in comparison with the main maxima, is low (≈ 1/22).
On condition 
,
the waves sent by each slit will be canceled out by interference and will appear additional minimums .
The number of slots determines the light flux through the grating. The more of them, the more energy is transferred by the wave through it. Besides than more number slots, the more additional minima fit between neighboring maxima. Consequently, the highs will be narrower and more intense (Figure 9.8).
From (9.4.3) it can be seen that the diffraction angle is proportional to the wavelength λ. This means that the diffraction grating decomposes white light into components, and rejects light with a longer wavelength (red) at a larger angle (unlike a prism, where everything happens the other way around).
This property of diffraction gratings is used to determine the spectral composition of light (diffraction spectrographs, spectroscopes, spectrometers).
One of the well-known effects that confirm the wave nature of light are diffraction and interference. Main area their applications are spectroscopy, in which diffraction gratings are used to analyze the spectral composition of electromagnetic radiation. The formula that describes the position of the main maxima given by this lattice is discussed in this article.
What are the phenomena of diffraction and interference?
Before considering the derivation of the formula for a diffraction grating, one should become familiar with the phenomena due to which this grating is useful, that is, with diffraction and interference.
Diffraction is the process of changing the motion of the wave front when it encounters an opaque obstacle on its way, the dimensions of which are comparable to the wavelength. For example, if you pass through a small hole sunlight, then on the wall one can observe not a small luminous point (which should have happened if the light propagated in a straight line), but a luminous spot of some size. This fact testifies to the wave nature of light.
Interference is another phenomenon that is unique to waves. Its essence lies in the imposition of waves on each other. If the waveforms from multiple sources are matched (coherent), then a stable pattern of alternating light and dark areas on the screen can be observed. The minima in such a picture are explained by the arrival of waves in given point in antiphase (pi and -pi), and the maxima are the result of waves hitting the considered point in one phase (pi and pi).
Both phenomena described were first explained by an Englishman when he investigated the diffraction of monochromatic light by two thin slits in 1801.
The Huygens-Fresnel principle and far and near field approximations
The mathematical description of the phenomena of diffraction and interference is a non-trivial task. Finding its exact solution requires complex calculations involving the Maxwellian theory electromagnetic waves. Nevertheless, in the 1920s, the Frenchman Augustin Fresnel showed that, using Huygens' ideas about secondary sources of waves, one can successfully describe these phenomena. This idea led to the formulation of the Huygens-Fresnel principle, which currently underlies the derivation of all formulas for diffraction by obstacles of arbitrary shape.
Nevertheless, even with the help of the Huygens-Fresnel principle, to solve the problem of diffraction in general view does not succeed, therefore, when obtaining formulas, some approximations are resorted to. The main one is a flat wave front. It is this waveform that must fall on the obstacle so that a number of mathematical calculations can be simplified.
The next approximation is the position of the screen where the diffraction pattern is projected relative to the obstacle. This position is described by the Fresnel number. It is calculated like this:
Where a is the geometric dimensions of the obstacle (for example, a slot or a round hole), λ is the wavelength, D is the distance between the screen and the obstacle. If for a particular experiment F<<1 (<0,001), тогда говорят о приближении дальнего поля. Соответствующая ему дифракция носит фамилию Фраунгофера. Если же F>1, then near field approximation or Fresnel diffraction takes place.
The difference between Fraunhofer and Fresnel diffraction lies in the different conditions for the phenomenon of interference at small and large distances from the obstacle.
The derivation of the formula for the main maxima of the diffraction grating, which will be given later in the article, involves the consideration of Fraunhofer diffraction.
Diffraction grating and its types
This grating is a plate of glass or transparent plastic a few centimeters in size, on which opaque strokes of the same thickness are applied. The strokes are located at a constant distance d from each other. This distance is called the lattice period. Two other important characteristics of the device are the lattice constant a and the number of transparent slits N. The value of a determines the number of slits per 1 mm of length, so it is inversely proportional to the period d.
There are two types of diffraction gratings:
- Transparent, as described above. The diffraction pattern from such a grating results from the passage of a wave front through it.
- Reflective. It is made by applying small grooves to a smooth surface. Diffraction and interference from such a plate arise due to the reflection of light from the tops of each groove.
Whatever the type of grating, the idea of its effect on the wave front is to create a periodic perturbation in it. This leads to the formation of a large number of coherent sources, the result of the interference of which is a diffraction pattern on the screen.
The basic formula of a diffraction grating
The derivation of this formula involves considering the dependence of the radiation intensity on the angle of its incidence on the screen. In the far-field approximation, the following formula for the intensity I(θ) is obtained:
I(θ) = I 0 *(sin(β)/β) 2 * 2 , where
α = pi*d/λ*(sin(θ) - sin(θ 0));
β = pi*a/λ*(sin(θ) - sin(θ 0)).
In the formula, the width of the slit of the diffraction grating is denoted by the symbol a. Therefore, the factor in parentheses is responsible for the diffraction by one slit. The value of d is the period of the diffraction grating. The formula shows that the factor in square brackets where this period appears describes the interference from the set of grating slots.
Using the above formula, you can calculate the intensity value for any angle of incidence of light.
If we find the value of the intensity maxima I(θ), then we can conclude that they appear under the condition that α = m*pi, where m is any integer. For the maximum condition, we get:
m*pi = pi*d/λ*(sin(θ m) - sin(θ 0)) =>
sin (θ m) - sin (θ 0) \u003d m * λ / d.
The resulting expression is called the formula for the maxima of the diffraction grating. The m numbers are the order of diffraction.
Other ways to write the basic formula for the lattice
Note that the formula given in the previous paragraph contains the term sin(θ 0). Here, the angle θ 0 reflects the direction of incidence of the front of the light wave relative to the grating plane. When the front falls parallel to this plane, then θ 0 = 0 o . Then we get the expression for the maxima:
Since the grating constant a (not to be confused with the slit width) is inversely proportional to the value of d, the formula above can be rewritten in terms of the diffraction grating constant as:
To avoid errors when substituting specific numbers λ, a and d into these formulas, you should always use the appropriate SI units.
The concept of the angular dispersion of the grating
We will denote this value by the letter D. According to the mathematical definition, it is written as follows:
The physical meaning of the angular dispersion D is that it shows by what angle dθ m the maximum will shift for the diffraction order m if the incident wavelength is changed by dλ.
If we apply this expression to the lattice equation, then we get the formula:
The dispersion of the angular diffraction grating is determined by the formula above. It can be seen that the value of D depends on the order m and the period d.
The greater the dispersion D, the higher the resolution of a given grating.
Grating resolution
Resolution is understood as a physical quantity that shows by what minimum value two wavelengths can differ so that their maxima appear separately in the diffraction pattern.
The resolution is determined by the Rayleigh criterion. It says: two maxima can be separated in a diffraction pattern if the distance between them is greater than the half-width of each of them. The angular half-width of the maximum for the grating is determined by the formula:
Δθ 1/2 = λ/(N*d*cos(θ m)).
The resolution of the grating in accordance with the Rayleigh criterion is:
Δθ m >Δθ 1/2 or D*Δλ>Δθ 1/2 .
Substituting the values of D and Δθ 1/2 , we get:
Δλ*m/(d*cos(θ m))>λ/(N*d*cos(θ m) =>
Δλ > λ/(m*N).
This is the formula for the resolution of a diffraction grating. The greater the number of strokes N on the plate and the higher the order of diffraction, the greater the resolution for a given wavelength λ.
Diffraction grating in spectroscopy
Let us write out once again the basic equation of maxima for the lattice:
It can be seen here that the more the wavelength falls on the plate with strokes, the greater the values of the angles will appear on the screen maxima. In other words, if non-monochromatic light (for example, white) is passed through the plate, then the appearance of color maxima can be seen on the screen. Starting from the central white maximum (diffraction zero order), then maxima will appear for shorter waves (purple, blue), and then for longer ones (orange, red).
Another important conclusion from this formula is the dependence of the angle θ m on the order of diffraction. The larger m, the larger the value of θ m . This means that the colored lines will be more separated from each other at the maxima for a high diffraction order. This fact was already consecrated when the grating resolution was considered (see the previous paragraph).
The described capabilities of the diffraction grating make it possible to use it to analyze the emission spectra of various luminous objects, including distant stars and galaxies.
Problem solution example
Let's show how to use the diffraction grating formula. The wavelength of light that falls on the grating is 550 nm. It is necessary to determine the angle at which first-order diffraction appears if the period d is 4 µm.
Convert all data to SI units and substitute into this equality:
θ 1 \u003d arcsin (550 * 10 -9 / (4 * 10 -6)) \u003d 7.9 o.
If the screen is at a distance of 1 meter from the grating, then from the middle of the central maximum, the line of the first order of diffraction for a wave of 550 nm will appear at a distance of 13.8 cm, which corresponds to an angle of 7.9 o .
DEFINITION
Diffraction grating is the simplest spectral instrument. It contains a system of slits that separate opaque spaces.
Diffraction gratings are divided into one-dimensional and multidimensional. A one-dimensional diffraction grating consists of parallel light-transparent sections of the same width, which are located in the same plane. Transparent areas separate opaque gaps. With these gratings, observations are made in transmitted light.
There are reflective diffraction gratings. Such a grating is, for example, a polished (mirror) metal plate, on which strokes are applied with a cutter. The result is areas that reflect light and areas that scatter light. Observation with such a grating is carried out in reflected light.
The grating diffraction pattern is the result of the mutual interference of waves that come from all the slits. Therefore, with the help of a diffraction grating, multipath interference of coherent light beams that have undergone diffraction and that come from all slits is realized.
Grating period
If we denote the width of the slot on the gratings as a, the width of the opaque section - b, then the sum of these two parameters is the grating period (d):
The period of a diffraction grating is sometimes also called the diffraction grating constant. The period of a diffraction grating can be defined as the distance over which the lines on the grating are repeated.
The diffraction grating constant can be found if the number of grooves (N) that the grating has per 1 mm of its length is known:
The period of the diffraction grating is included in the formulas that describe the diffraction pattern on it. So, if a monochromatic wave is incident on a one-dimensional diffraction grating perpendicular to its plane, then the main intensity minima are observed in the directions determined by the condition:
where is the angle between the normal to the grating and the direction of propagation of the diffracted rays.
In addition to the main minima, as a result of mutual interference of light rays sent by a pair of slits, they cancel each other out in some directions, resulting in additional intensity minima. They arise in directions where the difference in the path of the rays is an odd number of half-waves. The additional minima condition is written as:
where N is the number of slits of the diffraction grating; takes any integer value except 0. If the lattice has N slots, then between the two main maxima there is an additional minimum that separates the secondary maxima.
The condition for the main maxima for the diffraction grating is the expression:
The value of the sine cannot exceed one, therefore, the number of main maxima (m):
Examples of problem solving
EXAMPLE 1
| Exercise | A beam of light passes through a diffraction grating with a wavelength of . A screen is placed at a distance L from the grating, on which a diffraction pattern is formed using a lens. It is obtained that the first diffraction maximum is located at a distance x from the central one (Fig. 1). What is the grating period (d)? |
| Solution | Let's make a drawing. The solution of the problem is based on the condition for the main maxima of the diffraction pattern: By the condition of the problem, we are talking about the first main maximum, then . From Fig. 1 we get that:
From expressions (1.2) and (1.1) we have:
We express the desired period of the lattice, we get:
|
| Answer |
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) 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 quantities by 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 Belarusian study guide L. A. Latotina, V. Ya. Chebotarevsky in mathematics for grade IV, essentially the same two rules for rounding numbers are given. They are formulated as follows: "When rounding decimal fraction up 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 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 stored digit remains unchanged if it is even, and amplifies (increases by 1) if it is odd."
In view of the existence of various rounding rules for numbers, the rounding rules should be decimal numbers explicitly formulate 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) the main maximum and the symmetrical arrangement of the remaining main maximums relative to it total 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 on initial stage when testing in Belarus was optional and carried out by individual educational institutions at your own risk as an alternative to the usual individual written-oral form of entrance examinations.
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.
Solution
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.
Solution
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.
Solution
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])?
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( 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.
Solution
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".
Along with 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 micron. 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.
Solution
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.
Solution
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 ... .
Solution
Instead of "light with a wavelength λ"" need "light wavelength λ "" . Style, style and more style!
Because
then, taking into account the fact that X is const, a d 2 >di,
According to the formula (4, b)
Consequently, ∆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 ... .
Solution
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 ... .
Solution
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
