Angular diameter. How to measure the diameter of stars Angular diameter of the diffraction disk of a telescope formula

A smaller spot size does not allow the phenomenon of diffraction of electromagnetic waves to be obtained.

The diffraction limit was discovered in 1873 by Ernst Abbe.

The minimum diffraction limit is given by the formula d min = λ/(2 n), where λ is the length of the electromagnetic wave in vacuum, n- refractive index of the medium. Sometimes the diffraction limit is understood not as linear, but as angular size, determined by the formula ψ min = 1.22λ/ D(Rayleigh criterion, proposed in 1879), where D- aperture of the optical device.

The value of the diffraction limit in optics and technology

The diffraction limit imposes restrictions on the characteristics of optical devices:

• An optical microscope is not able to distinguish objects whose size is smaller than the value λ/(2 n sinθ), where θ is the so-called aperture angle (for good microscopes θ is close to 90°, and therefore the maximum resolution is close to the diffraction limit λ/(2 n)).
• When manufacturing microcircuits using photolithography, the minimum size of each element of the microcircuit cannot be less than the diffraction limit, which limits the improvement of the technological process.
• The principle of operation of an optical disk is to read information with a focused laser beam, so the diffraction limit imposes a limit on the maximum density of information.
• The resolution of the telescope cannot be greater than ψ min (that is, two point light sources located at an angular distance less than ψ min will be observed as one source). However, the resolution of terrestrial optical telescopes is limited not by the diffraction limit, but by atmospheric distortions (the diffraction limit of the largest telescopes is on the order of 0.01 arcseconds, but due to atmospheric distortions, the actual resolution usually does not exceed 1 second). At the same time, the resolution of radio telescopes and radio interferometers, as well as space telescopes, is limited precisely by the diffraction limit. In addition, new speckle techniques, such as the lucky exposure technique, make it possible to reach the diffraction limit even for large ground-based optical instruments through computerized post-processing of large arrays of observations.

Methods for reducing the diffraction limit

• Diffraction limit d min is proportional to the wavelength, therefore, it can be reduced by using shorter wavelength radiation. For example, the use of a violet laser (λ = 406 nm) instead of a red one (λ = 650 nm) made it possible to increase the capacity of optical disks from 700 MB () to 25 GB (Blu Ray); the transition to short-wave (ultraviolet) lasers allows us to constantly improve technological production standards microcircuits, the use of the X-ray range makes it possible to increase the resolution of microscopes by orders of magnitude (see X-ray microscope).
• The diffraction limit is inversely proportional to the refractive index of the medium. Therefore, it can be significantly reduced by placing the object in a transparent medium with a high refractive index. It is used in optical microscopy (see Immersion) and in photolithography (see Immersion lithography).
• The angular diffraction limit ψ min is inversely proportional to the aperture diameter, so the resolution can be increased by increasing the telescope aperture. However, in practice, the resolution of large telescopes is limited not by the diffraction limit, but by atmospheric distortions, as well as defects in the mirror geometry (or uneven composition of the lens for refractors), so the diffraction limit is important only for radio telescopes and space optical telescopes. In radio astronomy, resolution can be increased by using

Above, we considered light rays as geometric lines, and their intersections as mathematical points. However, this geometric representation is only suitable as a first approximation. The image that actually appears during the refraction and reflection of light is noticeably different from the geometric image that exists only in our imagination.

Examining the image of a star formed by the lens through a strong eyepiece, we notice that it is not a point, as required by the geometric diagram just discussed, but looks like a circle surrounded by several concentric rings, the brightness of which quickly decreases towards the periphery (Fig. 8). But this bright circle is not the true disk of the star, but the visible result of the phenomenon of light diffraction.

Rice. 8. View of images of luminous points of different brightness when they

viewing at the focal point of the lens using a strong eyepiece,

The bright central circle is called the diffraction disk, and the rings surrounding it are called diffraction rings. As the theory shows, the apparent angular diameter of the diffraction disk depends on the wavelength of the light (i.e., on the color of the incident rays) and on the diameter of the lens. This dependence is expressed by the following formula:

where p is the angular radius of the diffraction disk (at

take it from the center of the lens), D is the diameter of the free hole of the lens (in centimeters) and K is the wavelength of light (in centimeters). This expression gives the angular radius of the disk in radians; to convert it to degrees (arcseconds), it must be multiplied by the radian value in seconds. Hence,

p = 1.22^206,265 arcseconds.

At this angle, the radius of the diffraction disk is visible from the center of the lens; at the same angle it is projected from the center of the lens onto the celestial sphere. Its angular diameter will, of course, be twice as large. As we know (p. 20), this is equivalent to as if the true disk of the observed star had such an angular diameter.

The linear radius of the diffraction disk is found by the formula

g = p/, from where g - 1.22 7.V.

Thus, the angular dimensions of the diffraction pattern of the image are determined by the diameter of the lens and the wavelength of light (the color of the rays) and do not depend on /, and the linear dimensions depend on the relative focus and wavelength of light, but do not depend on D. Similarly, on the same quantities The sizes of the diffraction rings surrounding the central disk also depend. From the fact that the size of the rings depends on the wavelength of light, it is clear that in the case of white light they should be colored in rainbow colors; in fact, one can notice that the inner edges of the rings are blue and the outer edges are red (since the wavelength blue rays are shorter than the wavelength of red ones).

From these few pieces of information we can draw conclusions that are of great importance for working with a telescope: 1) the larger the diameter of the lens, the finer the details discernible with its help; 2) for each lens there is the smallest angular distance between two luminous points (for example, stars), which can still be distinguished separately using this lens; this smallest angular distance is called the limiting angle of resolution or the resolving angle and is the fundamental characteristic of the lens by which its resolving power is assessed

force. The smaller the limiting angle of resolution, the higher the resolving power of the lens.

The real value of the resolving power will become quite clear to us if we observe double stars with small angular distances between the components. If the images of stars at the focus of the lens were points, then at an arbitrarily small distance they would be observed as separate; in a strong enough eyepiece we would be able to see two separate points. But in reality, thanks to diffraction, the images of stars are not dots, but circles; and if so, then at a certain minimum distance their images will touch each other, and with a further decrease in the distance between the components of the opp, more and more overlapping each other, they will merge into one slightly oblong speck (Fig. 9). Really existing two

Rice. 9. Images of two stars merge if the angular distances between them are less than the resolving power of the telescope.

individual stars will appear as one, and no eyepiece will be able to see two images. The only way to see two such close stars separately is to use a lens with a large free aperture, since on will depict them as circles of a smaller angular size.

Let us now substitute the wavelength of light into the formula expressing the angular radius of the diffraction disk, taking green-yellow rays (to which the eye is most sensitive) with an average wavelength X = l = 0.00055 mm:

JT (arc seconds)

or, rounding up,

P = "77 (arc seconds),

where D is expressed in millimeters.

Using the same substitution, we obtain the value for the linear radius of the diffraction disk (for the same rays)

g = 1.22-0.00055-V = 0.00007 V mm = 0.07 V µm.

These numbers speak for themselves. No matter how small the luminous point is, its angular radius, when viewed through a lens with a free hole diameter of 140 mm, cannot be less than 1"; it will therefore appear as a circle with a diameter of 2". If we remember that the true angular diameter of stars rarely exceeds thousandths of a second, then it becomes clear how far from the truth the idea of ​​​​an object given by such a lens, although a telescope with a lens with a diameter of 140 logs is already one of the rather powerful instruments. It is appropriate to point out here that the angular radius of the diffraction disk given by

200-inch reflector (D - 5000 lt), equal to yes

Yes, 0.63 is exactly the value of the largest known true angular diameter of the star.

The angular diameter of the diffraction disk does not depend on the focal length, and its linear diameter is determined by the relative aperture of the lens. With the same 140-lens lens at a relative aperture of 1:15, the linear diameter of the diffraction disk will be

2g = 2-0.00067-15 yes 0j02 mm yes 20 microns.

Without going into the details of the theory, which would take us too far, let's say that the actual value of the limiting angle of resolution is somewhat less than the angular radius of the diffraction disk. The study of this issue leads to the conclusion that the measure of permitted

angle, one can practically take the fraction -g- (provided that the brilliance of the components of the double star is equal). Thus, a lens with a free aperture diameter of 120 mm can, to the limit, separate a double star with a distance of components of equal magnitude of 1". On the surface of Mars during the eras of great oppositions

(the angular diameter of the disk is about 25"), with the help of such a lens one can still distinguish two objects lying from each other at a distance of "/25 of the apparent diameter of the planet's disk, which corresponds to approximately 270 km; On the Moon, objects located at a distance of two kilometers from each other can be separately visible.

Let us now consider the relationship between resolving power and magnification. We have already said that no matter how great the magnification, it cannot reveal anything additional beyond the resolving power; No matter how hard we try to enlarge the image - with an eyepiece or by lengthening the focal length - we will not discover new details, but will only increase the apparent size of the diffraction disks. No magnification, no matter how strong, can separate a double star with a component distance of 0",5, if the lens diameter is less than 240 mm. Therefore, numerous attempts (occasionally resurrected even now) to construct “supertelescopes” based on the use of very strong ocular magnifications.The limit of resolving power is determined by the very nature of light (light wavelengths), and it can be moved only by increasing the free opening of the lens, i.e. increasing its diameter.

If strong magnification as a means of increasing resolving power beyond a certain limit is useless, then, as is clear to everyone, it should not be too small, otherwise the details of the image will seem so small that the eye will not be able to distinguish them and the lens will not be used to its fullest. power.

The human eye as an optical system is, of course, also limited to a certain resolving power. Applying telescope theory to it and remembering that for the eye D is 6 mm (i.e. the pupil diameter), we get

the value of the resolution angle is ^r - 20". In reality, however,

the eye has less resolving power due to a number of reasons (optical imperfections of the lens and internal media of the eye, the structure of the retina, etc.). As we have seen, we can assume that the normal human eye is capable of distinguishing an angular distance of 2", i.e. from a distance of 25 cm it will separately see two points spaced 0.15 mm apart.

Thus, the image created by the lens must be magnified using the eyepiece, but at least as many times as the resolving power of the lens is greater than the resolving power of the eye. Only then will the eye see the smallest details accessible to the lens at an angle sufficient to be able to confidently distinguish them. If we accept that the permissible angle for the eye is 120", then what has been said could be written as a simple equality

-

where tr is the required required magnification, and gr is the angle resolved by the lens.

Because

120^D [mm)"

then after substitution we will have

An interesting conclusion emerges: the magnification that allows the eye to discern all the smallest details accessible to the telescope lens is numerically equal to the diameter of the free opening of the lens, expressed in millimeters. This magnification is called resolving magnification. If we remember that the smallest useful magnification is equal to the ratio of the diameters of the lens and the pupil of the eye

^in = and that b = 6 mm, then we get an important relationship between tL1 and t:

t D C"

Therefore, the resolving magnification is equal to six times the smallest useful magnification. In other words, it corresponds to an exit pupil six times smaller than the pupil of the eye, i.e., having a diameter of 1 mm. It can be expressed in terms of the focal length of the eyepiece and the relative focus of the lens (V). Knowing

that j- - D, and J. == N1D. we get 12

whence /2 = V, i.e., the focal length of the eyepiece, expressed in millimeters, giving the resolving magnification, is equal to the relative focus of the lens. From here it is easy to understand that the smaller the relative focus of the lens (i.e., the larger its relative aperture), the more eyepieces are needed, and vice versa.

The given numerical ratios, derived on the basis of geometric optics, turn out to be not entirely accurate when tested by life, that is, by the practice of observing through a telescope. In fact, it turns out that the resolving magnification is 1.4 times greater than that found from our formulas. Therefore, the formula needs to be given this form:

tr - 1.4D = 8.4m.

The focal length of the eyepiece giving the resolving magnification is found from the relation

Consequently, the exit pupil of a telescope equipped with an eyepiece providing resolving magnification will not be equal to 1 mm yj, but ~ = 0.7 mm.

These corrections introduced by practice do not mean at all that the geometric theory on the basis of which the calculations are made is incorrect. The fact is that she simply does not take into account a number of circumstances that are not within her control and, first of all, arising from the characteristics of the eye. The eye is not only an optical instrument, but also an organ of a living body, which has many properties related to the so-called physiology of vision.

Of course, all our calculations are correct only if the observer has normal visual acuity, that is, eyes with a maximum resolution angle reaching our accepted value of 120." Many people think that myopia harms observations through a telescope. This is absolutely not true, since myopia has nothing to do with the resolving power of the eye. The only difference between a myopic eye and a normal eye in this case is that it needs a slightly different focusing, namely: a myopic person will need to move the eyepiece slightly towards the main focus of the lens. In connection with this turns out to be a short-sighted observer

even in a more advantageous position, since it sees the image from a slightly larger angle. True, this advantage when using a strong eyepiece is very insignificant in comparison with what the myopic eye gains when simply viewing close objects.

Now let's consider the effect of light diffraction on the brightness of the image. We know that in reality the image of a luminous point is not a geometric point, but a diffraction disk surrounded by diffraction rings. Light collected by a lens from a luminous point, for example from a star, is therefore distributed over a certain area rather than concentrated at one point. It follows, firstly, that the brightness of the star's image in a telescope is less than what would be expected, since part of its light is distributed along the diffraction rings, and, secondly, that the brightness of the star's image decreases with increasing magnification. Obviously, this decrease in brightness begins with a resolving increase, when the diffraction disks of stars become visible. It is therefore not surprising that very faint stars dim noticeably at the highest magnifications.

Research shows that about 15% of a star's light is distributed among the diffraction rings, and 85% is concentrated in the central diffraction ring. Here, in turn, the light is not distributed evenly, but is concentrated towards the center, which somewhat compensates for the decrease in the brightness of the input image as the magnification of the telescope increases.

In this chapter, we briefly examined the principles underlying the operation of a telescope (refractor or reflector). These principles directly follow from the basic laws of image formation by lenses or mirrors. Starting from the next chapter, we will turn to a real telescope with its advantages and disadvantages arising from the design features and technical implementation. We will take into account the influence of external conditions, features of the observed object, etc. But the initial concepts that we examined in this chapter will continuously serve as the basis for many conclusions, so we will have to return to them several times. The telescope builder and observer should not forget about them in their daily work.

The case of light diffraction with an obstacle having an open small part of the 1st Fresnel zone is of particular interest for practice. The diffraction pattern in this case m = R 2 L λ ≪ 1 or R 2 ≪ L λ is observed at large distances. When R = 1 m m, λ = 550 n m, then the distance L will be more than two meters. Such rays drawn to a distant point are considered parallel. This case is considered as diffraction in parallel rays or Fraunhofer diffraction.

Definition 1

The main condition for Fraunhofer diffraction– this is the presence of Fresnel zones passing through the point of the wave, which are flat relative to each other.

When a collecting lens is located behind an obstacle to the passage of rays at an angle θ, they converge at some point in the plane. This is shown in Figure 3. 9 . 1 . It follows that any point in the focal plane of a lens is equivalent to a point at infinity in the absence of a lens.

Figure 3. 9 . 1 . Diffraction in parallel rays. The green curve is the intensity distribution in the focal plane (the scale is increased along the x-axis).

A Fraunhofer diffraction pattern located at the focal plane of the lens is now available. Based on geometric optics, the focal point must have a lens with a point image of a distant object. The image of such an object is blurred due to diffraction. This is a manifestation of the wave nature of light.

An optical illusion does not produce a point-by-point image. If a Fraunhofer diffraction with a circular hole of diameter D has a diffraction image consisting of an Airy disk, then it accounts for about 85% of the light energy with surrounding light and dark rings. This is shown in Figure 3. 9 . 2. The resulting spot is taken as the image of a point source and is considered as Fraunhofer diffraction by an aperture.

Definition 2

To determine the radius of the central spot of the focal plane of the lens, the formula r = 1.22 λ D F is used.

The lens frame has the property of diffraction of light if rays fall on it, that is, it acts as a screen. Then D is denoted as the diameter of the lens.

Figure 3. 9 . 2. Diffraction image of a point source (circular hole diffraction). About 85% of the light energy falls into the central spot.

Diffraction images are very small in size. The central bright spot in the focal plane with a lens diameter D = 5 cm, focal length F = 50 cm, wavelength in monochromatic light λ = 500 nm has a value of about 0.006 mm. Strong distortion is masked in cameras, projectors by due to imperfect optics. Only high-precision astronomical instruments can realize the diffraction limit of image quality.

Diffraction blurring of two closely spaced points can give the result of observing a single point. When an astronomical telescope is set to observe two nearby stars with an angular distance ψ, defects and aberrations are eliminated, causing the focal plane of the lens to produce diffraction images of the stars. This is considered to be the diffraction limit of the lens.

Figure 3. 9 . 3. Diffraction images of two nearby stars in the focal plane of a telescope lens.

The above figure explains that the distance Δ l between the centers of diffraction images of stars exceeds the value of the radius r of the central bright spot. This case allows you to perceive the image separately, which means it is possible to see two closely located stars at the same time.

If you decrease the angular distance ψ, then overlap will occur, which will not allow you to see two close stars at once. At the end of the 19th century, J. Rayleigh proposed to consider the resolution conditionally complete when the distance between the centers of the images Δ l is equal to the radius r of the Airy Disk. Figure 3. 9 . 4 . shows this process in detail. The equality Δ l = r is considered the Rayleigh solution criterion. It follows that Δ l m i n = ψ m i n ċ F = 1, 22 λ D F or ψ m i n = 1, 22 λ D.

If the telescope has an objective diameter D = 1 m, then it is possible to resolve two stars when located at an angular distance ψ m i n = 6, 7 ċ 10 – 7 rad (for λ = 550 n m). Since the resolution cannot be greater than the value ψ m i n , the limitation is made using the diffraction limit of the space telescope, and due to atmospheric distortions.

Figure 3. 9 . 4 . Rayleigh solution limit. The red curve is the distribution of total light intensity.

Since 1990, the Hubble Space Telescope has been launched into orbit with a mirror having a diameter of D = 2.40 m. The maximum angular resolution of the telescope at a wavelength λ = 550 nm is considered to be ψ m i n = 2.8 ċ 10 – 7 r a d. The operation of a space telescope does not depend on atmospheric disturbances. You should enter the value R, which is the reciprocal of the limit angle ψ m i n.

Definition 3

In other words, the quantity is called telescope power and is written as R = 1 ψ m i n = D 1, 22 λ.

To increase the resolving power of the telescope, increase the size of the lens. These properties apply to the eyes. Its operation is similar to that of a telescope. The pupil diameter d z r acts as D. From here we assume that d з р = 3 mm, λ = 550 n m, then for the maximum angular resolution of the eye we accept the formula ψ g l = 1, 22 λ d з р = 2, 3 ċ 10 − 4 r a d = 47 " " ≈ 1 " .

The result is assessed using the resolution of the eye, which is performed taking into account the size of the light-sensitive elements of the retina. We conclude: a light beam with diameter D and wavelength λ, due to the wave nature of light, experiences diffraction broadening. The angular half-width φ of the beam is of the order of λ D, then the recording of the full width of the beam d at a distance L will take the form d ≈ D + 2 λ D L.

In Figure 3. 9 . 5 . It is clearly visible that when moving away from the obstacle, the light beam transforms.

Figure 3. 9 . 5 . A beam of light that expands due to diffraction. Area I – the concept of a ray of light, the laws of geometric optics. Region II – Fresnel zones, Poisson spot. Region III – diffraction in parallel rays.

The image shows the angular divergence of the beam and its decrease with increasing transverse dimension D. This judgment applies to waves of any physical nature. It follows that in order to send a narrow beam to the Moon, you first need to expand it, that is, use a telescope. When the laser beam is directed into the eyepiece, it travels the entire distance inside the telescope with a diameter of D.

Figure 3. 9 . 6. Resolution of the laser beam using a telescopic system.

Only under such conditions will the beam reach the surface of the Moon, and the radius of the spot will be written as
R ≈ λ D L, where L is denoted as the distance to the Moon. We take the value D = 2.5 m, λ = 550 n m, L = 4 ċ 10 6 m, we obtain R ≈ 90 m. If a beam with a diameter of 1 cm was directed, its “exposure” on the Moon would be in the form of a spot with with a radius 250 times larger.

A microscope is used to observe closely located objects, so the resolution depends on the linear distance between close points. The location of the subject should be near the front focus of the lens. There is a special liquid that is used to fill the space in front of the lens, which is clearly shown in Figure 3. 9 . 7. A geometrically conjugate object located in the same plane with its enlarged image is viewed using an eyepiece. Each point is blurred due to light diffraction.

Figure 3. 9 . 7. Immersion liquid in front of the microscope lens.

Definition 4

Microscope lens resolution limit was defined in 1874 by G. Helmholtz. This formula is written:

l m i n = 0.61 λ n · sin α .

The sign λ is required to indicate the wavelength, n - for the refractive index of the immersion liquid, α - to indicate the aperture angle. The quantity n · sin α is called the numerical aperture.

High-quality microscopes have an aperture angle α, which is close to the limit value α ≈ π 2. According to the Helmholtz formula, the presence of immersion allows one to improve the resolution limit. Suppose that sin α ≈ 1, n ≈ 1, 5, then l m i n ≈ 0, 4 λ.

It follows that a microscope does not provide the full ability to view any details with dimensions much smaller than the light wavelength. The wave properties of light affect the limit of image quality of an object that can be obtained using any optical system.

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Picture 1.

The most important quantity characterizing a lens is the ratio of the diameter of the lens entrance aperture to its focal length, which is called the relative aperture.

The amount of light collected by the lens from the star (point source) will depend only on the entrance hole (~D 2). The situation is different with objects that have noticeable angular dimensions, for example, with planets. In this case, the apparent brightness of the image will decrease, while when observing point objects, ~ D 2 will increase. In fact, as the focal length F increases, the linear dimensions of the image of such a luminary increase proportionally. In this case, the amount of light collected by the lens at a constant D remains the same. The same amount of light is therefore distributed over a larger area of ​​the image, which grows ~ F 2 . Thus, when F increases (or, what is the same: when A decreases) by half, the image area quadruples. The amount of light per unit area, which determines the brightness of the image, decreases in the same ratio. Therefore, the image will become dim as the relative aperture decreases.

Ocular magnification will have exactly the same effect, reducing the brightness of the image in the same ratio as reducing the relative aperture A of the lens.

Therefore, for observing the most extended objects (nebulae, comets), a weak magnification is preferable, but, of course, not lower than the lowest useful one. It can be significantly increased when observing bright planets, and especially the Moon.

Telescope magnification. If we denote the focal length of the lens by F and the focal length of the eyepiece by f, then the magnification M is determined by the formula:

The maximum permissible increase in a calm state of the atmosphere does not exceed 2D, where D is the diameter of the inlet.

Exit pupil diameter. The observed object is clearly visible through the telescope only if the eyepiece is installed at a strictly defined distance from the focus of the lens. This is the position in which the focal plane of the eyepiece is aligned with the focal plane of the lens. Bringing the eyepiece to this position is called focusing or focusing. When the telescope is brought into focus, the rays from each point on the object emerge from the eyepiece parallel (for a normal eye). Light rays from star images formed by the focal plane of the lens are converted by the eyepiece into parallel beams.

f F D d

The area where the light beams of stars intersect is called exit pupil. By pointing the telescope at a bright sky, we can easily see the exit pupil by bringing a screen made of a piece of white paper to the eyepiece. By zooming in and out of this screen, we will find a position in which the light circle is the smallest and at the same time most distinct. It is easy to understand that the exit pupil is nothing more than the image of the entrance hole of the lens formed by the eyepiece. From Figure 2. it is clear that

The last ratio allows you to determine the magnification given by the telescope if neither the focal length of the lens nor the focal length of the eyepiece is known.

The exit pupil concentrates all the light collected by the lens. Therefore, by obscuring part of the exit pupil, we seem to be obscuring part of the lens. This leads to one of the most important rules: The exit pupil should not be larger than the pupil of the observer's eye, otherwise some of the light collected by the lens will be lost.

From the definition of the exit pupil it follows that its size is smaller and the closer it is to the eyepiece, the shorter the focal length of the eyepiece (the “stronger” the eyepiece), and vice versa.

Let us determine the magnification given by an eyepiece that forms an exit pupil equal to the pupil of the eye (the smallest useful or equivalent magnification m):

where d is the diameter of the pupil of the eye or

The size of the field of view. The angle at which the eyepiece aperture is visible to the observer is called angular field of view eyepiece, in contrast to the angular field of view of the telescope, which represents the angular diameter of the circle in the sky visible through the telescope.

The telescope's field of view is equal to the eyepiece's field of view divided by the magnification.

Telescope resolution. Due to the phenomenon of diffraction at the edges of the lens, stars are visible through the telescope in the form of diffraction disks surrounded by several rings of decreasing intensity. Angular diameter of the diffraction disk:

where l is the light wavelength and D is the lens diameter. Two point objects with an apparent angular distance Q are at the limit of separate visibility, which determines the theoretical resolving power of the telescope. Atmospheric jitter reduces the telescope's resolution to:

Resolution determines the ability to distinguish between two adjacent objects in the sky. A telescope with greater resolution allows you to better see two objects that are close together, such as the components of a binary star. You can also see the details of any single object better.

When angular resolution is low, objects appear as a single blurry spot. As resolution increases, the two light sources will become distinguishable as separate objects.

DEFINITION

Diffraction grating- this is the simplest spectral device, consisting of a system of slits (areas transparent to light) 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 equal-width gaps that are opaque to light. Reflective diffraction gratings are considered the best. They consist of a set 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 diffraction pattern on a grating is the result of mutual interference of waves coming from all slits. Using a diffraction grating, multi-beam interference of coherent beams of light that have undergone diffraction and coming from all slits is realized.

A characteristic of a diffraction grating is its period. The period of the diffraction grating (d) (its constant) is a 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 a length of 0 is incident perpendicular to the plane of the diffraction grating. Since the slits of the grating are located at equal distances from each other, the differences in the path of the rays () coming from two adjacent slits for 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 out in some directions. As a result, additional intensity minima arise. They appear in those directions where the difference in the path of the rays is an odd number of half-waves. The condition for additional minima is the formula:

where N is the number of slits of the diffraction grating; — integer values ​​other than 0. If the grating has N slits, then between the two main maxima there is an additional minimum that separates the secondary maxima.

The condition for the main maxima for a diffraction grating is:

The value of the sine cannot be greater than one, then the number of main maxima is:

Examples of solving problems on the topic “Diffraction grating”

EXAMPLE 1

Exercise	A monochromatic beam of light with wavelength θ 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 diffraction grating constant 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 angle of deflection of the rays, which corresponds to the diffraction maximum number m: According to the conditions of the problem, since the angle of deflection of the rays can be considered small (), we assume that: From Fig. 1 it follows that: Let's 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 will find the number of maxima that our diffraction grating can give. To do this we use the formula: where we assume that for . Then we get: