What is light measured in? Intensity, pressure and momentum of an electromagnetic wave

It can vary greatly, and visually we are not able to determine the degree of illumination, since the human eye is endowed with the ability to adapt to different lighting conditions. Meanwhile, the lighting intensity is extremely importance in a wide variety of fields of activity. For example, we can take the process of film or video filming, as well as, for example, growing indoor plants.

The human eye perceives light from 380 nm ( purple) up to 780 nm (red). Best of all, we perceive waves with a wavelength that is just not the most suitable for plants. Lighting that is bright and pleasant to our eyes may not be suitable for plants in a greenhouse, which may not receive the waves important for photosynthesis.

Light intensity is measured in lux. On a bright sunny afternoon in our middle lane it reaches about 100,000 lux, in the evening it drops to 25,000 lux. In a dense shadow, its value is tenths of these values. indoor intensity solar lighting much less, because the light is weakened by trees and window panes. The brightest lighting (on the south window in summer right behind the glass) is at best 3-5 thousand lux, in the middle of the room (2-3 meters from the window) - only 500 lux. This is the minimum light necessary for the survival of plants. For normal growth, even unpretentious ones require at least 800 lux.

We cannot determine the intensity of light by eye. To do this, there is a device, the name of which is a luxmeter. When buying it, you need to clarify the wave range measured by it, because. The capabilities of the device, although wider than the capabilities of the human eye, are still limited.

Light intensity can also be measured with a camera or a photometer. True, you will have to recalculate the received units into suites. To carry out the measurement, you need to put in the place of measurement White list paper and point a camera at it, the sensitivity of which is set to 100, and the aperture to 4. Having determined the shutter speed, its denominator should be multiplied by 10, the resulting value will approximately correspond to the illumination in lux. For example, with a shutter speed of 1/60 sec. lighting around 600 lux.

If you are fond of growing flowers and caring for them, then, of course, you know that light energy is vital for plants for normal photosynthesis. Light affects the growth rate, direction, development of the flower, size and shape of its leaves. With a decrease in light intensity, all processes in plants slow down proportionally. Its amount depends on how far the light source is, on the side of the horizon to which the window is facing, on the degree of shading street trees, from the presence of curtains or blinds. The brighter the room, the more actively the plants grow and the more they need water, heat and fertilizer. If the plants grow in the shade, then they require less maintenance.

When shooting a movie or TV show, lighting is very important. High-quality shooting is possible with illumination of about 1000 lux, achieved in a television studio with the help of special lamps. But acceptable image quality can be obtained with less lighting.

The intensity of lighting in the studio before and during shooting is measured using exposure meters or high-quality color monitors that are connected to the video camera. Before shooting, it is best to walk the exposure meter around the entire set in order to identify dark or overly lit areas of it in order to avoid negative phenomena when viewing the footage. In addition, by properly adjusting the lighting, you can achieve additional expressiveness of the scene being shot and the desired directorial effects.

Consider an elementary area with area , located in a space filled with radiation from different sources. We will characterize the orientation of the site in space by the vector of the normal to its surface.

Important property intensity: this value characterizes the radiative properties of the source and does not depend on how far the elementary area is placed from it. Let's move the platform some distance. Indeed, as distance increases r before the source, the power of radiation passing through the area drops as r2, but the solid angle at which the source is seen also falls according to the same law. An elementary area can be combined with an observer, or it can be represented as being on the surface of the source. The intensity will be the same.

Definition. Radiation intensity is the power of light energy (radiation flux per unit time) passing through an area of ​​a unit section, located perpendicular to the chosen direction in a unit solid angle.

Candela– (CANDLE INTERNATIONAL until 1970) a unit of intensity (luminous intensity) equal to the luminous intensity of such a point source that emits luminous flux of one lumen inside a unit solid angle (steradian), that is, 1cd \u003d 1lm / sr

The intensity of radiant energy has the dimension - w/sr, erg/s*sr

It is also necessary to take into account the orientation of the site in space. In general, if the angle between the normal and the chosen direction is q, then

where = is the element of the solid angle.

The solid angle at which the source is seen is expressed by the equality:

where S is the area cut out by a cone on a sphere of radius r. When the solid angle is 1.

This value is called steradian. All space has a solid angle equal to 4p.

In this way, source intensity is the radiation flux within a solid angle equal to a steradian.

Definition. A source is said to be isotropically radiating if its intensity does not depend on direction in space.

From (2.1) one can obtain the power of radiation passing through a unit area. To do this, we integrate the intensity over the solid angle.

For an isotropic radiation field, we obtain the total flux through the area by the formula = 0. For an isotropically radiating infinite area, integration over the hemisphere gives the flux

Illumination.

Consider the flow from the source at the place of observation. In the absence of absorption, the flux decreases with distance as due to a decrease in the solid angle at which the source is seen. Therefore, the flux can be considered as illumination at the observation point created by the source.

Definition. Illuminance E is the luminous flux per unit area.

Taking into account (2.2), we obtain:

If the platform bounding the cone is located at an angle q to the normal, then in general view we can write the expression for the area illumination in the form:

Lux is taken as a unit of illumination - when a flow equal to 1 lumen passes through a 1m2 area. 1lx \u003d 1lm / m 2

Illumination in energy units - W / cm 2, erg / sec * cm 2

From a point source, the telescope can only register the radiation flux, not the intensity. Let us consider radiation from a star of radius R, which can be represented as a spherically symmetric isotropic source located at a distance r. The directly measured flux from the star will be:

where is the intensity at the point of the receiver (telescope), and = is the solid angle at which the star is visible. The flux per unit surface from a star for isotropic intensity is simply = . In the absence of absorption = . Therefore, for the measured quantity, we find:

= (2.7)

Since , then the transition from a directly measured value to intensity is possible only if the angular diameter R/r of the source is known, that is, if it is not perceived as a point.

1. Addition of light waves from natural sources Sveta.

2. Coherent sources. Light interference.

3. Obtaining two coherent sources from one point source of natural light.

4. Interferometers, interference microscope.

5. Interference in thin films. Illumination of optics.

6. Basic concepts and formulas.

7. Tasks.

Light is electromagnetic in nature, and the propagation of light is the propagation of electromagnetic waves. All optical effects observed during the propagation of light are associated with an oscillatory change in the intensity vector electric field E who is called light vector. For each point in space, the light intensity I is proportional to the square of the amplitude of the light vector of the wave arriving at this point: I ~ E m 2 .

20.1. Addition of light waves from natural light sources

Let us find out what happens when the two light waves with the same frequencies and parallel light vectors:

In this case, the expression for the light intensity is obtained

When obtaining formulas (20.1) and (20.2), we did not consider the question of the physical nature of light sources that create vibrations E 1 and E 2. According to modern concepts, individual molecules are elementary sources of light. The emission of light by a molecule occurs when it passes from one energy level to another. The duration of such radiation is very short (~10 -8 s), and the moment of radiation is a random event. In this case, a time-limited electromagnetic pulse with a length of about 3 m is formed. Such an pulse is called train.

The natural sources of light are bodies heated to high temperatures. The light of such a source is a collection of a huge number of trains emitted by different molecules at different times. Therefore, the average value of cosΔφ in formulas (20.1) and (20.2) is obtained zero, and these formulas take the following form:

The intensities of natural light sources at each point in space add up.

The wave nature of light is not manifested in this case.

20.2. coherent sources. Light interference

The result of the addition of light waves will be different if the phase difference for all trains arriving at a given point has constant value. This requires the use of coherent light sources.

coherent are called light sources of the same frequency, ensuring the constancy of the phase difference for waves arriving at a given point in space.

Light waves emitted by coherent sources are also called coherent waves.

Rice. 20.1. Addition of coherent waves

Consider the addition of two coherent waves emitted by sources S 1 and S 2 (Fig. 20.1). Let the point for which the summation of these waves is considered be removed from the sources by distances s 1 And s2 respectively, and the media in which the waves propagate have different refractive indices n 1 and n 2 .

The product of the length of the path traveled by the wave and the refractive index of the medium (s * n) is called optical path length. The absolute value of the difference optical lengths called optical path difference:

We see that when adding coherent waves, the magnitude of the phase difference at a given point in space remains constant and is determined by the optical path difference and the wavelength. At the points where the condition

cosΔφ = 1, and formula (20.2) for the intensity of the resulting wave takes the form

In this case, the intensity takes on the maximum possible value.

For points where the condition

Thus, during the addition of coherent waves, a spatial redistribution of energy occurs - at some points, the energy of the wave increases, while at others it decreases. This phenomenon is called interference.

Light interference - the addition of coherent light waves, as a result of which a spatial redistribution of energy occurs, leading to the formation of a stable pattern of their amplification or weakening.

Equalities (20.6) and (20.7) are the conditions for maximum and minimum interference. It is more convenient to write them in terms of the path difference.

Maximum intensity interference is observed when the optical path difference is equal to an integer number of wavelengths (even number of half waves).

The integer k is called the order of the interference maximum.

Similarly, the minimum condition is obtained:

Minimum intensity during interference is observed when the optical path difference is equal to odd the number of half waves.

Wave interference is especially pronounced when the wave intensities are close. In this case, in the region of the maximum, the intensity is four times greater than the intensity of each wave, and in the region of the minimum, the intensity is almost zero. An interference pattern is obtained from bright light bands separated by dark gaps.

20.3. Obtaining two coherent sources from one point source of natural light

Before the invention of the laser, coherent light sources were created by splitting a light wave into two beams that interfered with each other. Let's consider two such methods.

Young's method(Fig. 20.2). An opaque barrier with two small holes is placed on the path of a wave coming from a point source S. These holes are the coherent sources S 1 and S 2 . Since the secondary waves emanating from S 1 and S 2 belong to the same wave front, they are coherent. Interference is observed in the area of ​​overlap of these light beams.

Rice. 20.2. Obtaining coherent waves by the Young method

Usually, holes in an opaque barrier are made in the form of two narrow parallel slots. Then the interference pattern on the screen is a system of light stripes separated by dark gaps (Fig. 20.3). Light bar corresponding to

Rice. 20.3. Interference pattern corresponding to Young's method, k is the order of the spectrum

zero-order maximum is located in the center of the screen in such a way that the distances to the slots are the same. To the right and to the left of it are the first-order maxima, and so on. When slits are illuminated with monochromatic light light stripes have the corresponding color. When using white light, maximum zero order It has White color, and the rest of the maxima have iridescent coloring, since the maxima of the same order for different lengths waves are formed in different places.

Lloyd's Mirror(Fig. 20.4). A point source S is located at a small distance from the surface of a flat mirror M. Direct and reflected beams interfere. Coherent sources are the primary source S and its imaginary image in the mirror S 1 . Interference is observed in the overlap region of the direct and reflected beams.

Rice. 20.4. Obtaining coherent waves using Lloyd's mirror

20.4. Interferometers, interference

microscope

The action is based on the use of light interference interferometers. Interferometers are designed to measure the refractive indices of transparent media; to control the shape, microrelief and deformation of the surfaces of optical parts; to detect impurities in gases (used in sanitary practice to control the purity of air in rooms and mines). Figure 20.5 shows a simplified diagram of the Jamin interferometer, which is designed to measure the refractive indices of gases and liquids, as well as to determine the concentration of impurities in the air.

Rays of white light pass through two holes (Young's method), and then through two identical cuvettes K 1 and K 2 filled with substances with different refractive indices, one of which is known. If the refractive indices were the same, then White the zero-order interference maximum would be located in the center of the screen. The difference in refractive indices leads to the appearance of an optical path difference when passing through the cuvettes. As a result, the maximum of the zeroth order (it is called achromatic) is shifted relative to the center of the screen. The second (unknown) refractive index is determined from the displacement value. We give without derivation a formula for determining the difference between the refractive indices:

where k is the number of bands by which the achromatic maximum has shifted; l- cuvette length.

Rice. 20.5. Ray path in the interferometer:

S - source, narrow slit illuminated by monochromatic light; L is the lens in which the source is in focus; K - identical cuvettes of length l; D - diaphragm with two slits; E-screen

Using the Jamin interferometer, it is possible to determine the difference in refractive indices to the sixth decimal place. This high accuracy makes it possible to detect even small air pollution.

interference microscope is a combination of an optical microscope and an interferometer (Fig. 20.6).

Rice. 20.6. The path of rays in an interference microscope:

M - transparent object; D - diaphragm; O - microscope eyepiece for

observations of interfering rays; d - object thickness

Due to the difference in the refractive indices of the object M and the medium, the rays acquire a path difference. As a result, a light contrast is formed between the object and the environment (in monochromatic light) or the object becomes colored (in white light).

This device is used to measure the concentration of dry matter, the size of transparent uncolored micro-objects that are non-contrast in transmitted light.

The path difference is determined by the thickness d of the object. The optical path difference can be measured with an accuracy of hundredths of a wavelength, which makes it possible to quantitatively study the structure of a living cell.

20.5. Interference in thin films. Enlightenment of optics

It is well known that gasoline stains on the surface of water or the surface of a soap bubble have an iridescent color. The transparent wings of dragonflies also have iridescent colors. The appearance of color is due to the interference of light rays reflected

Rice. 20.7. Reflection of rays in a thin film

from the front and back sides of the thin film. Let's consider this phenomenon in more detail (Fig. 20.7).

Let beam 1 of monochromatic light fall from air onto the front surface of the soap film at some angle α. At the point of incidence, the phenomena of reflection and refraction of light are observed. The reflected beam 2 returns to the air. The refracted beam is reflected from the rear surface of the film and, refracted on the front surface, exits into the air (beam 3) parallel to beam 2.

Coming through optical system eyes, rays 2 and 3 intersect on the retina, where their interference occurs. Calculations show that for a soap film in air environment, the path difference between beams 2 and 3 is calculated by the formula

The difference is due to the fact that when light is reflected from the optical denser environment, its phase changes by π, which is equivalent to a change in the optical path length of beam 2 by λ/2. When reflected from a less dense medium, no phase change occurs. In a film of gasoline on the surface of water, reflection from a denser medium occurs twice. Therefore, the addition λ/2 appears in both interfering beams. When a path difference is found, it is destroyed.

Maximum interference pattern is obtained for those angles of view (α) that satisfy the condition

If we looked at a film illuminated with monochromatic light, we would see several bands of the corresponding color, separated by dark gaps. When the film is illuminated with white light, we see interference maxima of various colors. The film thus acquires an iridescent color.

The phenomenon of interference in thin films is used in optical devices that reduce the proportion of light energy reflected by optical systems and increase (due to the law of conservation of energy), therefore, the energy supplied to the recording systems - a photographic plate, an eye.

Illumination of optics. The phenomenon of light interference is widely used in modern technology. One such application is the "enlightenment" of optics. Modern optical systems use multi-lens objectives with a large number of reflective surfaces. Light loss due to reflection can reach 25% in a camera lens and 50% in a microscope. In addition, multiple reflections degrade the quality of the image, for example, a background appears that reduces its contrast.

To reduce the intensity of reflected light, the lens is covered with a transparent film, the thickness of which is equal to 1/4 of the wavelength of light in it:

where λ P - the length of the light wave in the film; λ is the wavelength of light in vacuum; n is the refractive index of the film substance.

Usually they are guided by the wavelength corresponding to the middle of the spectrum of the light used. The film material is selected so that its refractive index is less than that of the lens glass. In this case, formula (20.11) is used to calculate the path difference.

The main part of the light falls on the lens at small angles. Therefore, we can put sin 2 α ≈ 0. Then formula (20.11) takes the following form:

Thus, the rays reflected from the front and rear surfaces films are out of phase and almost completely cancel each other out during interference. This takes place in the middle part of the spectrum. For other wavelengths, the intensity of the reflected beam also decreases, although to a lesser extent.

20.6. Basic concepts and formulas

End of table

20.7. Tasks

1. What is the spatial extent L of the train of waves formed during the time t of the emission of an atom?

Solution

L \u003d c * t \u003d 3x10 8 m / cx10 -8 s \u003d 3 m. Answer: 3m.

2. The difference between the wave paths from two coherent light sources is 0.2 λ. Find: a) what is the phase difference in this case, b) what is the result of interference.

3. The difference in the path of waves from two coherent light sources at some point of the screen is δ = 4.36 μm. What is the result of interference if the wavelength λ is: a) 670; b) 438; c) 536 nm?

Answer: a) the minimum b) maximum; c) an intermediate point between the maximum and minimum.

4. A soap film (n = 1.36) is exposed to white light at an angle of 45°. At what minimum film thickness h will it acquire a yellowish tint? (λ = 600 nm) when viewed in reflected light?

5. A soap film with a thickness h = 0.3 μm is illuminated by white light incident perpendicular to its surface (α = 0). The film is viewed in reflected light. Refractive index soap solution is equal to n = 1.33. What color will the film be?

6. The interferometer is illuminated with monochromatic light from λ = 589 nm. Cuvette length l= 10 cm. When the air in one cell was replaced by ammonia, the achromatic maximum shifted by k = 17 bands. The refractive index of air n 1 = 1.000277. Determine the refractive index of ammonia n 1.

n 2 = n 1 + kλ/ l = 1,000277 + 17*589*10 -7 /10 = 1,000377.

Answer: n 1 = 1.000377.

7. Thin films are used to brighten optics. How thick should the film be in order to transmit light of wavelength λ = 550 nm without reflection? The refractive index of the film is n = 1.22.

Answer: h = λ/4n = 113 nm.

8. What is it in appearance distinguish enlightened optics? Answer: Since it is impossible to extinguish the light of all lengths at the same time

waves, then they achieve the extinction of light corresponding to the middle of the spectrum. The optics become purple.

9. What role does a coating with an optical thickness of λ/4 deposited on glass play if the refractive index of the coating substance more refractive index of glass?

Solution

In this case, the half-wave is lost only at the film-air interface. Therefore, the path difference is λ instead of λ/2. In this case, the reflected waves intensify rather than extinguish each other.

Answer: the coating is reflective.

10. Rays of light incident on a thin transparent plate at an angle α = 45° color it when reflected in green color. How will the color of the plate change when the angle of incidence of the rays changes?

At α = 45°, the interference conditions correspond to the maximum for green rays. As the angle increases, the left side decreases. Therefore, the right side must also decrease, which corresponds to an increase in λ.

As the angle decreases, λ will decrease.

Answer: as the angle increases, the color of the plate will gradually change towards red. As the angle decreases, the color of the plate will gradually change towards purple.

Thus, in geometric optics, a light wave can be considered as a beam of rays. The rays, however, by themselves determine only the direction of propagation of light at each point; the question remains about the distribution of light intensity in space.

Let us single out an infinitesimal element on one of the wave surfaces of the considered beam. It is known from differential geometry that every surface has at each of its points two, generally speaking, different principal radii of curvature.

Let (Fig. 7) be the elements of the principal circles of curvature drawn on a given element of the wave surface. Then the rays passing through points a and c will intersect each other at the corresponding center of curvature and the rays passing through b and d will intersect at another center of curvature.

At given angles of opening, the rays emanating from the length of the segments are proportional to the corresponding radii of curvature (i.e., the lengths and); the area of ​​the surface element is proportional to the product of the lengths, i.e., proportional. In other words, if we consider an element of the wave surface, limited by a certain number of rays, then when moving along them, the area of ​​this element will change proportionally.

On the other hand, the intensity, i.e., the density of the energy flux, is inversely proportional to the surface area through which a given amount of light energy passes. Thus, we conclude that the intensity

This formula must be understood in the following way. On each given ray (AB in Fig. 7) there are certain points and , which are the centers of curvature of all wave surfaces that intersect this ray. The distances u from the point O of the intersection of the wave surface with the beam to the points are the radii of curvature of the wave surface at the point O. Thus, formula (54.1) determines the light intensity at the point O on a given beam as a function of the distances to certain points on this beam. We emphasize that this formula is unsuitable for comparing intensities in different points the same wave surface.

Since the intensity is determined by the square of the field modulus, to change the field itself along the beam, we can write:

where in the phase factor, R can be understood as both and the quantities differ from each other only by a constant (for a given ray) factor, since the difference , the distance between both centers of curvature, is constant.

If both radii of curvature of the wave surface coincide, then (54.1) and (54.2) have the form

This is, in particular, always the case when light is emitted from a point source (the wave surfaces are then concentric spheres, and R is the distance to the light source).

From (54.1) we see that the intensity becomes infinite at points, i.e., at the centers of curvature of the wave surfaces. Applying this to all rays in the beam, we find that the intensity of light in a given beam goes to infinity, generally speaking, on two surfaces - the locus of all centers of curvature of the wave surfaces. These surfaces are called caustics. In the special case of a beam of rays with spherical wave surfaces, both caustics merge into one point (focus).

Note that, according to the properties of the locus of centers of curvature of a family of surfaces known from differential geometry, the rays touch the caustics.

It must be borne in mind that (for convex wave surfaces) the centers of curvature of the wave surfaces may turn out to lie not on the rays themselves, but on their extensions beyond the optical system from which they emanate. In such cases one speaks of imaginary caustics (or imaginary foci). In this case, the intensity of light does not go to infinity anywhere.

As regards the intensification of the intensity to infinity, in reality, of course, the intensity at the points of the caustic becomes large, but remains finite (see the problem in § 59). The formal conversion to infinity means that the approximation of geometrical optics becomes, in any case, inapplicable near caustics. Related to the same circumstance is the fact that the change in phase along the ray can be determined by formula (54.2) only in sections of the ray that do not include points of contact with the caustics. Below (in § 59) it will be shown that, in fact, when passing by the caustic, the phase of the field decreases by . This means that if in the section of the beam before it touches the first caustic, the field is proportional to the factor - the coordinate along the beam), then after passing by the caustic, the field will be proportional. The same will happen near the point of contact of the second caustic, and beyond this point the field will be proportional

Let us now calculate the total energy emitted by the charge during acceleration. For generality, let us take the case of arbitrary acceleration, however, assuming that the motion is nonrelativistic. When the acceleration is directed, say, vertically, electric field radiation is equal to the product of the charge and the projection of the retarded acceleration, divided by the distance. Thus, we know the electric field at any point, and from here we know the energy passing through the unit area for.

The value is often found in formulas for the propagation of radio waves. Its reciprocal can be called vacuum impedance (or vacuum resistance); it is equal to . Hence the power (in watts per square meter) is the mean square of the field divided by 377.

Using formula (29.1) for the electric field, we obtain

, (32.2)

where is the power at , radiated at an angle . As already noted, inversely proportional to the distance. By integrating, we obtain from here the total power radiated in all directions. To do this, first multiply by the area of ​​the strip of the sphere, then we get the energy flow in the interval of the angle (Fig. 32.1). The area of ​​the strip is calculated as follows: if the radius is , then the thickness of the strip is , and the length is , since the radius of the annular strip is . So the area of ​​the strip is

(32.3)

Figure 32.1. The area of ​​a ring on a sphere is .

Multiplying the flux [power by , according to formula (32.2)] by the area of ​​the strip, we find the energy radiated in the interval of angles and ; then you need to integrate over all angles from to :

(32.4)

When calculating, we use the equality and as a result we get . Hence finally

Several remarks need to be made about this expression. First of all, since there is a vector, then in the formula (32.5) means , i.e., the square of the length of the vector. Secondly, the formula (32.2) for the flow includes an acceleration taken with the delay taken into account, i.e., the acceleration at the time when the energy passing through the surface of the sphere was radiated. The thought may arise that the energy was actually radiated at exactly the indicated moment in time. But this is not entirely correct. The moment of emission cannot be determined exactly. It is possible to calculate the result of only such a movement, for example, oscillations, etc., where the acceleration finally disappears. Consequently, we can only find the total energy flux over the entire period of oscillation, which is proportional to the average square of the acceleration over the period. Therefore, in (32.5) should mean the time average of the squared acceleration. For such a movement, when the acceleration at the beginning and at the end vanishes, the total radiated energy is equal to the time integral of expression (32.5).

Let's see what formula (32.5) gives for an oscillating system, for which the acceleration has the form . The average for the period from the square of the acceleration is (when squaring, you must remember that in fact, instead of the exponent, its real part, the cosine, should be included, and the average from gives):

Consequently,

These formulas were obtained relatively recently - at the beginning of the 20th century. These are wonderful formulas, they were of great historical significance, and it would be worth reading about them in old books on physics. True, a different system of units was used there, and not the SI system. However, in the final results relating to electrons, these complications can be eliminated by using next rule correspondences: the value where is the charge of an electron (in coulombs), used to be written as . It is easy to verify that in the SI system the value is numerically equal to , since we know that And . In what follows, we will often use the convenient notation (32.7)

If this numerical value is substituted into the old formulas, then all other quantities in them can be considered defined in the SI system. For example, formula (32.5) previously had the form . And the potential energy of a proton and an electron at a distance is or , where SI.