Light refraction- a phenomenon in which a ray of light, passing from one medium to another, changes direction at the boundary of these media.
Refraction of light occurs according to the following law:
The incident and refracted rays and the perpendicular drawn to the interface between the two media at the point of incidence of the ray lie in the same plane. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant value for two media:
,
Where α
- angle of incidence,
β
- refraction angle,
n - a constant value independent of the angle of incidence.
When the angle of incidence changes, the angle of refraction also changes. The greater the angle of incidence, the greater the angle of refraction.
If light comes from an optically less dense medium to a more dense medium, then the angle of refraction is always less than the angle of incidence: β < α.
A ray of light directed perpendicular to the interface between two media passes from one medium to another without refraction.
absolute refractive index of a substance- a value equal to the ratio of the phase speeds of light (electromagnetic waves) in vacuum and in a given environment n=c/v
The quantity n included in the law of refraction is called the relative refractive index for a pair of media.
The value n is the relative refractive index of medium B with respect to medium A, and n" = 1/n is the relative refractive index of medium A with respect to medium B.
This value, other things being equal, is greater than unity when a beam passes from a denser medium to a less dense medium, and less than unity when a beam passes from a less dense medium to a denser medium (for example, from a gas or from a vacuum to a liquid or solid). There are exceptions to this rule, and therefore it is customary to call a medium optically more or less dense than another.
A ray falling from airless space onto the surface of some medium B is refracted more strongly than when falling on it from another medium A; The refractive index of a ray incident on a medium from airless space is called its absolute refractive index.
(Absolute - relative to vacuum.
Relative - relative to any other substance (the same air, for example).
The relative indicator of two substances is the ratio of their absolute indicators.)
Total internal reflection- internal reflection, provided that the angle of incidence exceeds a certain critical angle. In this case, the incident wave is completely reflected, and the value of the reflection coefficient exceeds its highest values for polished surfaces. Reflection coefficient at full internal reflection does not depend on wavelength.
In optics, this phenomenon is observed for a wide range of electromagnetic radiation, including the X-ray range.
IN geometric optics the phenomenon is explained within the framework of Snell's law. Considering that the angle of refraction cannot exceed 90°, we obtain that at an angle of incidence whose sine is greater than the ratio of the smaller refractive index to the larger index, electromagnetic wave should be fully reflected on the first Wednesday.
In accordance with the wave theory of the phenomenon, the electromagnetic wave still penetrates into the second medium - the so-called “non-uniform wave” propagates there, which decays exponentially and does not carry energy with it. The characteristic depth of penetration of an inhomogeneous wave into the second medium is of the order of the wavelength.
Laws of light refraction.
From all that has been said we conclude:
1 . At the interface between two media of different optical densities, a light ray changes its direction when passing from one medium to another.
2. When a light beam passes into a medium with a higher optical density, the angle of refraction is less than the angle of incidence; When a light ray passes from an optically denser medium to a less dense medium, the angle of refraction is greater than the angle of incidence.
The refraction of light is accompanied by reflection, and with an increase in the angle of incidence, the brightness of the reflected beam increases, and the refracted beam weakens. This can be seen by carrying out the experiment shown in the figure. Consequently, the reflected beam carries with it more light energy, the greater the angle of incidence.
Let MN- the interface between two transparent media, for example, air and water, JSC- incident ray, OB- refracted ray, - angle of incidence, - angle of refraction, - speed of light propagation in the first medium, - speed of light propagation in the second medium.
The first law of refraction sounds like this: the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant value for these two media:
, where is the relative refractive index (the refractive index of the second medium relative to the first).
n 2β=90˚
At light does not pass into the 2nd medium, being completely reflected.
6. Light wavelength
(in any environment)
(in a vacuum)
The frequency of light does not depend on the properties of the medium.
When moving from one medium to another, the wavelength of light changes, because the speed of light changes.
– light wavelength
– frequency of light radiation
The color of light perceived by the eye depends on frequency and does not depend on wavelength.
Examples of problem solving
266. What is the angular height of the Sun above the horizon if a flat mirror was used to illuminate the bottom of the well with solar rays, tilting it at an angle of 25º to the vertical?
vertical
25˚ *Sun
well bottom
Find the anglex.
1) According to the law of light reflection ;
Answer: the angular height of the Sun above the horizon is 40º.
267. The speed of light propagation in a certain liquid is 240·10 3 km/s. A light beam falls from the air onto the surface of this liquid at an angle of 25°. Determine the angle of refraction of the beam.
268. A ray of light falls on the interface between two transparent media at an angle of 35° and is refracted at an angle of 25°. What will happen angle is equal refraction if the beam falls at an angle of 50°?
269. Determine the angle of incidence of a ray in air on the surface of water if the angle between the refracted and reflected rays is 90°.
| Given: 1 Wed. air 2 avg. water φ=90˚ | Solution: n 1 air n 2φ water According to the law of refraction According to the law of reflection From the drawing it can be seen that Let us substitute the expression through : into the formula for the law of refraction: That. Using the table of tangents we find Answer: |
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270. The maximum angle of incidence when a beam passes from turpentine into air is 41°51’. What is the speed of light propagation in turpentine?
| Given: 1st medium – turpentine 2nd medium – air | Solution:  Hence: because m/s, then m/s Answer: km/s |
271. A ray of light passes from methyl alcohol into the air. Will this ray escape into the air if it hits a surface at an angle of 45°?
| Given: 1st medium – methylene alcohol 2nd medium – air | Solution: Let us find the limiting angle of total reflection for two given media using the formula:  Let's compare the angle of incidence of the beam with the limiting angle Answer: light will come out of methyl alcohol into the air, refraction is observed at the interface. |
272. The speed of light propagation in glass (in a light crown) for red light is 199·10 3 km/s, for violet light 196·10 3 km/s. Determine the refractive index of glass for red and violet light.
| Given: km/s km/s | Solution: With– speed of light in vacuum  |
273. Determine the thickness of a plane-parallel plate with a refractive index of 1.7 if a ray of light, passing through this plate, is shifted by 2 cm. The angle of incidence of the beam on the plate is 50°.
| Given: glass air d= 2 cm | 1. Let's use the law of refraction and find the angle.  2. From this we find the length of the side AB=x: see 3. From we find h: see Answer: plate thickness 4.5 cm. |
Tasks for independent decision
274. A light beam is incident on a plane mirror at an angle of 20°. How will the angle between the incident and reflected rays change if the ray hits the mirror at an angle of 35°? (increase by 30°)
275. Flat mirror rotated around an axis passing through the point of incidence of the ray perpendicular to the plane in which the incident and reflected rays lie. At what angle was the mirror turned if the beam reflected from it turned by 42°? How much will the angle between the incident and reflected rays change? (at 21°; at 42°)
276. A flat circular mirror can rotate around its vertical diameter. At a distance of 1.2 m from the mirror, a flat screen hangs on the wall, parallel to the plane of the mirror. A horizontal ray of light hits the center of the mirror at an angle of 12° and is reflected onto the screen. Determine how far the light spot on the screen will move when the mirror is rotated 15°. (at 82 cm)
277. Why does the direction of a light ray change when it passes from one transparent medium to another?
278. There are two transparent liquids in a glass, between which there is a sharp horizontal boundary. Using a beam, how can one determine in which of these liquids the speed of light propagation is slower?
279. How many times is the speed of light propagation in diamond less than in crystalline sugar? (1.55 times)
280. Why is it difficult to hit a fish when shooting at it with a gun from the shore if it is located at a depth of several tens of centimeters from the surface of the water? (a person sees a displaced virtual image)
281. In what cases is light, passing from one transparent medium to another, not refracted? (n 1 = n 2 ; )
282. At the bottom of an empty vessel (see picture) lies a mirror. How will the path of the reflected beam change as the vessel is filled with water? (will move to the right parallel to the original direction)
283. The diver determined the angle of refraction of the beam in water. It turned out to be equal to 32°. At what angle do the light rays fall to the surface of the water? (45°)
284. While in the water, the scuba diver established that the direction to the Sun makes an angle of 28° with the vertical. When he emerged from the water, he saw that the Sun was lower above the horizon. Determine by what angle the direction to the Sun has changed for the scuba diver. (at 11°)
285. Determine by what angle the light beam will deviate from its initial position when passing from air to glass with an index of 1.5, if the angle of incidence is 25°; 65°. (at 9°; at 28°)
286. There is a pebble at the bottom of the stream. The boy wants to hit him with a stick. While aiming, the boy holds the stick in the air at an angle of 45°. At what distance from the pebble will the stick stick into the bottom of the stream if its depth is 32 cm? (12 cm)
287. A ray of light passes from water into glass with a refractive index of 1.7. Determine the angle of incidence of the beam if the angle of refraction is 28°. (37°)
288. A beam of light passes from glycerin into water. Determine the angle of refraction of the beam if the angle of incidence at the interface between two media is 30°. (33.5°)
289. The speed of light propagation in the first transparent medium is 225,000 km/s, and in the second – 200,000 km/s. A ray of light falls on the interface between these media at an angle of 30° and passes into the second medium. Determine the angle of refraction of the beam. (26°)
290. A ray of light falls from air onto the surface of a liquid at an angle of 40° and is refracted at an angle of 24°. At what angle of incidence of the beam will the angle of refraction be equal to 20°? (33°)
291. A beam of light passes from glycerin into the air. What will be the angle of refraction of the beam if it hits the interface between two media at an angle of 22°? (33.5°)
292. Determine the angle of refraction of a ray when passing from air to ethyl alcohol, if the angle between the reflected and refracted rays is 120°. (25°)
293. A rectangular glass plate 4 cm thick has a refractive index of 1.6. The beam falls on its surface at an angle of 55°. Determine the displacement of the beam relative to the original direction after it exits the plate into the air. (by 1.9 cm)
294. Draw the further course of the rays (see figure) falling at the points A And B from a source located at the bottom of a vessel into which water is poured.
295. Knowing the speed of light in a vacuum, find the speed of light in a diamond. (1.28 10 8 m/s)
296. Compare the speed of light in ethyl alcohol and carbon disulfide. (1.2 times more in alcohol)
297. Determine the limiting angle at the transition of rays from diamond to crystalline sugar. (40°)
298. Calculate the maximum angles of incidence for water, crystalline sugar, and diamond. (49°; 40°; 24°)
299. Find the refractive index of ruby if the maximum angle of total reflection for ruby is 34°? (1,8)
300. Ray SN falls on a straight triangular glass prism BAC perpendicular to the face AB. Will the beam be refracted at the edge? A.C. at the point of his fall M or he will experience total reflection, if the angle BAC=30°? (will happen)
301. It has been experimentally established that the refractive index of water for red light is 1.329, for violet light – 1.344. Determine the speed of light propagation in water. How much faster does red light travel in water than violet light? (Refractive indices are given for the maximum wavelength of red and the minimum wavelength of violet.) (2.26 10 8 m/s; 2.24 10 8 m/s; by 0.02 10 8 m/s)
302. Knowing the speed of light in a vacuum, calculate the speed of light in water and glass. (230 mm/s; 190 mm/s)
303. What vibration frequencies correspond to the extreme red (μm) and extreme violet (μm) rays of the visible part of the spectrum? (390 THz; 750 THz)
304. What is the speed of light in water if the wavelength at 440 THz is 0.51 µm? (220 mm/s)
305. The refractive index of water for light with a wavelength of 0.76 microns in a vacuum is 1.329, and for light with a wavelength of 0.4 microns it is 1.344. For which rays the speed of light in water is greater? (for reds)
306.
Water is illuminated with red light, for which the wavelength in air is 0.7 microns. What will be the wavelength in water? What color does a person see when he opens his eyes underwater? (0.53 µm; red) 6. Photon momentum – photon energy
The refractive index of the medium relative to vacuum, i.e. for the case of the transition of light rays from vacuum to medium, is called absolute and is determined by formula (27.10): n=c/v.
When calculating, absolute refractive indices are taken from tables, since their value is determined quite accurately through experiments. Since c is greater than v, then The absolute refractive index is always greater than unity.
If light radiation passes from a vacuum into a medium, then the formula of the second law of refraction is written as:
sin i/sin β = n. (29.6)
Formula (29.6) is often used in practice when rays pass from air to a medium, since the speed of light propagation in air differs very little from c. This can be seen from the fact that the absolute refractive index of air is 1.0029.
When a ray goes from a medium into a vacuum (into air), then the formula of the second law of refraction takes the form:
sin i/sin β = 1 /n. (29.7)
In this case, the rays, when leaving the medium, necessarily move away from the perpendicular to the interface between the medium and vacuum.
Let's find out how to find the relative refractive index n21 from the absolute refractive indices. Let light pass from a medium with absolute exponent n1 to a medium with absolute exponent n2. Then n1 = c/V1 andn2 = c/v2, from:
n2/n1=v1/v2=n21. (29.8)
The formula for the second law of refraction for such a case is often written as follows:
sin i/sin β = n2/n1. (29.9)
Let us remember that by Maxwell's theory absolute exponent refraction can be found from the relation: n = √(με). Since for substances that are transparent to light radiation, μ is practically equal to unity, we can assume that:
n = √ε. (29.10)
Since the frequency of oscillations in light radiation is of the order of 10 14 Hz, neither dipoles nor ions in the dielectric, which have a relatively large mass, do not have time to change their position with such a frequency, and dielectric properties substances under these conditions are determined only by the electronic polarization of its atoms. This is precisely what explains the difference between the value ε=n 2 from (29.10) and ε st in electrostatics. So, for water ε = n 2 = 1.77, and ε st = 81; for the ionic solid dielectric NaCl ε = 2.25, and ε st = 5.6. When a substance consists of homogeneous atoms or non-polar molecules, that is, it contains neither ions nor natural dipoles, then its polarization can only be electronic. For similar substances, ε from (29.10) and ε st coincide. An example of such a substance is diamond, which consists only of carbon atoms.
Note that the value of the absolute refractive index, in addition to the type of substance, also depends on the oscillation frequency, or on the wavelength of the radiation . As the wavelength decreases, as a rule, the refractive index increases.