Comprehensive review of college entrance examination-reflection and refraction of light ● Knowledge network ● College entrance examination sites

Examination requirements:

The knowledge points require to explain the straight-line propagation of light, the reflection of umbra and penumbra ⅰ, the law of reflection, the method of plane mirror imaging ⅱ, the refraction of light, the law of refraction, the refractive index, the total reflection and the critical angle ⅱ, the prism ⅰ of optical fiber, and the dispersion ⅰ of light.

"Reflection and refraction of light" is based on the straight-line propagation of light. This paper mainly discusses the basic laws of light in reflection and refraction phenomena-reflection law and refraction law, corresponding plane mirror, drawing method and specific application, in which the concept of refractive index, total reflection phenomenon and optical path drawing are the focus of this chapter. In addition, the concept of light and the optical path diagram made by light are important methods and tools to analyze and solve geometric optics problems, which should also be highly valued in review.

In recent years, the analysis and calculation of refractive index and total reflection phenomenon is the most frequent in the proposition of "reflection and refraction of light" in college entrance examination. In addition, plane mirror imaging drawing also appears in college entrance examination questions from time to time.

● Factor analysis

☆ Linear propagation of light:

1. light source: an object that can emit light is called a light source. The light source emits light and converts other forms of energy into light energy.

2. The straight-line propagation of light:

(1) ray: The geometric line representing the propagation direction and path of light is called ray. Mark an arrow on the lamp to indicate the direction of light propagation.

(2) Medium: The substance in which light energy propagates is called medium, also called medium. The propagation of light can be carried out in a vacuum, relying on the special substance of electromagnetic field.

(3) Linear propagation of light:

Light travels in a straight line in the same uniform medium.

Evidence: Shadow, solar eclipse and lunar eclipse, pinhole imaging.

The process of light propagation is also the process of energy transmission.

3. The formation of the shadow:

(1) Definition: When the light emitted by a point light source shines on an opaque object, the surface of the object facing the light is illuminated, and a dark area is formed behind the backlight surface, which is the shadow of the object. The shaded area is surrounded by light from the light source and tangent to the surface of the illuminated object.

(2) Classification: umbra and penumbra.

Umbra: The area on the light source where the light emitted by all the luminous points can't reach. For the same object, the size of its umbra area is related to the size of the light emitting surface of the light source and the distance between the light source and the object: when the distance between the light source and the object is constant, the larger the light emitting surface of the light source, the smaller the umbra of the object; The smaller the luminous surface of the light source, the larger the umbra of the object. When the luminous surface of the light source is fixed, the smaller the distance between the light source and the object, the larger the umbra area of the object; The greater the distance from the light source to the object, the smaller the umbra area of the object.

Penumbra: The light emitted by some luminous points on the light source shines, while the areas where the light emitted by other luminous points does not shine become penumbra.

Both umbra and penumbra are the result of straight-line propagation of light.

4. Eclipse and eclipse: When the eclipse occurs, the sun, the moon and the earth are in a straight line, and the earth is in the middle. As shown in the picture on the left below, when the moon completely enters the umbra (A) of the earth, a total lunar eclipse is formed. When the moon enters the boundary between the umbra and the penumbra of the earth, a partial lunar eclipse is formed.

When the solar eclipse occurs, the sun, the moon and the earth are also on the same straight line, with the moon in the middle. When people on the earth in the umbra (a) of the moon can't see the whole luminous surface of the sun, this is a total solar eclipse; People in the penumbra of the moon (B) can't see the luminous surface on the side of the sun, which is a partial solar eclipse; People who are located in the extended space (C) of the umbra of the moon can't see the middle of the luminous surface of the sun, but can only see the surrounding annular surface, which is an annular eclipse. As shown in the figure on the right below. ☆ Light speed:

1. Light speed: the speed at which light travels.

(1) the speed of light in vacuum: the speed of light with different frequencies in vacuum is the same, which is c = 3.0×108m/s. 。

(2) The speed of light in air is approximately equal to c = 3.0×108m/s..

(3) The speed of light is less than c in other media, and its magnitude is related not only to the nature of the media, but also to the frequency of light (this is different from mechanical waves, whose speed is only determined by the nature of the media, such as density, elasticity, temperature, etc.).

2. Light years:

(1) Definition: The distance traveled by light in a vacuum in one year is called light years.

Note: Light years are not units of time, but units of length.

(2) Size: 1 light year = c T = 9.46× 10 15m.

light reflex

1. Light reflection phenomenon:

(1) When light hits the interface between one medium and another, part of the light returns to this medium, which is called light reflection.

(2) The reflection law of light: reflected light, incident light and interface normal are in the same plane, and reflected light and incident light are located on both sides of the normal respectively; The reflection angle is equal to the incident angle.

In the reflection phenomenon, the optical path is reversible.

2. Plane mirror imaging;

Optical characteristics of (1) plane mirror: only the propagation direction of the beam is changed, but the properties of the beam are not changed. As shown in the following figure: the incident beam is a parallel beam, and the reflected beam is still a parallel beam;

The incident beam is a convergent beam, and the reflected beam is still a convergent beam;

The incident beam is a divergent beam, and the reflected beam is still a divergent beam.

(2) Imaging characteristics of the plane mirror: the image behind the plane mirror is an equidistant virtual image, and the object image is symmetrical about the mirror surface.

That is: the orientation relationship between image and object: up and down can not be reversed, left and right should be interchanged.

3. Common problems of flat mirror imaging

Analysis method of (1) object or plane mirror moving problem

When the object or plane mirror is moving in translation: if the mirror is not moving, the object speed is V, and the vertical mirror is moving, the image speed is V, and the vertical mirror is moving in the opposite direction to the object; If the mirror is moving and the object is not moving, when the speed of the mirror is V, the speed of the image to the object is 2v, and the direction is the same as that of the mirror.

When the plane mirror rotates by a slight angle θ around a certain point on the mirror surface, the normal also rotates by the angle θ, and the reflected light deflects by 2θ.

(2) Plane mirror imaging drawing method

Reflection law method: make any two incident rays from an object point and make its reflected rays according to the reflection law. The intersection of the opposite extension lines of two reflected rays is the virtual image point.

Symmetry method: the vertical line from the clipping point to the mirror, and the other side of the mirror intersects with the point whose distance from the object point to the mirror is the same as the virtual image point.

Note: the drawing should be standardized, the light should draw an arrow indicating the direction, and the solid line and dotted line should be clear; Symmetry method can only be used to determine the position of the image, and light must be added when making the light path.

(3) the field of view of the plane mirror:

Looking at a virtual image through a flat mirror is like looking at an object outside the window through a "window" as big as a flat mirror. The specific observation range is defined by the line connecting the image point and the edge of the flat mirror.

refraction of light

1. Definition: When light enters another medium from one medium or propagates in the same inhomogeneous medium, the phenomenon of direction deflection is called light refraction.

2. Illustration: As shown in the following figure, AO is incident light, O is incident point, OB is reflected light, and OC is refracted light. (1) incident angle: The angle I between incident light and normal is called incident angle.

(2) Refraction angle: The included angle r between refracted light and normal is called refraction angle.

3. The law of refraction:

Content: Refracted light lies in the plane determined by incident light and normal. The refracted ray and the incident ray are located on both sides of the normal respectively, and the sine of the incident angle is directly proportional to the sine of the refraction angle. This is the law of refraction of light, also called the law of sneer (sneer is a Dutch mathematician).

4. Refractive index:

When (1) light is injected into a medium from vacuum, the ratio of sine of incident angle to sine of refraction angle is called the refractive index of this medium. The refractive index is expressed by. Namely:

(2) The refractive index of the medium is equal to the ratio of the propagation speed c of light in vacuum to the propagation speed v of light in the medium, namely:

Because of the argument

5. Optical hydrophobic media and optical dense media:

(1) concept:

Optical hydrophobic medium: the medium with smaller refractive index of the two media is called optical hydrophobic medium.

Optical dense medium: the medium with higher refractive index of the two media is called optical dense medium.

(2) Understanding:

Optical hydrophobic medium and optical dense medium are relative. Only for the given two kinds of media can we talk about optically hydrophobic media and optically dense media. There is no absolute optical density medium.

The definitions of optically hydrophobic media and optically dense media are based on refractive index and have nothing to do with other properties of media (such as density).

When the light enters the dense medium from the hydrophobic medium, the refraction angle is smaller than the incident angle; When the light enters the light-hydrophobic medium from the light-dense medium, the refraction angle is greater than the incident angle.

☆ Total reflection and critical angle

1. Total reflection: When light enters light-sparse medium from light-dense medium, the refraction angle is greater than the incident angle. When the incident angle increases to a certain angle, the refraction angle is equal to 90, and the refracted light disappears completely. All the incident light is reflected into the original medium, which is called total reflection.

2. Critical angle:

(1) Definition: When light is emitted from an optically dense medium to an optically hydrophobic medium, the incident angle when the refraction angle is equal to 90 is called the critical angle. Represented by the letter C, the critical angle refers to the minimum incident angle when total reflection occurs when light is emitted from a dense medium to a hydrophobic medium, and it is the critical state of total reflection. When light is emitted from an optically dense medium into an optically hydrophobic medium:

If the incident angle I < c, total reflection will not occur, and there will be both reflection and refraction.

If the incident angle i≥C, a total reflection image appears.

(2) Calculation of critical angle:

3. Conditions for total reflection: light is emitted from the light dense medium to the light hydrophobic medium, and the incident angle is greater than or equal to the critical angle. When drawing the optical path diagram and solving practical problems by using the refraction law of light, it is first necessary to judge whether total reflection will occur. When it is determined that total reflection does not occur, the incident angle or refraction angle is determined according to the refraction law.

☆ prism and light dispersion

1. prism:

(1) Definition: The transparent body where planes intersect is called a prism. What is usually made is a prism with a triangular cross section, that is, a prism.

(2) Function: change the direction of light propagation; Split light.

2. Light passing through the prism:

(1) The law of light deflection by the prism is as shown in the following figure: the light passing through the prism will be deflected to the bottom of the prism; Prism should change the propagation direction of light without changing the properties of light beam.

The parallel beam is still a parallel beam after passing through the prism;

The divergent beam is still divergent after passing through the prism;

The convergent beam is still a convergent beam after passing through the prism.

The included angle between emergent light and incident light is called deflection angle.

(2) prism imaging:

As shown in the following figure, the image of the object seen through the prism is an upright virtual image, and the position of the image is offset to the top corner of the prism. 3. Total reflection prism

(1) Definition: A prism with an isosceles right triangle is called a total reflection prism.

(2) The optical route is controlled by a reflective prism, as shown in the figure below. (3) Advantages of total reflection prism: Total reflection prism and plane mirror have the same effect in changing the optical path. In contrast, the total reflection prism has clearer imaging and less light energy loss.

4. Scattering of light

(1) Dispersion of light: The phenomenon of decomposing polychromatic light into monochromatic light is called dispersion of light.

After passing through the prism, white light is decomposed into seven colors: red, orange, yellow, green, blue, indigo and purple.

(2) correctly understand the dispersion of light:

The color of light is determined by its frequency. Among the monochromatic lights that make up white light, the frequency of red light is the smallest, and that of violet light is the largest. The frequency of light is constant in different media.

The propagation speed of colored light with different frequencies is the same in vacuum, which is C = c=3× 108 m/s/s, but in other media, the speed is different. In the same medium, the speed of violet light is the smallest and the speed of red light is the largest.

The refractive index of different colors of light in the same medium is different. Generally, the higher the frequency, the greater the refractive index in the medium. So white light refracts into the medium, violet light refracts the most seriously and red light refracts the least.

Because when colored light propagates in the medium, the speed of light will change, so will the wavelength. In different media, the same color light with large refractive index has small speed and short wave length; Light with low refractive index has high speed and wavelength. Light with different colors in the same medium has high refractive index, low speed of light and short wavelength; Low-frequency refractive index, high speed of light and large wavelength.

● Excellent topics and lectures.

Example 1. The point light source S is located in front of the plane mirror, as shown in the figure below. Assuming that the light source is stationary, the plane mirror translates to the light source along the OS direction at a speed v, and the angle between the mirror surface and the OS direction is 30. Then the image S' of the light source will () A. move to S along the connecting line of S at a speed of 0.5v

Move along the line of s at speed v to s.

C, move to s at a speed v along the connecting line of s.

D. move to s along the line of s at a speed of 2v.

Analysis:

Because the object and the image are symmetrical about the mirror surface, the line between the object and the image is always perpendicular to the plane mirror whether the plane mirror moves near or away from the light source, so the image S' can only move along the line of S. In addition, the approaching speed of the plane mirror along the S direction is V' = VSIN 30 = 0.5V, so the approaching speed of the plane mirror along the S direction is V' = V' = 0.5V, so the approaching speed of the plane mirror along the S direction is V 1 = V'+V' = V, which shows that option B is correct.

Comments:

This topic mainly investigates the imaging law of plane mirror. When there is relative motion between the plane mirror and the light source, it is the key to solve this kind of problem to correctly understand the dynamic picture of the image point movement. In addition, this problem can also be determined by the displacement relationship between the mirror and the image movement.

Example 2. As shown in the following figure (a), the luminous point S moves in a straight line at a uniform speed from point A along the connecting line of AB, with the speed v = m/s, and there is an axis O perpendicular to the paper at the distance L=3m from the starting point A, and the plane mirror MN can rotate around the axis O. In order to make the luminous point S always on the connecting line of PO parallel to AB, try to find the elapsed time T = 65438+. Analysis:

According to the meaning of the question, the angle between the plane mirror and OA is θ1= 45, and the luminous point reaches point C after t =1s.

, then

The included angle between OP reverse extension line and OC is β = 60.

According to the imaging characteristics of plane mirror, MN should be on the bisector of β.

So the plane mirror turned 15.

Comments:

This problem is a synthesis of plane imaging, uniform linear motion and mathematics. In the training, we should clear our thinking, draw a good imaging light path diagram and apply mathematical geometry knowledge to solve it.

Example 3. As shown in the figure below, MN is a horizontal plane mirror and PQ is a vertical ruler. Try to draw the area where the human eye can see the AB part on the ruler imaged in the mirror, and write the drawing steps. Analysis:

What the human eye sees is that the light from AB enters the convergence point of the reverse extension line of the eye after being reflected by the plane mirror.

(1) According to the characteristics of symmetry, make AB look like A'B' in the mirror.

(2) If a is used as the two boundary rays AM and AN and the corresponding reflected rays MA 1 and NA2, a virtual image A' can be seen in the area sandwiched by MA 1 and NA2.

(3) If the boundary rays BM, BN and the corresponding reflected rays MB 1, NB2 pass through B, a virtual image B' can be seen in the area sandwiched by MB 1 and NB2.

Therefore, in the male * * * part of the two regions (as shown by diagonal lines in the above figure), A 1MNB2, that is, the range where the complete image of AB can be seen, A'B' can be seen at the same time.

Comments:

Looking at a virtual image through a flat mirror is like looking at an object outside the window through a "window" as big as a flat mirror. In this problem, it is equivalent to hollowing out the mirror MN into a "window". The specific observation range is defined by the line connecting the image point and the edge of the flat mirror.

Example 4. As shown in the figure below, AB stands for flat mirror, P 1P2 stands for horizontal meter scale (scale surface faces flat mirror), and M and N are screens, which are parallel to each other. Ab on the screen MN represents a vertical seam (that is, there is light between ab). Someone's eyes are close to the small hole S of the meter ruler (see the picture below), and part of the scale of the meter ruler can be seen through the plane mirror. Try to draw the visible part with a triangle on the diagram of this problem, and draw this part on P 1P2 for analysis:

If S is a point light source, the range that the light it emits can shine on the meter scale after being reflected by the plane mirror is known as the reversibility of the light path, that is, the range that the human eye can see through the plane mirror and the slit ab. The light path is shown in the following figure. Comments:

When looking for the topic of "human eye observation range", it is simpler and easier to use the method of "taking human eye as light source and finding illumination range", so we should use it boldly in the future. What the human eye sees from the flat mirror is actually the image and scale of the screen. The correct conclusion can be drawn by taking the images of the screen and ruler first, and then determining the range, but it is much more complicated than drawing by using the reversibility of the light path. Please try the latter drawing method.

Example 5. As shown in the figure below, a small ball is close to the front of the point light source S, thrown horizontally to the left, and just landed at the angle A, then the movement of the center of the ball shadow on the vertical wall is () A. Uniform linear movement.

B. Variable acceleration linear motion with gradually increasing acceleration

C. uniformly accelerating linear motion

D. Variable acceleration linear motion with gradually decreasing acceleration

Analysis:

As shown in the figure below, when t = 0, the ball will be thrown horizontally, and P0 is its shadow. After time t, the ball reaches Q, and P is its shadow, and the rectangular coordinate system as shown in the figure is established. According to the law of horizontal throwing motion, x=v0t and y=gt2, it can be seen that L is the distance from the throwing point to the vertical wall.

Solved:

So the motion of the shadow is uniform, and its speed is

Comments:

Solving such problems cannot be taken for granted, but should be solved according to the law and regulations.

Example 6. As shown in the following figure, the polychromatic light beam PO containing only yellow light and purple light enters the glass semi-cylinder in the air along the radial direction, and then is divided into two beams OA and OB, which are emitted in the direction shown in the figure. Then () A. OA is a yellow light, OB is a purple light, B. OA is a purple light, and OB is a yellow light.

C.OA is yellow light and OB is polychromatic light. D. OA is purple light, and OB is polychromatic light.

Analysis:

Because n purple > n yellow, the critical angle C purple.

Comments:

When discussing the propagation law of light at the interface between two media, we should first judge whether total reflection has occurred, and then apply the corresponding law to discuss and solve it.

Expand:

As shown in the following figure, A and B are both glass plates with uniform thickness, and the included angle between them is φ, and a thin beam enters from point P with incident angle θ, θ >; φ. It is known that this light beam consists of red light and blue light. Then when the beam passes through the b plate. () A. The propagation direction is deflected to the left by φ angle relative to the direction of incident light.

B, the propagation direction is deflected by φ angle to the right relative to the direction of incident light.

C. Red light is to the left of blue light.

D. Red light is on the right of blue light.

Analysis:

According to the meaning of the question, A and B are both parallel glasses. When the light passes through the parallel flat glass, the propagation direction of the light remains unchanged, but it deviates to the side. Because the refractive index of light is related to the frequency of light, the refractive index of light with high frequency is also large for the same medium. The blue refractive index nb is greater than the red refractive index nr, that is, nb > NR. According to the law of refraction, there are

It is known that the refraction angle rr of red light is greater than the refraction angle rb of blue light.

When the mixed beam of blue light and red light is incident on the lower surface of the A glass plate, it is refracted by the glass, and the red light and blue light are separated. When it exits from the upper surface of the A glass plate, it is parallel to the original incident light, but the red light and blue light are separated. Similarly, when the red light and blue light enter the B glass plate and then exit, the red light and blue light are still parallel to the incident light when they enter the A plate, but the distance between the two beams of light is further increased. Therefore, options a and b are wrong. See the figure below. Because the refraction angle of red light is greater than that of blue light, the position of red light is on the right side of blue light, and option D is correct.

Example 7. The cross section of a semi-cylindrical glass brick with radius r is shown in the following figure, with O as the center. Light I enters the glass in the radial direction from point A, and then totally reflects at point O. The other light II is parallel to light I, enters the glass brick from the highest point B, and then refracts to point D on MN. What is the measured refractive index of glass brick? Analysis:

Let the incident angle and refraction angle of ray Ⅱ at ⊿bOd be I and R, respectively.

According to the law of refraction, there is, that is.

In addition, rays I and II are parallel, and total reflection occurs at O point, so, so, we get.

Comments:

To solve this kind of problem, we should grasp the key of refraction law and the condition of total reflection. When analyzing and studying the light path, it is often assumed that a certain light can meet the requirements of the problem, and then the light path diagram of its reflection, refraction or total reflection is drawn accordingly, so as to infer or solve it.

Example 8. What is the apparent depth of a pool, actually H, looking down at the vertical water surface? (Let the refractive index of water be n)

Analysis:

As shown below, two refracted rays are made from the bottom of the water S, one perpendicular to the water surface and the other incident at a small angle (the opening angle of the eyes to the spot is very small). The intersection of the extension lines of these two refracted rays is the image of S you see. As can be seen from the figure, the depth of the image becomes shallow.

Comments:

In the analysis and calculation of refraction problem, we should first draw the light path diagram (including incident light and refraction line), then try to draw a right triangle, and then use geometric knowledge and refractive index formula for analysis.

Example 9. Transparent optical materials, with AB as an end face, establish a plane rectangular coordinate system as shown in the left figure below, assuming that the refractive index of optical materials decreases uniformly along the positive direction of Y axis (that is, the BA direction). If a beam of light PO is emitted from the air to point O at a certain incident angle θ and enters the optical material, then the possible propagation path of light in the optical material is the () analysis in the right figure below:

Because the refractive index of optical material decreases uniformly along the positive direction of Y axis, when light continues to propagate after entering the optical material, the refraction angle is greater than the incident angle, and the light will gradually deviate from the normal, and the incident angle will gradually increase when entering the upper layer from the lower layer of the optical material. When the incident angle is greater than the critical angle, total reflection begins to enter the lower layer from the upper layer. Similarly, because the refractive index of the material increases gradually along the negative direction of the Y axis, the refraction angle is smaller than the incident angle, and the light should gradually approach the normal, so the propagation path can be regarded as D.

Comments:

The refractive index of optical materials is uniform, which can be regarded as the superposition of several layers of materials with different refractive indexes, so it is convenient to analyze the general direction of optical path by using the refraction law of light. This approximate treatment method is also very common in the discussion of physical problems and has practical significance.

● Feedback exercises

First, multiple choice questions

1. The following figure shows the cross section of a right-angle prism, ∠ BAC = 90, ∠ Abc = 60, and the parallel beamlets enter the prism from point O in the direction perpendicular to the bc plane. It is known that the refractive index of prism material is n =, and if the reflected light of the original incident light on the bc plane is not considered, there will be light () A. emerging from the ab plane.

B. Emission from AC surface

C, emitting from bc plane and obliquely crossing bc plane.

D, emitted from bc plane and perpendicular to bc plane.

2. As shown in the figure below, the right-angle prism AB is made of glass with a critical angle of 42, where ∠ b =15 and ∠ c = 90. When the light is incident perpendicular to the AC plane, the times of total reflection in the prism are () a.2 times b.3 times c.4 times d.5 times.

3. As shown in the figure below, the principle of the laser level controller is to fix the laser beam AO to irradiate the horizontal level at the incident angle i, and the reflected light OB irradiates the horizontally placed light screen. The light signal on the screen is converted into an electrical signal, which is input into the control system to control the height of the liquid level. If the light spot moves to the right on the screen and hits point B', the height of the liquid level changes to () A. The liquid level decreases.

B. Liquid level rise

C. Liquid level drop

D. liquid level rise

4. On the surface of the water, people can judge () by observing the small colored lights of red, orange, yellow, green, blue, indigo and purple in the water directly below and seeing them lined up at the same depth under the water.

A. The actual position of purple light is the deepest, deeper than the position seen.

B. The actual position of the purple lamp is deepest and shallowest than that seen.

C the actual position of the red light is the deepest, deeper than the position you see.

D. the actual position of the red light is deeper and shallower than it looks.

Optical fiber communication is a modern communication means, which can provide large capacity, high speed and high quality communication services. At present, China is vigorously building a high-quality broadband optical fiber communication network. The following statement is true ()

A. Optical fiber communication uses light as the carrier to transmit information.

B. Optical fiber transmits optical signals by using the diffraction principle of light.

C. Optical fiber uses the principle of optical dispersion to transmit optical signals.

D. At present, the widely used optical fiber is a very fine special glass fiber.

6. A beam of polychromatic light is emitted from the air to a parallel plane glass brick and refracted into two monochromatic lights A and B. As we all know, the frequency of A light is less than that of B light. Which of the following light path diagrams may be correct? () 7. Place a plane mirror in the center of the cylinder, facing the point light source S on the cylinder wall. The plane mirror rotates around the cylindrical axis O at an angular velocity ω from the position shown in the figure below. During its rotation of 45, the angular velocity of the image motion formed by the point light source in the mirror is ω 1, and the angular velocity of the reflected light spot motion is ω2, so the following relationship is correct: () A. ω 60.

B.ω 1=ω，ω2=2ω

C.ω 1=2ω，ω2=2ω

D.ω 1=ω，ω2=ω

8. After the rain clears, people often see rainbows in the sky, which is a phenomenon when the sun shines on water droplets in the air. When explaining this phenomenon, it is necessary to analyze the light path after the light enters the water drop. A thin beam enters a water drop, which can be regarded as a ball with radius r, the vertical distance from the center o of the ball to the incident light is d, and the refractive index of water is n (1). Draw the optical path diagram of the beam entering the water drop and exiting from the water drop after a reflection.

(2) Find the deflection angle of this beam of light from shooting to shooting.

Feedback exercise answers:

1. Answer: BD

Tip:

When the incident angle is 60 at the ab interface and greater than the critical angle of 45, the incident light will be totally reflected on the ab plane, and when the incident angle is 30 at the ac plane and less than the critical angle of 45, the incident light will be refracted and reflected on the ac interface. Refracted light will be emitted from the ac plane and reflected light will be emitted from the bc plane. According to the geometric relationship, when the incident angle is 0, it will exit vertically. BD is correct. 2. Answer: B

Tip:

The first incident angle on AB plane is 75, the second incident angle on BC plane is 60, the third incident angle on AB plane is 45, and the fourth incident angle on BC plane is 30. The first three events are greater than the critical angle of 42, and B is correct.

3. Answer: D

Tip:

As shown in the figure, after the liquid level rises, the laser beam AO is reflected and emitted to point B'. According to the reflection law and geometric relationship, bb' = δ S = O' D, δ O' OD is an isosceles triangle, and its height is the height h of liquid level rise, so δ S/2 = h× Tani, so the height of liquid level rise is. 4. A: A.

Tip:

When people are on the water, they see small colored lights underwater as their images. If the image height is the same, it means that the outgoing light direction is the same, then the incident angle of underwater violet light is the smallest, and the position is also the lowest, which is lower than the actual position.

5. answers: a and D.

Tip:

Optical fiber uses the principle of total reflection of light to transmit optical signals, so options B and C are wrong. According to the book knowledge, options A and D are correct.

6. Answer: B

Tip:

After any light passes through the parallel glass brick, the transmitted light must be parallel to the original light direction, so C and D are wrong; Because the frequency of A is less than that of B, we can know that the refractive index of A is less than that of B, and the refractive angle of A is greater than that of B, so A is wrong in B. ..

7. A: A.

Tip:

You can use the special value method to make the plane mirror rotate 45 for comparison, and you can know that the correct option is A.

8.