Shanghai senior one math book answers.

Unit 10: linear propagation and reflection of light

First, the golden knowledge points:

1, linear propagation of light;

2. umbra and penumbra;

3. solar eclipse;

4. The law of reflection;

5. Imaging law of plane mirror;

6. Apply the law of reflection to analyze the problem.

Second, the main points revealed:

(a), the linear propagation of light:

1, light source: the luminous object is called light source. A light source is a device that converts other forms of energy into light energy.

2. Linear propagation of light;

(1) medium: A substance in which light can propagate is called a medium.

(2) Light travels in a straight line in the same uniform medium. Aperture imaging is formed by the linear propagation of light in the air.

(3) Light: A straight line used to indicate the direction of light beam propagation when studying the behavior of light. Note: Light is not real, but an ideal model.

(2) Shadows, eclipses and the speed of light.

(1) umbra and penumbra: Behind the backlight surface of an opaque object. The area where light can't shine at all is called umbra. The area where only part of the light shines is called penumbra.

Note: the size of umbra area is related to the luminous surface of light source. The larger the luminous surface, the smaller the umbra area.

(2) Analysis of the causes of solar and lunar eclipses:

When the shadow of the moon falls on the earth, a solar eclipse can be observed on the earth. (The umbra has a total solar eclipse, and the penumbra has a partial solar eclipse. See the example. )

The moon enters the umbra of the earth, and an eclipse can be observed on the ground (see example).

Note: whether it is a solar eclipse or a lunar eclipse, it generally refers to what is observed on the earth.

(3), the speed of light:

The propagation speed of light in vacuum is 3x108m/s.

[Example 1] Regarding solar and lunar eclipses, the following statements are correct:

A, you can see the total solar eclipse in the umbra of the moon; B, you can see the partial solar eclipse in the penumbra of the moon;

C. When the moon enters the penumbra of the earth, a partial lunar eclipse can be seen;

D when the moon completely enters the umbra of the earth, you can see the total lunar eclipse.

[analysis and answer] the correct options are a, b and d.

This problem needs to understand the principle that light travels in a straight line to form shadows. By making a schematic diagram of light propagation, we can see that in the umbra area of the moon, it is impossible for any light from the sun to enter, so we can see the total solar eclipse (we can't see the sun, as if it had been eaten by the "Tiangou"). In the penumbra of the moon, only part of the sunlight enters, so a partial solar eclipse can be seen in the penumbra (only part of the sun is "eaten" by the moon).

So options a and b are correct.

In the penumbra of the moon, only the light near the edge of the sun shines, so the annular eclipse can be seen in the penumbra.

As can be seen from Figure B, when the moon enters the penumbra A and penumbra C of the earth, the whole moon is still bright, just a little darker than the penumbra, so there will be no eclipse on the earth.

Only when a part of the moon enters the umbra B of the earth can we see the partial solar eclipse of the moon, so C is wrong;

When the moon enters the umbra of the earth, a total lunar eclipse occurs (the moon can't be seen at this time), so the D option is correct.

[Summary] When analyzing solar and lunar eclipses, the key is to distinguish who "ate" who, that is, to distinguish the positional relationship between the sun, the moon and the earth. When the moon is between the sun and the earth, the "sun" is "eaten", that is, an eclipse occurs; When the earth is between the sun and the moon, the moon may be eaten, that is, an eclipse may occur. Both solar and lunar eclipses are seen on the earth.

(3) reflection law, plane mirror

1. reflection law: reflected light is on the same plane as incident light and normal line, reflected light and incident light are located on both sides of normal line respectively, and the reflection angle is equal to the incident angle.

Note: ① Incident angle and reflection angle refer to the included angle between incident light and reflected light and normal;

② In the reflection phenomenon, the optical path is reversible. For example, if A sees B in the mirror, then B can certainly see A in the mirror.

2, specular reflection and diffuse reflection:

Characteristics; Parallel light is still parallel light after specular reflection, and it is reflected in all directions after diffuse reflection. Whether it is specular reflection or diffuse reflection, the reflection of each kind of light obeys the law of reflection.

3, flat mirror:

(1) Regarding the virtual image formed by the plane mirror, we can determine a luminous point S ..

Position, because the light emitted by the luminous point S enters the eyes, making

We think that the luminous point S is located at the intersection of these rays (as shown in the figure). then

Yao, when the light beam emitted by the luminous point S is reflected by the plane mirror, all

The reverse extension line of the reflected light also intersects with the same point S, and we will feel it.

It seems that these reflected rays are emitted from point S (pictured).

S, which is the "virtual image" of the luminous point S.

(2) Plane mirror imaging law: flat and positive virtual image.

(3) Plane mirror imaging drawing method:

① reflection law method: any two rays emitted from an object point are directed to a plane mirror, and the reflected rays are made by reflection law, so that the intersection of the opposite extension lines of the two reflected rays is a virtual image point.

② Symmetry method: the vertical line from the clipping point to the mirror, and the point on the other side of this vertical line where the object point is equidistant from the mirror is the virtual image point.

[Example 2] Draw the mirror MN emitted by the light source point S in the right picture.

After reflection, the light passing through point P needs drawing steps.

[Analysis and Solution] Because the light emitted from point S is reflected by mirror MN.

After the reflected light passes through point P, its backward extension line must pass through the image S'

Point. The steps are as follows:

(1) is the symmetry point of point s to MN, and s and s are image points;

(2) Connect the intersection of S'P and MN at point A (S and A should be indicated by dotted lines);

③ If SA is connected, AP is the reflected light of incident light SA. (pictured)

[Example 3] The point light source S is placed in front of the plane mirror MN. If MN does not move, the light source S moves straight to the right at a speed of 2m/s, making an angle of 600 with the mirror. As shown in the right figure, the image of the light source in the mirror will be:

A, translate to the right along the OS straight line at a speed of 4m/s;

B, downward translation perpendicular to OS at a speed of 2m/s;

C, moving linearly from the other side of the mirror to point O at a speed of 2m/s;

D. See the image approaching S2 on S 1 at a speed of 2m/s. 。

[analysis and answer] the correct options are c and D.

Make the image S' (the symmetry method can be used), and make the image S 1 after the position S 1 and the time T. According to the geometric symmetry, SSl=S, S 1', so the image S will move linearly along the S and O lines at the speed of V = 2m/s. 。

So c is correct.

According to the symmetry relation, in t time, the distances SA and S'A' along the direction perpendicular to the mirror surface are equal, so they are close to each other.

The distance change is X=SA+S'A'=2SA, and the approaching speed of the image relative to the object is:

VSS' = x/t = 2sa/t = 2v0cos300 = v0 = 2 (m/s), so d is correct.

[Summary] It is the key to solve the practical problems of plane mirror imaging by drawing the light path diagram correctly, applying the imaging law of plane mirror (image and object are symmetrical about mirror surface) and combining with geometric knowledge analysis.

(D) the application of reflection method

1, the method of analyzing and discussing the phenomenon of light reflection by using the law of light reflection;

(1) Make the reflected light path map correctly (the image points can be determined by symmetry method first, and then the light is made);

(2) Applying geometric knowledge to analyze the relationship between variables.

2. The field of view of the plane mirror: the light emitted by the object can enter the range of the observer's eyes after being reflected by the plane mirror, that is,

The limited range of reflected light at the edge of a plane mirror.

[Example 4] One person stands by the pool (as shown in the figure) and draws a light path diagram.

Explain that he saw the range of trees across the pool from the reflected light on the water surface and thought

Find out the drawing steps. Where point A is the human eye, and point B and point C are the edges of the pool.

[Analysis and solution] Method 1: According to the imaging law of flat mirror, it is up to people to decide.

The range that the eye can see determines the range it can see. The steps are as follows:

① Draw the virtual image of the tree by using the symmetry law of the object image about the mirror;

(2) Connect AB and extend, and at E, the point meets the image;

(3) Connect alternating current and extend. If it intersects the image at point d, then E'D' D' means seeing it.

The range of virtual image;

(4) using the symmetry of the object image to find e and d? Corresponding luminous points e and d;

⑤ Connect DC and EB to get the light path diagram, and DE means to see the tree through water.

The range of (pictured).

Method 2: Using the principle of reversible optical path, the image of human eyes is regarded as point light.

Light source, the part it illuminates is the visible light range. The steps are as follows:

(1) Image an object called the human eye with an object image;

(2) Connect a' c and extend it, meet the tree at point D, and point D is to see it.

The highest point of the tree;

(3) Connect a'b and extend it, and meet the tree at point E, which is visible.

The lowest point of the tree;

(4) Connect BA and CA to get a reflection line, and DE means to see the tree through the water.

The range of (pictured).

Third, the good question is for you:

(1) Preview questions for this lesson:

1, which of the following statements is true:

A, light always travels in a straight line; B, light always travels in a straight line in the same medium;

C pinhole imaging is formed by the linear propagation of light; Light always travels in a straight line in the same homogeneous medium. 2. The following statement is correct:

A, the light source can emit an infinite number of beams; B, light does not actually exist;

C, light is an abstract expression of a thin beam; D, light is a straight line used to indicate the direction of light propagation.

3. The following statements about the reflection of light are wrong:

A, reflected light and incident light are on the same plane, and the reflected light and incident light are separated on both sides of the normal;

B, specular reflection means that the reflection angle is equal to the incident angle, which belongs to regular reflection; The reflection angle of diffuse reflection is not equal to the incident angle,

It is a reflection without a clear direction;

C. When the included angle between the incident light and the interface is 300, the reflection angle must be 300;

D when the incident light forms an angle of 900 with the interface, the reflection angle is 00.

4. Which of the following phenomena is caused by diffuse reflection:

A, you can see the light source from different directions; B, seeing the image of the object through the water surface;

C, reading a book at night (under the lamp) will see a dazzling luster on the paper;

D, you can see the images on the movie screen from different directions.

5. The following statement is correct:

A. The light emitted by a point light source is reflected by a plane mirror and converged at one point to form a virtual image;

B, the image of an object in a plane mirror must be an isometric, upright virtual image, and the image and the object are symmetrical about the mirror;

C, the divergent beam is still a divergent beam after being reflected by the plane mirror;

D, plane mirror can change the direction of light propagation, but can't change the parallel or non-parallel relationship between two beams of light.

Reference answer:

1、C2、ABCD 3、BC 4、D 5、BCD

(2) Basic questions:

1, the propagation speed of light in vacuum is _ _ m/s, if Sirius is 8.7 light years away from the earth, then it is from the earth.

_ _ _ _ _ _ _ _ _ _ kilometers.

2. If the plane mirror bypasses the incident point and is perpendicular to the plane axis determined by the incident ray and the normal, the plane mirror reflects a beam of light.

Turning the angle θ, the reflected light will:

A, keep the original position; B, rotation angle θ; C, turning through an angle of 2θ; D. rotate by an angle of 4θ.

3. As shown in the figure, the incident light AO can propagate from the ground.

The plane shot vertically at a horizontal plane 30 meters above the ground.

On the mirror, if you turn the mirror around the horizontal axis by an angle of 150, it will be the other way around.

The light will be reflected to point B far away from the light source _ _ _ _ _ _ _.

Reference answer:

1、3 108,8.23 10 13 ; 2、C； 3、 10 ;

(3) Application problems:

1, the light emitted by the light source will form a shadow when it strikes an opaque object. Judging the area of the light source and the size of the shadow, the correct ones are:

A, the larger the light source area, the larger the penumbra area; B, the light source area is zero, and the penumbra is not zero;

C, the larger the light source area, the larger the umbra area; D the size of the shadow has nothing to do with the area of the light source.

2, put a pencil in front of the flat mirror, if you want to make the pencil perpendicular to its image, the pencil should be _ _ _ _ degrees with the mirror.

Angle, when the pencil and the plane mirror form an angle of α, the included angle between the pen and the image should be _ _ _ _ _ _ _ _.

3. If someone stands 2m in front of a flat mirror and wants to observe the complete image of the 9m-high object at 16m in front of the mirror, the minimum length of the mirror is _ _ _ _ _ m. 。

Reference answer:

1、A； 2、450,2α; 3、 1

(4) Improve questioning:

1. Draw the optical path diagram of plane mirror imaging (as shown in the figure). To write a drawing step:

2. A circular reservoir with a radius of 5m is filled with water, and the water surface is flush with the ground.

There is a lamp hanging in the center of the pool 3 meters from the water surface, which can keep your eyes high from the ground.

1.8m How far away from the pool can people still see the image of the lamp in the water?

3. In the picture on the right, try to draw a complete image of AB within the range of human eyes.

Reference answer:

1, [analysis and solution]

Drawing steps:

Left: (1) According to the symmetry of the object and the image, S is made.

Such as s';

(2) then making two incident rays at will;

(3) According to the reverse extension line of the reflected light, it must pass.

Like a point, make two reflected rays;

Right: (1) If the object AB is regarded as two point light sources, the images of point A and point B are made respectively;

(2) Then follow the same steps as shown in the left figure;

(3) Connect the images A' and B' of A and B to form the image A'B' of AB.

2. [Analysis and Answer]

The pool surface is equivalent to a flat mirror, and the optical path diagram of reflected light is shown in the figure.

If a person is in a certain position, he can only see the light in the water through the edge of the pool.

In the image, when the light from the light source just reflects through the edge of the pool.

Enter people's sight. According to the law of reflection, α = β is determined by the following formula

Geometric analysis shows that SBO is similar to DCO, so it has

H/H = X/R, so x=hr/H= 1.8 5/3= 3 (m).

Therefore, it is the key to solve this kind of problem to make a correct reflection light path diagram and combine the reflection law for geometric analysis.

3. [Analysis and solution] The image seen by the human eye is the convergence point of the backward extension line of the light from AB after being reflected by the plane mirror. Firstly, the image AB' of AB is made by symmetric method. The two reflected rays ma 1 and NA2 that hit the boundary of the mirror at point A can be seen in the range of ma 1 and NA2.

To image point a,; Then make two reflected rays from point B to the boundary of the mirror.

Mb 1 and NB2, and the image point b can be seen in the range of MB 1 and NB2.

Obviously, the eye is in the male part of the above two areas (that is, between MAl and NB2).

You can see the complete image of AB (pictured).

[Summary] In order to see the image of an object in a flat mirror, it is necessary to use boundary light. The common part of the area defined by the boundary light is the observation range of the complete image. It can be simply described as the upper limit of the connecting line between the upper end of the mirror and the upper end of the image, and the lower limit of the connecting line between the lower end of the mirror and the lower end of the image.

(5) After-school martial arts venues:

1, shadowless lamp is a combination of multiple large-area light sources, about its lighting effect; The correct way is:

A, no shadow; B, there is an umbra; C, no umbra; D, no penumbra.

2. The picture shows a schematic diagram of a solar eclipse with people in it.

There you can see a partial solar eclipse:

A and a;

b，b；

C and c;

D I can't see it.

There is a small hole in the linoleum roof of the shed, through which the sunlight falls to the ground to form a circular spot. This phenomenon shows that:

A, the shape of the hole must be round; B. the shape of the sun is round;

C, the spot on the ground is the image of the sun; Light travels in a straight line.

4. The characteristics of plane mirror imaging are:

A, the image is located behind the mirror, which is an upright virtual image; B, the image distance behind the mirror is equal to the object distance in front of the mirror;

C, the size of the image is equal to the size of the object; D, the color of the image is the same as the color of the object;

E. If the object is moved closer to the mirror, the image will gradually become larger.

5, people standing in front of a flat mirror, about his image, the correct statement is:

A, a flat mirror, no matter how placed, can see his full-length image;

B, if a person is away from the plane mirror at speed v, then the person sees his image moving in the opposite direction at speed v;

C, people can image anywhere in front of the plane mirror on the other side;

D. People can see their own images anywhere in front of the flat mirror.

6, a person standing in front of the plane mirror 1.5 m, this person in front of the mirror is 0.5 m, the distance between people and images is:

a、 1m； b、 1.5m； c、2m； 2.5 meters.

7. The incident light forms a 45-degree angle with the plane mirror. If the plane mirror rotates to increase the incident angle of 5, the incident light and the reflected light will

The included angle is:

a、700 B、750 C、800 D、 1000。

8. When a person faces a vertical plane mirror from a distance, the size of his image in the mirror changes as follows:

A, gradually getting bigger; B, gradually getting smaller; C, the size is unchanged; D, I can't be sure.

9. As shown in the picture, there is a hair in front of the plane mirror MN.

Spot S, please use it as a starting point for drawing. The light spot s is in the plane mirror.

The optical path diagram of the image shows that the range of the image is indicated by diagonal lines.

To write a chart:

10. It is known that the width of the river channel is d, and the banks of A and B are higher than the water level hA and hB respectively. One wants to measure the height of a tree by observing the reflection of A on the bank of B, such as d=30m, hA=2.0m, HB = 1.0m, and the height of the observer's eyes from the ground is H =1.8m..

Reference answer:

1、C； 2、B； 3、BCD4、ABCD5、C； 6、C； 7、D； 8、C；

9. [Analysis and Answer]

Drawing steps:

1, according to the symmetry between the object and the image, make the image of the luminous point S';

2. According to the edge part of the plane mirror, make two straight lines from the luminous point S..

Incident light;

3. Two reflected rays of incident light are generated;

4. Then the area between the two reflected rays is the required area.

10, [analysis and answer]:

Draw a picture, according to the figure, d1+D2 = d.

Let the tree height be x, then according to similar triangles, we can get: