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Geometric Optics - Practice Questions (38)

Question 1: 1. Figure a beam of red light and a beam of violet light to the appropriate angle of incidence to th...

1. Figure a beam of red light and a beam of violet light to the appropriate angle of incidence to the semicircular glass block, its emitted rays are from the center of the circle $O$ point along the $O P$ direction of the shot, as shown in the figure, then ( ) ![](/images/questions/phys-geometric-optics/image-001.jpg)

  • A. A. $A O$ is red light, which takes a short time to pass through the glass block
  • B. B. $A O$ is violet light, which takes a long time to pass through the glass block
  • C. C. $B O$ is red light, which takes a long time to pass through the glass block
  • D. D. $B O$ is violet light, which takes a short time to pass through the glass block

Answer: A

Solution: From the light path diagram, $B O$ light has a stronger refractive ability than $A O$ light, proving that the refractive index of $B$ light is greater than the refractive index of $A$ light, and so is red light and $B O$ is violet light, and light travels faster in the medium $v = \frac { c } { n }$, so $A O$ travels faster in the medium, and it is known from $t = \frac { l } { v }$ that $t = \frac { l } { v }$ is $A O$. $A O$ takes less time to travel through the glass block, so choose A.

Question 2: 2. As shown in the figure, the complex color light consisting of two monochromatic colors is divided...

2. As shown in the figure, the complex color light consisting of two monochromatic colors is divided into $a , b$ two beams after passing through a sufficiently large rectangular transparent material, then ( ) ![](/images/questions/phys-geometric-optics/image-002.jpg)

  • A. A. In this transparent material, the propagation speed of $a$ light is less than the propagation speed of $b$ light
  • B. B. When encountering obstacles, $a$ light is more likely to be significantly diffracted
  • C. C. The refractive index of this transparent material for both types of light $n _ { a } < n _ { b }$
  • D. D. When total reflection occurs from this transparent material into air, the critical angle of $a$ light is large

Answer: A

Solution: AC. According to the law of refraction $$ n = \frac { \sin i } { \sin r } $$ The refractive index of this transparent material for these two types of light can be known as $$ n _ { a } > n _ { b } $$ According to $$ v = \frac { c } { n } $$ INLINE_FORMULA_4]], the speed of propagation of $b$ light is less than that of $b$ light in the transparent material, so A is correct and C is wrong; B. The longer the wavelength, the easier it is for obvious diffraction to occur. $b$ Light has a larger wavelength and is more likely to undergo obvious diffraction, so B is wrong; D. According to the relationship between critical angle and refractive index $$ n = \frac { 1 } { \sin C } $$ it can be seen that $b$ light has a larger critical angle, so D is wrong.

Question 3: 3. three transparent media stacked together, and parallel to each other, a beam of light in the inte...

3. three transparent media stacked together, and parallel to each other, a beam of light in the interface of the two media I and II after total reflection, shot to the interface of the two media II and III, refraction occurs as shown in the figure, set the rate of light in the three media $v _ { 1 } , v _ { 2 } , v _ { 3 }$, then the size of their relationship is () ![](/images/questions/phys-geometric-optics/image-003.jpg)

  • A. A. $v _ { 1 } > v _ { 2 } > v _ { 3 }$
  • B. B. $v _ { 1 } > v _ { 3 } > v _ { 2 }$
  • C. C. $v _ { 1 } < v _ { 2 } < v _ { 3 }$
  • D. D. $v _ { 2 } > v _ { 1 } > v _ { 3 }$

Answer: B

Solution: According to the question, total reflection of light occurs at the interface of two media I and II, which means that the refractive index of I is less than the refractive index of II, i.e. $n _ { 1 } < n _ { 2 }$ . Refraction occurs at the interface of two media II and III and the angle of refraction is greater than the angle of incidence, indicating that the refractive index of II is greater than the refractive index of III, i.e. $n _ { 2 } > n _ { 3 }$ . Comparing medium I with III, the refractive index of medium I is less than the refractive index of medium III, i.e., there is $n _ { 1 } < n _ { 3 }$. So there is $n _ { 2 } > n _ { 3 } > n _ { 1 }$ and according to the formula for the rate of light in these three media $v = c / n$, the speed of light is directly proportional to the refractive index, then $v _ { 1 } > v _ { 3 } > v _ { 2 }$. Therefore, choose B.

Question 4: 4. A beam of red light from the air into a medium, the interface on both sides of the light and the ...

4. A beam of red light from the air into a medium, the interface on both sides of the light and the interface angle were $45 ^ { \circ }$ and $60 ^ { \circ }$, as shown. Then the refractive index of the medium for this red light is ( ) ![](/images/questions/phys-geometric-optics/image-004.jpg)

  • A. A. $\frac { \sqrt { 2 } } { 2 }$
  • B. B. $\frac { \sqrt { 6 } } { 3 }$
  • C. C. $\frac { \sqrt { 6 } } { 2 }$
  • D. D. $\sqrt { 2 }$

Answer: D

Solution: According to the question, the angle of incidence and the angle of refraction are $$ \theta = 90 ^ { \circ } - 45 ^ { \circ } = 45 ^ { \circ } , \alpha = 90 ^ { \circ } - 60 ^ { \circ } = 30 ^ { \circ } $$ Then the refractive index is $$ n = \frac { \sin \theta } { \sin \alpha } = \sqrt { 2 } $$

Question 5: 6. The following statements about total reflection are correct ()

6. The following statements about total reflection are correct ()

  • A. A. Total reflection of light from a light-dense medium to a light-sparse medium is not possible.
  • B. B. Total reflection may occur when light is directed from a light-sparse medium to a light-dense medium.
  • C. C. Total reflection of light from a medium with a large refractive index to a medium with a small refractive index is not possible.
  • D. D. Total reflection may occur when light is directed from a medium with a small velocity of propagation to a medium with a large velocity of propagation.

Answer: D

Solution: The condition for total reflection is that light is directed from a light-dense medium to a light-sparse medium, and the medium with a large refractive index relative to the small refractive index of the medium is a light-dense medium, the same kind of light propagation in different media, propagation speed of the medium is small relative to the propagation speed of the medium is light-dense medium.

Question 6: 7. The rising bubbles in the fish tank is bright, at this moment the light to the bubbles in the wat...

7. The rising bubbles in the fish tank is bright, at this moment the light to the bubbles in the water on the light path may be correct ( )

  • A. A. ![](/images/questions/phys-geometric-optics/image-001.jpg)
  • B. B. ![](/images/questions/phys-geometric-optics/image-002.jpg)
  • C. C. ![](/images/questions/phys-geometric-optics/image-003.jpg)
  • D. D. ![](/images/questions/phys-geometric-optics/image-004.jpg)

Answer: A

Solution: Light rays are directed from a light-dense medium into a light-sparse medium, and according to the law of refraction $n = \frac { \sin i } { \sin r }$, it is known that the light rays are at a greater angle to the normal in the bubble, and total reflection may occur when the angle of incidence is greater.

Question 7: 8. The following statements about the contributions of physicists are correct ( )

8. The following statements about the contributions of physicists are correct ( )

  • A. A. Faraday introduced the concept of fields
  • B. B. Coulomb's earliest measured values of meta-charge
  • C. C. Thomas Young summarized the law of refraction of light after analyzing a large amount of experimental data
  • D. D. Auster proposed to study magnetic fields by visualizing magnetic lines of induction

Answer: A

Solution: A. Faraday proposed the concept of field, so A is correct; B.Milligan first measured the value of meta-charge, so B is wrong; C. Snell in the analysis of a large number of experimental data, summarized the law of refraction of light, so C error; D. Faraday proposed to visualize the magnetic field through the image of magnetic lines to study, so D error.

Question 8: 9. A prism top angle $\theta = 41.30 ^ { \circ }$, a beam of white light with a large angle of incid...

9. A prism top angle $\theta = 41.30 ^ { \circ }$, a beam of white light with a large angle of incidence from one side of the prism into the prism, through the prism from the other side of the shot, in the formation of the light screen from the red to violet colored light band (as shown). When the incident angle $i$ gradually reduced to zero in the process of colored light band changes are: ( ) ![](/images/questions/phys-geometric-optics/image-005.jpg) | Color Light | Purple | Blue | Green | Yellow | Orange | Red | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | Refractive Index | 1.532 | 1.528 | 1.519 | 1.517 | 1.514 | 1.513 | | Critical Angle | $40.75 ^ { \circ }$ | $40.88 ^ { \circ }$ | $41.17 ^ { \circ }$ | $41.23 ^ { \circ }$ | $41.34 ^ { \circ }$ | $41.37 ^ { \circ }$ | [INLINE_FORMULA_7]]

  • A. A. The violet light was the first to disappear, and finally only the orange and red light remained
  • B. B. The violet light was the first to disappear, and finally only yellow, orange and red light remained
  • C. C. The red light was the first to disappear, and finally only violet, indigo and blue light remained
  • D. D. The red light was the first to disappear, and finally only violet, indigo, blue and green light remained

Answer: A

Solution: From the table data, the refractive index of violet light is the largest, the critical angle is the smallest, when the incident angle $\theta _ { 1 }$ gradually reduced to zero in the process, the refractive angle decreases, the angle of incidence of the light to the right side of the prism increases, the violet light's angle of incidence is the first to reach the critical angle, the total reflection occurs, and it disappears first; when the incident angle $\theta _ { 1 }$ When the angle of incidence $\theta _ { 1 }$ decreases to zero, the angle of incidence of the light to the right side of the prism is equal to $\alpha = \theta = 41.30 ^ { \circ }$, which is smaller than the critical angle of the red light and the orange light, so the two kinds of light do not undergo total reflection and can still be shot on the light screen. Therefore, only red and orange light remain on the light screen. Therefore, A is correct, BCD is wrong.

Question 9: 10. It is known that the refractive index of water, crystal, glass and carbon disulfide are $1.33 , ...

10. It is known that the refractive index of water, crystal, glass and carbon disulfide are $1.33 , 1.55 , 1.60$ and 1.63 respectively, if the light propagates according to the following ways, total reflection may occur ( )

  • A. A. Shooting from crystal into glass
  • B. B. Carbon disulfide injection from water
  • C. C. From the glass into the water
  • D. D. From water into crystal

Answer: C

Solution: Light only from a light-dense medium into a light-sparse medium can occur total reflection. A. Light from a crystal into a glass is from a light-sparse medium into a light-dense medium, and total reflection is not possible, so option A does not fit the question; B. Light from water into carbon disulfide is from a sparse medium into a dense medium, total reflection is not possible, so option B does not fit the question; C. Light from the glass into the water, is light from the light-dense medium into the light-sparse medium, it is possible to occur total reflection, so option C is consistent with the meaning of the question; D light from the water into the crystal, is light from the light sparse medium into the light dense medium, it is not possible to occur total reflection, so option D does not fit the question.

Question 10: 11. As shown in the figure, the ice sculpture in the ice world shines with various colors, Xia Xia s...

11. As shown in the figure, the ice sculpture in the ice world shines with various colors, Xia Xia students found that the ice surface is equipped with various colors under the lights, the ice is also distributed in the bubble, then ( ) ![](/images/questions/phys-geometric-optics/image-006.jpg)

  • A. A. Red light in a vacuum has a shorter wavelength than green light
  • B. B. Red light travels faster than green light in an ice cube
  • C. C. Lights below the ice surface that appear to be deeper than their actual location
  • D. D. The bubbles in the ice appear brighter because of total reflection when light is incident on the ice from inside the bubbles.

Answer: B

Solution: A. Based on $$ c = \lambda f $$ In a vacuum, the wave speeds of red and green light are the same, and the frequency of green light is large and the wavelength is short, so A is wrong; B. According to $$ v = \frac { c } { n } $$ Red light is smaller than the frequency of green light, the refractive index is small, the wave speed is large, so B is correct; C.The light from the lamp below the ice surface passes through the ice first and then enters the air, and the angle of refraction in the air is greater than the angle of incidence in the ice, and the light path is shown in the figure below, along the reverse extension of the light path in the air, the position of the lamp is seen to be shallower than the actual position; ![](/images/questions/phys-geometric-optics/image-007.jpg) Therefore, C is wrong ; D. Ice bubbles look brighter is due to the light incident from the outside of the bubble to the air bubble and ice at the interface of total reflection occurs in the formation of light from the air to the ice can not occur total reflection, so D error.

Question 11: 12. As shown in the figure, a thin beam of monochromatic light shoots from the $P$ point on a semici...

12. As shown in the figure, a thin beam of monochromatic light shoots from the $P$ point on a semicircular glass block to the center of the circle $O$ in the direction of radius, just at the glass The total reflection occurs on the bottom surface of the glass block and then shoots out from the glass block, the reflection angle of the monochromatic light on the lower surface of the glass block is $53 ^ { \circ }$, take $\sin 53 ^ { \circ } = \frac { 4 } { 5 }$, the refractive index of the glass block for the monochromatic light is ( ) ![](/images/questions/phys-geometric-optics/image-008.jpg)

  • A. A. $\frac { 4 } { 3 }$
  • B. B. $\frac { 5 } { 3 }$
  • C. C. $\frac { 5 } { 4 }$
  • D. D. $\frac { 3 } { 5 }$

Answer: C

Solution: According to the refractive index and the critical condition for total reflection, we can see that $$ n = \frac { 1 } { \sin C } = \frac { 1 } { \sin 53 ^ { \circ } } = \frac { 5 } { 4 } $$

Question 12: 13. A prism has the cross-section shown in the figure and a reflective coating on the $\angle C = 90...

13. A prism has the cross-section shown in the figure and a reflective coating on the $\angle C = 90 ^ { \circ } , \angle B = 60 ^ { \circ } , A C$ side. A fine beam of light enters the prism from the $O$ side perpendicular to the $A B$ side, and is reflected by the $A C$ side into the $B C$ side, where it undergoes total reflection in the $B C$ side. The refractive index of the prism is then the refractive index of the prism. Then the refractive index of the prism is ![](/images/questions/phys-geometric-optics/image-009.jpg)

  • A. A. 2
  • B. B. $\sqrt { 2 }$
  • C. C. $\frac { 2 \sqrt { 3 } } { 3 }$
  • D. D. $\sqrt { 3 }$

Answer: C

Solution: Make a light path diagram as shown ![](/images/questions/phys-geometric-optics/image-010.jpg) According to the geometric relationship, the angle of incidence of the light ray reaching the $A C$ plane is $\alpha = 30 ^ { \circ }$, then the angle of reflection is $30 ^ { \circ }$, and according to the geometric relationship, the angle of incidence of the light ray in the $B C$ plane is , since the ray of light happens to undergo total reflection in the $B C$ plane, then $$ n = \frac { 1 } { \sin i } = \frac { 2 \sqrt { 3 } } { 3 } $$

Question 13: 14. As shown in the figure for a transparent medium made of prism cross-section, this cross-section ...

14. As shown in the figure for a transparent medium made of prism cross-section, this cross-section is a radius $R$ semicircle, $O$ for the semicircle of the center of the circle, the diameter of the $M N$ along the straight direction. A beam of monochromatic light enters the medium at an angle of incidence $\alpha$ from a point $A$ on the diameter $M N$ in the paper plane, and enters the medium from a point $B$ on the arcuate side of the circle at a level equal to the center $O$ of the circle. FORMULA_7]] point on the circular curved surface at a height equal to the center of the $B$ point first shot out of the prism. The speed of propagation of this monochromatic light in a vacuum is known to be $c , O A = \frac { \sqrt { 3 } } { 3 } R$, and the time of propagation of the light from $A$ to $B$ is $\frac { 2 R } { C }$, then the following statement is correct ( ) ![](/images/questions/phys-geometric-optics/image-011.jpg)

  • A. A. The speed of propagation of this monochromatic light in the medium is $\frac { C } { 2 }$
  • B. B. The sine of the angle of refraction of the beam at point $A$ is $\frac { \sqrt { 3 } } { 3 }$
  • C. C. The angle of incidence of this monochromatic light at point $A$ is equal to $60 ^ { \circ }$
  • D. D. The refractive index of a transparent medium for this monochromatic light is 2

Answer: C

Solution: A. According to the geometric relationship $$ A B = \sqrt { R ^ { 2 } + \left( \frac { \sqrt { 3 } } { 3 } R \right) ^ { 2 } } = \frac { 2 \sqrt { 3 } } { 3 } R $$ Then the speed of propagation of this monochromatic light in the medium is $$ v = \frac { A B } { t } = \frac { \frac { 2 \sqrt { 3 } } { 3 } R } { \frac { 2 R } { c } } = \frac { \sqrt { 3 } } { 3 } c $$ Therefore, A is wrong; B. According to the geometrical relationship, the sine of the angle of refraction of the beam at $A$ point is $$ \sin r = \frac { O A } { A B } = \frac { 1 } { 2 } $$ Therefore, B is wrong; CD. According to $$ v = \frac { c } { n } = \frac { \sqrt { 3 } } { 3 } c $$ the refractive index can be obtained as $$ n = \sqrt { 3 } $$ According to the law of refraction it is obtained $$ n = \frac { \sin \alpha } { \sin r } $$ can be obtained $$ \sin \alpha = \frac { \sqrt { 3 } } { 2 } $$ Then the angle of incidence of this monochromatic light at the point $A$ is $$ \alpha = 60 ^ { \circ } $$ Therefore, C is correct and D is wrong.

Question 14: 15. A "light switch" as shown in the dotted area of the figure, its main components by two very clos...

15. A "light switch" as shown in the dotted area of the figure, its main components by two very close to the cross-section of the semi-circular cylindrical prisms, two half-cylindrical prisms can be rotated around the center of the $O$ point. Monochromatic light $a$ is directed from the left side along the radius to the center of the half-cylindrical prisms $O$, and if the light can come out from the right side, it is "on", otherwise it is "off". The refractive index of a prism to $a$ is known to be 1.5, and the angle between the light $a$ and the diameter of the semi-cylindrical prism $M N$ is $45 ^ { \circ }$. The following statement is correct ( ) ![](/images/questions/phys-geometric-optics/image-012.jpg)

  • A. A. Monochromatic light $a$ has 1.5 times the frequency in a prism than in a vacuum.
  • B. B. Monochromatic light $a$ has a wavelength in a prism that is 1.5 times greater than in a vacuum.
  • C. C. Clockwise rotation of the two halves of the cylindrical prism enables the "on" function.
  • D. D. Rotate both halves of the cylindrical prism counterclockwise to realize the "open" function.

Answer: D

Solution: A. The frequency is determined by the light source, which does not change the frequency, so A is wrong; B. According to $$ n = \frac { c } { v } , v = \lambda f $$ B. According to $$ n = \frac { c } { v } , v = \lambda f $$, the wavelength of monochromatic light $a$ in a vacuum is 1.5 times that in a prism, so B is wrong; C. At this time, the angle of incidence is $45 ^ { \circ }$, according to the meaning of the question can be seen $$ \sin C = \frac { 1 } { n } = \frac { 2 } { 3 } < \frac { \sqrt { 2 } } { 2 } $$ Critical angle is less than $45 ^ { \circ }$, so this time for the off state, clockwise rotation of the two halves of the cylindrical prism, the angle of incidence becomes larger, then the total reflection occurs, the "off" function, so C error; D. According to the analysis of the C option can be seen, counterclockwise rotation of the two halves of the cylindrical prism, the angle of incidence decreases, total reflection does not occur, can realize the "open" function, so D is correct.

Question 15: 16. As shown in the figure, a beam of complex-color natural light refracted through a glass prism, i...

16. As shown in the figure, a beam of complex-color natural light refracted through a glass prism, in the firm straight placed on the light screen received $a , b , c$ three monochromatic light, the following statements are correct ( ) ![](/images/questions/phys-geometric-optics/image-013.jpg)

  • A. A. $a , b , c$ Three monochromatic colors remain natural light
  • B. B. $a$ The wavelength of light is maximized when it propagates through a glass prism
  • C. C. If the photoelectric effect occurs when a metal is irradiated with $b$ light, then the photoelectric effect also occurs when the metal is irradiated with $c$ light.
  • D. D. If $b$ light coming from the end face of a bent glass rod at a certain angle of incidence happens to come out from the other end face, then let the beam of complex-colored light be directed to the end face of the rod at the same angle of incidence and then there will be two colors of light coming out from the other end face

Answer: D

Solution: A. Complex natural light refracted through a prism becomes monochromatic light, no longer natural light, is polarized light, so A error; B. $a$ light has the greatest degree of polarization, and it can be known that the refractive index of $a$ light is the greatest, then the frequency is the greatest, and according to ${ } ^ { c = \lambda f }$, it can be known that the wavelength of $a$ light is the shortest, so it's B wrong; C. The greater the frequency of light, the higher the energy, $a , b , c$ three monochromatic light, $a$ light frequency is the largest, $c$ light frequency is the smallest, if $b$ light irradiation of a metal can occur photoelectric effect, then use $b$ light irradiation of a metal can occur photoelectric effect, then use $a$ light frequency of the shortest. INLINE_FORMULA_7]] light can produce photoelectric effect, then $c$ light irradiation of the metal may not be able to produce photoelectric effect, so C is wrong; D. Total reflection occurs when the angle of incidence is greater than or equal to the critical angle, and according to the condition of total reflection $\sin C = \frac { 1 } { n }$, it can be seen that if $b$ light from the end face of a bent glass rod is projected into the rod with a certain angle of incidence and can be precisely projected out from the other end face, then the light can be projected into the end face of the rod at the same incident angle, and then the $b$ light will not be projected out from the other end face. INLINE_FORMULA_11]] has the greatest frequency, the greatest refractive index, the smallest critical angle for total reflection, and is more likely to undergo total reflection than $b$ and can be ejected; $c$ has the smallest frequency, the smallest refractive index, the greatest critical angle for total reflection, and, if the angle of incidence remains unchanged, it It cannot be ejected from the other side, so there will be two colors of light ejected from the other side, so D is correct.

Question 16: 17. With a beam of laser light directed obliquely into the liquid surface, the incident light, refle...

17. With a beam of laser light directed obliquely into the liquid surface, the incident light, reflected and refracted light as shown in the figure, through the measurement of the incoming Projected light and the liquid surface into $37 ^ { \circ }$, reflected light and refracted light perpendicular to each other, then the refractive index of the liquid is ( ) ![](/images/questions/phys-geometric-optics/image-014.jpg)

  • A. A. $\frac { 4 } { 3 }$
  • B. B. $\frac { 3 } { 4 }$
  • C. C. $\frac { 5 } { 4 }$
  • D. D. $\frac { 5 } { 3 }$

Answer: A

Solution: According to the question, the angle of incidence $i = 90 ^ { \circ } - 37 ^ { \circ } = 53 ^ { \circ }$ and the angle of refraction $r = 37 ^ { \circ }$, so the refractive index of the medium is $n = \frac { \sin i } { \sin r } = \frac { \sin 53 ^ { \circ } } { \sin 37 ^ { \circ } } = \frac { 4 } { 3 }$

Question 17: 18. There is a square pool filled with water, at the bottom of the central installation can be seen ...

18. There is a square pool filled with water, at the bottom of the central installation can be seen as a point light source of colored lights, colored lights can be alternately emitted a, b two monochromatic light, if a color light flashes on the water surface illuminated area is exactly tangent to the square water surface of the circle, the b-color light flashes can illuminate the entire square water surface area, do not take into account the multiple reflections of the light, then the a, b two kinds of color light in the water the propagation rate of the ratio is ( )

  • A. A. $2 : \sqrt { 3 }$
  • B. B. $\sqrt { 3 } : 2$
  • C. C. $\sqrt { 3 } : \sqrt { 5 }$
  • D. D. $\sqrt { 5 } : \sqrt { 3 }$

Answer: C

Solution: As shown in the figure, the length of the square is $2 a$, and the monochromatic light emitted by the point source at the bottom center of the pool at $O$ indicates that the light ${ } ^ { O O _ { 2 } }$ is totally reflected at ${ } ^ { O }$, and that the light ${ } ^ { O }$ is totally reflected at ${ } ^ { O }$. As shown in the figure. ![](/images/questions/phys-geometric-optics/image-015.jpg) At this point $\angle O O _ { 2 } O _ { 1 }$ is equal to the critical angle $C _ { 1 }$, then there is $\sin C _ { 1 } = \frac { a } { \sqrt { a ^ { 2 } + 4 a ^ { 2 } } } = \frac { 1 } { \sqrt { 5 } }$ When the point light source emits B monochromatic light, it can illuminate the whole water surface, indicating that the light ${ } ^ { O O _ { 4 } }$ is fully emitted at ${ } ^ { O _ { 4 } }$, then we have ${ } ^ { O _ { 4 } }$. Then we have $\sin C _ { 2 } = \frac { \sqrt { 2 } a } { \sqrt { ( \sqrt { 2 } a ) ^ { 2 } + 4 a ^ { 2 } } } = \frac { 1 } { \sqrt { 3 } }$ Combined with the full emission critical condition $\sin C = \frac { 1 } { n }$ The refractive index is $n = \frac { c } { v }$ The solution is $V _ { \text {甲 } } : V _ { 乙 } = \sqrt { 3 } : \sqrt { 5 }$ The ratio of the speed of propagation of color A and B light in water is ${ } ^ { \sqrt { 3 } } : \sqrt { 5 }$.

Question 18: 19. A, B two monochromatic light respectively through the same double-slit interference device to ob...

19. A, B two monochromatic light respectively through the same double-slit interference device to obtain their respective interference patterns, set the distance between the centers of the two adjacent bright fringes for $\Delta x$, if $\Delta x _ { \text {甲 } } > \Delta x _ { \text {乙 } }$, comparing the A, B two kinds of monochromatic light, then the following statements are correct ( )

  • A. A. A light in a vacuum has a shorter wavelength
  • B. B. A-light has less photon energy
  • C. C. The critical angle at which total reflection of A light occurs is smaller
  • D. D. A light can be polarized while B light cannot

Answer: B

Solution: A. According to the double-slit interference fringe spacing $\Delta x = \frac { L } { d } \lambda$, we can know that for the same experimental setup, the longer the wavelength, the larger the fringe spacing, and ${ } ^ { \Delta x _ { \text {甲 } } > \Delta x _ { \text {乙 } } }$ we can know that the wavelength of light A must be larger than the wavelength of light B. Therefore, A is wrong. B. According to $C = \lambda y$, we can know that the frequency of light A is smaller than that of light B, and the energy of photons is $E = h y$, so the energy of photons in light A must be smaller than that of photons in light B. Therefore, B is correct. C.By $\Delta x _ { \text {甲 } } > \Delta x _ { \text {乙 } }$ we know that the wavelength of light A must be greater than the wavelength of light B, then the same uniform medium, the refractive index of light A is less than the refractive index of light B, by the critical angle formula $\sin C = \frac { 1 } { n }$ light A occurs in total reflection of the critical angle is greater than the critical angle of light B occurs in total reflection of the critical angle. Therefore, C is wrong. D.Polarization is a unique phenomenon of transverse waves, and since light is a transverse wave, both A and B can be polarized. Therefore, D is wrong. 20.A Knowledge]The law of refraction, the refraction phenomenon of light [Detailed explanation] The principle of small hole imaging is that light travels in a straight line. A. "Hand shadow" game, the formation of the shadow is the propagation of light along the straight line, so A correct; B. water "folding chopsticks", is light from the water into the air, refraction, so B error; C. Mountain "reflection", belonging to the plane mirror imaging, the principle is the reflection of light, so C error; D. After the rain "rainbow", is the phenomenon of light dispersion, the principle is the refraction of light, so D error.

Question 19: 21. A beam of complex color light along the radius direction to a semicircular glass block, refracti...

21. A beam of complex color light along the radius direction to a semicircular glass block, refraction and divided into $a , b$ two beams of monochromatic light, the direction of propagation as shown in the figure. The following statements are correct ![](/images/questions/phys-geometric-optics/image-016.jpg)

  • A. A. The refractive index relationship of a glass block to $a , b$ is $n _ { a } < n _ { b }$
  • B. B. $a , b$ propagation velocity relationship in glass is $v _ { a } > v _ { b }$
  • C. C. Monochromatic light $a$ The critical angle of total reflection from glass to air is less than the critical angle of total reflection from glass to air for monochromatic light $b$.
  • D. D. From the double-slit interference fringe width formula: $\Delta x = \lambda \frac { L } { d }$, it can be seen that the spacing of the $a$ light interference fringes in an experiment with the same double-slit interferometer is wider than the $b$ light width, and the width of the interference fringes in an experiment with the same double-slit interferometer is wider than the $b$ light width.

Answer: C

Solution: A. From the graph, we know that $a$ light has a large deflection and a large refractive index, i.e. $$ \mathrm { n } _ { \mathrm { a } } > \mathrm { n } _ { \mathrm { b } } $$ Therefore, A is wrong; B. According to the propagation speed $v = \frac { c } { n }$, the refractive index of the wave speed is small, that is, $$ \mathrm { n } _ { \mathrm { a } } > \mathrm { n } _ { \mathrm { b } } $$. $$ \mathrm { v } _ { \mathrm { a } } < \mathrm { v } _ { \mathrm { b } } $$ Therefore, B is wrong; C. According to the critical angle formula $\sin C = \frac { 1 } { n }$, the critical angle of large refractive index is small, and the critical angle of $a$ is small, so C is correct; D. Because $a$ has a large refractive index and a large frequency, the wavelength is small, so $\Delta x = \lambda \frac { L } { d }$ can be known to use the same double-slit interferometer experiments with $a$ light interference fringes are narrower than the $b$ light, so the D D is wrong.

Question 20: 22. As shown in the figure, a rectangular transparent container filled with some kind of transparent...

22. As shown in the figure, a rectangular transparent container filled with some kind of transparent liquid, there is a round bubble in the liquid (air inside the bubble), a fine beam of light shot perpendicular to the left wall into the container, the fine beam of light in the bubble is divided into two beams of color light $M , N$, the following statements are correct ![](/images/questions/phys-geometric-optics/image-017.jpg)

  • A. A. The refractive index of liquid to $M$ light is greater than that to $N$ light.
  • B. B. $N$ Light travels faster in liquids than $M$ Light travels faster in liquids
  • C. C. Two colors of light pass through the same double-slit interference device, and the stripes of $N$ light are wider apart
  • D. D. $N$ The frequency of light is greater than $M$ The frequency of light is greater than $M$.

Answer: D

Solution: A. According to the figure, the degree of refraction of $N$ light is greater than that of $M$ light, then the refractive index of the liquid to $M$ light is smaller than that to $N$ light, so A is wrong; According to $$ v = \frac { c } { n } $$ B. Since the refractive index of a liquid for $M$ light is smaller than that for $N$ light, the propagation speed of $N$ light in a liquid is smaller than that of $M$ light in a liquid, and therefore B is wrong; CD. Since the refractive index of the liquid for $M$ light is smaller than that for $N$ light, the frequency of $N$ light is greater than that of $M$ light. FORMULA_14]] the wavelength of light is smaller than the wavelength of $M$ light. $$ \Delta x = \frac { L } { d } \lambda $$ According to $$ \Delta x = \frac { L } { d } \lambda $$, two colors of light pass through the same double-slit interference device, and the $N$ light has a narrower spacing between the stripes, so C is wrong and D is correct.

Question 21: 23. Conventional materials have a positive refractive index ($n > 0$). There are artificial material...

23. Conventional materials have a positive refractive index ($n > 0$). There are artificial materials designed for certain electromagnetic waves that can have a negative refractive index ($n < 0$), called negative refractive index materials. In the case of such materials in air, the angle of incidence $i$ and the angle of refraction still satisfy $\frac { \sin i } { \sin r } = n$, but the refracted rays are located on the same side of the normal line as the incident rays (the angle of refraction takes a negative value in this case). Now the air has a parallel upper and lower surfaces of the negative refractive index material, a beam of electromagnetic waves from its upper surface into the lower surface of the shot, if the material on the electromagnetic wave of the refractive index ${ } ^ { n = - \sqrt { 2 } }$, can correctly reflect the electromagnetic wave propagation path through the material schematic is ( )

  • A. A. ![](/images/questions/phys-geometric-optics/image-005.jpg)
  • B. B. ![](/images/questions/phys-geometric-optics/image-006.jpg)
  • C. C. ![](/images/questions/phys-geometric-optics/image-007.jpg)
  • D. D. ![](/images/questions/phys-geometric-optics/image-008.jpg)

Answer: C

Solution: A. The question gives the information that "the refracted light and the incident light are on the same side of the normal line", so A is wrong. BCD.By the law of refraction of light, the refractive index is $$ n = | - \sqrt { 2 } | = \frac { \sin \alpha } { \sin \beta } $$ The refractive index is $$ n = | - \sqrt { 2 } | = \frac { \sin \alpha } { \sin \beta } $$ on the same side of the angle of refraction is less than the angle of incidence, so C is correct, BD is wrong.

Question 22: 24. A rainbow is formed when sunlight enters a water droplet, is refracted once, then reflected at t...

24. A rainbow is formed when sunlight enters a water droplet, is refracted once, then reflected at the back of the droplet, and finally refracted again when it leaves the droplet. The following figure for the formation of the rainbow schematic, a beam of white light L from the left into the droplet, $a , b$ is the white light shooting $a , b$ is a white light into the water drop after a reflection and refraction of two monochromatic light lines. The [INLINE_FORMULA_0]] ![](/images/questions/phys-geometric-optics/image-018.jpg)

  • A. A. $a$ The frequency of light is less than the frequency of $b$ light.
  • B. B. $a , b$ The wavelength of light becomes longer as it passes from air into a water droplet.
  • C. C. From the same medium to air, $a$ light is more prone to total reflection than $b$ light.
  • D. D. Through the same double-slit interference device, the neighboring fringe spacing of $a$ light is larger than that of $b$ light

Answer: C

Solution: A. From the light path diagram, we can see that the degree of refraction of $a$ light is greater than that of $b$ light, and it can be seen that the refractive index of the water droplet is greater than that of $a$ light, so the frequency of $a$ light is greater than that of $b$ light, so A is wrong; A. From the light path diagram, we can see that the degree of refraction of $a$ light is greater than that of $b$ light. FORMULA_4]] is greater than the frequency of $b$ light, so A is wrong; B. $a , b$ light has a smaller wave speed and the same frequency when it enters a water droplet from air, according to $\lambda = \frac { v } { f }$. According to $\lambda = \frac { v } { f }$, the wavelength becomes shorter, so B is wrong; C. According to $\sin C = \frac { 1 } { n }$, the critical angle of ${ } _ { a }$ light is smaller than that of ${ } _ { b }$ light, that is, ${ } _ { a }$ light is easier to be totally reflected than ${ } _ { b }$ light. Total reflection occurs more easily than ${ } _ { b }$ light; D. Since the frequency of ${ } _ { a }$ is greater than that of ${ } _ { b }$, the wavelength of ${ } _ { a }$ is smaller than that of ${ } _ { b }$. _17]], the spacing between neighboring fringes of $a$ light interfered by the same double-slit interferometer is smaller than that of $b$ light, so D is wrong.

Question 23: 25. The following statements are correct

25. The following statements are correct

  • A. A. Light from a medium into another medium, light at the interface to change the direction of propagation of the phenomenon called light refraction
  • B. B. The refractive index is proportional to the sine of the angle of refraction.
  • C. C. In water, blue light travels faster than red light
  • D. D. Total reflection can occur if the angle of incidence is large enough

Answer: A

Solution: A. light from a medium into another medium, light at the interface to change the direction of propagation phenomenon called light refraction, so A is correct; B. The refractive index is related to the medium, and the angle of incidence and refraction angle has nothing to do, so B error; C. The refractive index of blue light is large, according to $n = \frac { c } { v }$ can be known, blue light propagation speed in water is smaller than the red light, so C error; D. light only from the light-dense medium to the light-sparse medium, and the angle of incidence is greater than the critical angle, total reflection can occur. Light incident from a light-sparse medium to a light-dense medium, regardless of the angle of incidence, will not occur total reflection, so D error.

Question 24: 26. Coated with a translucent film of glass as a wall instead of tinted glass, yellow and green ligh...

26. Coated with a translucent film of glass as a wall instead of tinted glass, yellow and green light can enter the room unimpeded, so that people inside the building can be a glimpse of the outside scene, while people outside can only be seen from the blue-violet reflected light of the sky and clouds, on the inside of the building is difficult to peek at the situation. About this phenomenon, the following statements are correct

  • A. A. The phenomenon is related to the interference of light
  • B. B. The phenomenon has the same cause as a mirage
  • C. C. The phenomenon is due to total reflection of blue-violet light
  • D. D. This phenomenon has the same cause as Poisson's bright spots

Answer: A

Solution: A. The working principle of the permeable film is the interference phenomenon of light. Light has wave-particle duality, that is, light waves and mechanical waves can be interfered with, in front of the lens coated with a layer of permeable film (generally calcium fluoride), when the light irradiated on this film, if the thickness of the film is equal to the incident light in the permeable film in a quarter of the wavelength, then the light reflected in the front and rear surfaces of the film will be interfered with the light to cancel each other out and reduce the reflected light, A correct; B. Mirage is due to air inhomogeneity, light in inhomogeneous media does not propagate in a straight line, the occurrence of light refraction and total reflection caused by, B error; C. Total reflection occurs when the conditions are light from a light-dense medium into a light-sparse medium, so the air (light-sparse medium) into the glass (light-dense medium) of blue and violet light can not occur total reflection, C error; D. Poisson bright spot is due to light diffraction and an optical phenomenon, is when monochromatic light irradiation diameter is less than or equal to the wavelength of the light source of the opaque plate, there will be a ring of light and dark diffraction stripes on the light screen after the plate, and in the center of the circle there will be a smaller bright spot, D error.

Question 25: 27. As shown in the figure, the correct statement about the following picture is ![](/images/questi...

27. As shown in the figure, the correct statement about the following picture is ![](/images/questions/phys-geometric-optics/image-019.jpg) Figure A ![](/images/questions/phys-geometric-optics/image-020.jpg) Figure B ![](/images/questions/phys-geometric-optics/image-021.jpg) Figure C ![](/images/questions/phys-geometric-optics/image-022.jpg) Figure D

  • A. A. As shown in Figure A, $3 D$ the principle of stereoscopic movies is the phenomenon of light polarization
  • B. B. As shown in Figure B, the colored stripes on the soap film are the phenomenon of light refraction
  • C. C. As shown in Figure C, the chopsticks appear to "break" because the total reflection phenomenon occurs.
  • D. D. As shown in Figure D, the principle of optical fibers is to use the interference phenomenon of light

Answer: A

Solution: A. Stereoscopic film is the use of the principle of polarization of light to see stereoscopic films need to wear special glasses, the glasses lens is made of two polarizers, they allow the vibration of light is not the same direction, A correct; B. Soap film on the colored stripes is the soap film on the two surfaces of the light reflected back to the results of interference, B error; C. Chopsticks appear "broken" phenomenon is due to the occurrence of light refraction, C error; D. The principle of optical fiber is to use the principle of total reflection of light, D error.

Question 26: 28. With respect to red and violet light, the following statements are correct ( Total reflectio...

28. With respect to red and violet light, the following statements are correct ( Total reflection must also occur

  • A. A. The frequency of red light is greater than the frequency of violet light
  • B. B. The speed of red light is less than the speed of violet light in the same glass
  • C. C. Red and violet light travel at equal speeds in a vacuum.
  • D. D. When red light and violet light are directed from glass into air at the same angle of incidence, if the violet light can just undergo total reflection, the red light

Answer: C

Solution: The wavelength of red light is greater than the wavelength of violet light, the longer the wavelength, the smaller the frequency, so the frequency of red light is smaller than the frequency of violet light, A error; the frequency of red light is small, the refractive index is small, according to $v = \frac { c } { n }$, the speed of propagation of red light in the glass block is large, according to $\sin C = \frac { 1 } { n }$ know, violet light is large in frequency, the critical angle is small, the violet light is just able to Total reflection occurs, the red light will not occur total reflection, BD error; any color of light in a vacuum propagation speed is equal to the speed of light, C correct.

Question 27: 29. In an experiment to observe total reflection of light, a laser pointer is used to illuminate the...

29. In an experiment to observe total reflection of light, a laser pointer is used to illuminate the center of a semicircular glass block at point $O$ from point $A$, and it is found that there are two thin beams of light at point $O B , O C$. When the incident beam $A O$ is turned clockwise by a small angle $\theta , O B , O C$ around the point $O$, it is also turned. Then ![](/images/questions/phys-geometric-optics/image-023.jpg)

  • A. A. Beam $O B$ rotates clockwise more than $\theta$
  • B. B. Beam $O C$ rotates counterclockwise by an angle less than $\theta$
  • C. C. Beam $O B$ Gradually brightening
  • D. D. Beam $O C$ Gradual dimming

Answer: A

Solution: A. From the figure, $O B$ is a refracted light ray, light from a light-dense medium to a light-sparse medium, the angle of incidence is less than the angle of refraction, when the incident beam $A O$ rotates clockwise around the $O$ point, $O B$ is refracted. The light also turns clockwise, but faster, so the beam $O B$ turns clockwise by an angle greater than $\theta$, A is correct; B. From the figure, we can see that the beam $O C$ is a reflected light ray, and the direction of rotation is opposite to the incident light ray, but the rotation speed is the same, so the angle of rotation of the beam $O C$ is equal to the angle of rotation of the beam $\theta , \mathrm { B }$, which is wrong; CD. When the incident light $A O$ rotates to a certain degree, the angle of incidence is equal to the critical angle, total reflection occurs, and the refracted rays will disappear, and in the process, the refracted rays $O B$ gradually become darker, and the reflected rays $O C$ gradually become brighter, CD is wrong. CD Error.

Question 28: 30. As shown in the figure, light from the $A$ point into the circular glass, and from the $B$ point...

30. As shown in the figure, light from the $A$ point into the circular glass, and from the $B$ point out, if the incident light relative to the deflection angle of the incident ray of $30 ^ { \circ } , \mathrm { AB }$ arc on the angle of the center of the circle for the $120 ^ { \circ }$. The following statement is correct () ![](/images/questions/phys-geometric-optics/image-024.jpg)

  • A. A. The refractive index of glass is ${ } ^ { \sqrt { 2 } }$
  • B. B. The refractive index of glass is $\sqrt { 3 }$
  • C. C. The angle of incidence of light at point A is $105 ^ { \circ }$
  • D. D. The refractive index of glass is $\frac { \sqrt { 2 } + \sqrt { 6 } } { 2 }$

Answer: A

Solution:

Question 29: 31. A beam of complex color light consisting of $a , b$ two monochromatic light is shot into the air...

31. A beam of complex color light consisting of $a , b$ two monochromatic light is shot into the air from the water and divided into two beams of $a , b$, the following is correct about the $a , b$ two beams of monochromatic light ( ) ![](/images/questions/phys-geometric-optics/image-025.jpg)

  • A. A. If the angle of incidence of the incident light SA in the water is increasing, the $a$ light undergoes total reflection first.
  • B. B. $a$ frequency is smaller than $b$ optical frequency.
  • C. C. $a$ Light in water has a greater frequency than $a$ light in air.
  • D. D. In water $a$ light travels at a smaller speed than $b$ light

Answer: B

Solution: From the question, the angle of incidence i of the two beams is the same, the angle of refraction $\mathrm { r } _ { \mathrm { a } } < \mathrm { r } _ { \mathrm { b } }$, and according to the law of refraction, we know that the refractive index $\mathrm { n } _ { \mathrm { a } } < \mathrm { n } _ { \mathrm { b } }$ is $\sin C = 1 / n$. From $\sin C = 1 / n$ we know that the critical angle of $a$ light is larger. If the angle of incidence of $S A$ in the water is increasing, the angle of incidence reaches the critical angle of $b$ first, and then total reflection of light b occurs first. If the refractive index of a light is small, then the frequency of a light is small, so B is correct. The frequency of light is determined by the light source, independent of the medium, then the frequency of a light in water is as large as the frequency of a light in air, so C is wrong. Equation $\mathrm { v } = \mathrm { c } / \mathrm { n }$ analyzes that the speed of a light in water is larger than the speed of b light. Therefore, D is wrong.

Question 30: 32. As shown in the figure, two different beams of monochromatic light $P$ and $Q$ are directed at a...

32. As shown in the figure, two different beams of monochromatic light $P$ and $Q$ are directed at a semicircular glass block, and their emitted rays are from the center of the circle at the $O$ point along the direction of the $O F$ direction, which can be seen ( ) ![](/images/questions/phys-geometric-optics/image-026.jpg)

  • A. A. The frequency of $P$ light is greater than that of $Q$ light.
  • B. B. $Q$ The time required for light to pass through this glass block is shorter than $P$ light
  • C. C. $P , Q$ Two beams of light are directed from water to air at the same angle of incidence, if $Q$ light can undergo total reflection, then $P$ light must be able to undergo total reflection as well.
  • D. D. If two beams of $P , Q$ monochromatic light are each allowed to pass through the same double-slit interference device, the $P$ light forms interference fringes with a larger spacing than that of the $Q$ light.

Answer: D

Solution: A. From the figure, $P$ light has a smaller degree of refraction, a smaller refractive index, and a smaller frequency, and A is wrong; B. From $v = \frac { C } { n }$, it can be seen that $P$ light has a greater speed of propagation in the glass, and $P$ light takes a shorter time to pass through the glass block than $Q$ light, which is B wrong; C. From $\sin C = \frac { 1 } { n }$, $P$ can be known, $P$ the critical angle for total reflection is larger, so $P , Q$ the two beams of light are shot from the water to the air at the same angle of incidence, and if $Q$ the light can be totally reflected, $Q$ the light can be totally reflected, and $P$ the light will be totally reflected, and $P$ the light will be completely reflected. INLINE_FORMULA_9]] light does not necessarily undergo total reflection, C is wrong; D. The refractive index of $P$ light is small, and the corresponding wavelength is large. It can be seen from $\Delta x = \frac { L } { d } \lambda$ that, if two beams of monochromatic light pass through the same double-slit interference device, the interference fringe spacing between the $P$ light is larger than that between the $P$ light, and the interference fringe spacing is smaller than that of the $Q$ light. The interference fringe spacing is larger than that of the $Q$ light, D is correct.

Question 31: 33. The following statements about the phenomenon of light are true

33. The following statements about the phenomenon of light are true

  • A. A. The oil film floating on the water surface is colored, which is a refraction phenomenon of light
  • B. B. Transmission-enhancing films on camera lenses, which utilize the polarization of light
  • C. C. The Poisson bright spot phenomenon is a diffraction phenomenon of light, which shows that light has volatility
  • D. D. The rainbow that appears in the sky after rain is an interference phenomenon of light

Answer: C

Solution: A. The oil film floating on the surface of the water is colored, which is a thin film interference phenomenon of light, so A is wrong; B. The camera lens on the transmittance film, the use of light interference, so B error; C. Poisson bright spot phenomenon is the diffraction of light, indicating that the light has volatility, so C is correct; D. Rainbow in the sky after the rain, is the phenomenon of light refraction, so D error.

Question 32: 34. The interference patterns obtained with monochromatic light $a , b$ of different frequencies are...

34. The interference patterns obtained with monochromatic light $a , b$ of different frequencies are shown in Figs. A and B. The following statements are not correct ![](/images/questions/phys-geometric-optics/image-027.jpg) A ![](/images/questions/phys-geometric-optics/image-028.jpg) B

  • A. A. $a$ Light is more likely to diffract significantly
  • B. B. Monochromatic $a$ light travels faster than $b$ light in water.
  • C. C. $a$ The frequency of light is greater than $b$ The frequency of light is greater than $b$.
  • D. D. The critical angle of $a$ light is greater than the critical angle of $b$ light when total reflection occurs from the same medium into air.

Answer: C

Solution: AC. From the formula for the spacing of the interference fringes, we have $$ \Delta x = \frac { l } { d } \lambda $$ Since the diagram and the formula show that the wavelength of $a$ light is greater than that of $b$ light, $a$ is more likely to diffract significantly. From the formula $$ c = \lambda v $$ It can be seen that the frequency of the $a$ light is smaller than that of the $b$ light, so A is correct and doesn't fit the question, and C is wrong and fits the question; BD . From the previous analysis, the frequency of $a$ light is smaller than that of $b$ light, so the refractive index of $a$ light is small. $$ n = \frac { c } { v } $$ It can be seen that monochromatic $a$ light travels faster than $b$ light in water. From the critical angle formula, we have $$ \sin C = \frac { 1 } { n } $$ Since the refractive index of $a$ light is small, the critical angle of $a$ light is large, i.e., when total reflection occurs when light is injected into air from the same medium, the critical angle of $a$ light is larger than that of $b$ light. The critical angle of $b$ light is greater than the critical angle of $b$ light.

Question 33: 35. As shown in the figure, a beam containing two monochromatic lights is directed along $P O$ into ...

35. As shown in the figure, a beam containing two monochromatic lights is directed along $P O$ into a semi-circular column, $O$ is the center of the semi-circle, and the refracted rays of light come out from $M , N$ at two points. The angle between $P O$ and the normal $\alpha = 45 ^ { \circ } , O M$ and the angle between $\beta = 60 ^ { \circ }$ and the interface $\beta = 60 ^ { \circ }$, the speed of light in a vacuum ${ } ^ { c } = 3 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s }$, then ![](/images/questions/phys-geometric-optics/image-029.jpg)

  • A. A. The refractive index of this medium for light from $M$ is $\sqrt { 3 }$
  • B. B. The speed of light from $M$ is $\frac { 3 \sqrt { 2 } } { 2 } \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s }$ in this medium.
  • C. C. The length of time it takes for the color light from the point $M$ to pass through this column
  • D. D. The frequency of color light emitted from point $N$ is lower than that from point $M$.

Answer: B

Solution: AB. By the law of refraction, the medium has an index of refraction for light coming from $M$. $$ n = \frac { \sin \alpha } { \sin \left( 90 ^ { \circ } - \beta \right) } = \sqrt { 2 } $$ Then the speed of light from $M$ in the medium is $$ v = \frac { c } { n } = \frac { 3 \sqrt { 2 } } { 2 } \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } $$ Therefore, A is wrong and B is correct; CD. From the figure, we know that the color light from point $N$ is deflected more, and its refractive index is larger, and from $v = \frac { c } { n }$, we know that the wave speed of the color light from point $N$ is smaller, and the distance that the color light travels in the column is equal, then the time taken for the color light from point $N$ to pass through the column is longer, and the color light with larger refractive index takes longer. FORMULA_7]], the color light from $N$ takes longer time to pass through the column, and the frequency of the color light with large refractive index is high, then the frequency of the color light from $N$ is higher than that of the color light from $M$, so CD is wrong.

Question 34: 36. A glass sphere with radius $R , O$ as the center and $A B$ as the diameter is shown in the figur...

36. A glass sphere with radius $R , O$ as the center and $A B$ as the diameter is shown in the figure, and on the left side of the sphere there is a hard direct The closing screen is tangent to the glass sphere at $A$. A ray of light $B M$ from point $B$ exits the glass sphere at point $M$, and the ray of light exits the glass sphere at point $A B$ parallel to ${ } ^ { Q }$, and shines on point ${ } ^ { Q }$ on the receiving screen. ] point on the receiving screen. Another ray ${ } ^ { B N }$ is totally reflected at ${ } ^ { N }$. Knowing that $\angle A B M = 30 ^ { \circ }$, the speed of light traveling in a vacuum is $c$, the following statement is correct ( ) ![](/images/questions/phys-geometric-optics/image-030.jpg)

  • A. A. The critical angle at which light undergoes total reflection in this glass sphere is greater than $45 ^ { \circ }$
  • B. B. The time required for light to travel from ${ } _ { B }$ to $Q$ is $\frac { 4 R } { C }$.
  • C. C. The firm distance from the point $N$ to $A B$ is $\frac { 2 \sqrt { 2 } } { 3 } R$
  • D. D. If the light source at point $B$ is red and is replaced by a green light source, the point $N$ on the arc where total reflection occurs is shifted to the right.

Answer: C

Solution: A. It is known that $\angle A B M = 30 ^ { \circ }$, the angle of incidence $\alpha = 30 ^ { \circ }$ of the ray $\alpha = 30 ^ { \circ }$ at the point $\alpha = 30 ^ { \circ }$ is known from the geometrical relation $B M$. Angle of refraction $\beta = 60 ^ { \circ }$ Then the refractive index of the glass is $n = \frac { \sin \beta } { \sin \alpha } = \frac { \sin 60 ^ { \circ } } { \sin 30 ^ { \circ } } = \sqrt { 3 }$ The sine of the critical angle $\sin C = \frac { 1 } { n } = \frac { \sqrt { 3 } } { 3 } < \frac { \sqrt { 2 } } { 2 }$ So $C < 45 ^ { \circ }$ Therefore, A is wrong; B. Light travels in glass with a speed $v = \frac { c } { n }$ Light travels from $B$ to $M$ in $t _ { 1 } = \frac { B M } { v } = \frac { 2 R \cos 30 ^ { \circ } } { \frac { c } { n } } = \frac { \sqrt { 3 } n R } { c } = \frac { 3 R } { c }$. The time when light travels from $M$ to $Q$ $t _ { 2 } = \frac { M Q } { c } = \frac { R - R \sin 30 ^ { \circ } } { c } = \frac { R } { 2 c }$. Then the time required for light to travel from $B$ to $Q$ is $t = t _ { 1 } + t _ { 2 } = \frac { 7 R } { 2 c }$. Therefore, B is wrong; C. From the question, we know that the critical angle is $C = \angle O N B$ Then $\sin C = \frac { 1 } { n } = \frac { \sqrt { 3 } } { 3 } , \cos C = \frac { \sqrt { 6 } } { 3 }$ Therefore, the firm distance $N$ from the point $A B$ to $d = 2 R \cos C \times \sin C = \frac { 2 \sqrt { 2 } } { 3 } R$ is $d = 2 R \cos C \times \sin C = \frac { 2 \sqrt { 2 } } { 3 } R$. C is correct; D. If the light source at the $B$ point is red, and it is replaced by a green light source, the refractive index will be larger, and the critical angle will be smaller according to $\sin C = \frac { 1 } { n }$, then the $N$ point on the arc which is fully reflected will be shifted to the left, so D is wrong.

Question 35: 37. A student used on the following parallel glass block to do the following experiments: will be a ...

37. A student used on the following parallel glass block to do the following experiments: will be a beam of red light and violet light composed of complex color light from the air on the upper surface of the glass block $P$ point, the light enters the glass block into the internal split into two beams of monochromatic light $a , b$, as shown in the figure. It is known that the incident light makes an angle of $\theta = 45 ^ { \circ }$ with the upper surface of the glass block, then the number of correct statements in the following is ( ) ![](/images/questions/phys-geometric-optics/image-031.jpg) (1) The propagation time of $a$ light in a glass block is greater than the propagation time of $b$ light (2) $a$ is violet light and $b$ is red light. (3) $a , b$ The two beams of light are parallel when the light comes from the lower surface of the glass block (4) $a$ light has a higher energy than $b$ light (5) When $a , b$ light passes through an obstacle, $a$ light is more likely to diffract than $b$ light. (6) If two-slit interference experiments are carried out separately with $a , b$ light, the fringe spacing of $a , b$ light may be equal.

  • A. A. 6
  • B. B. 5
  • C. C. 4
  • D. D. 3

Answer: B

Solution: According to the law of refraction $$ n = \frac { \sin \left( 90 ^ { \circ } - \theta \right) } { \sin r } $$ Since the angle of refraction of $a$ light is smaller, the index of refraction of $a$ light is larger, so $a$ is violet light and $b$ is red light, so (2) is correct; According to the geometric relationship, $a , b$ light from the lower surface of the glass block when the angle of incidence are equal to the angle of refraction, then $a , b$ light from the lower surface of the glass block when the angle of refraction are equal to the angle of incidence, i.e., the two beams of light are parallel, so (3) is correct; according to the expression of the energy of the photon $$ \varepsilon = h v $$ Since $a$ is violet light and $b$ is red light, and the frequency of $a$ light is greater than that of $b$ light, the $a$ light has an energy greater than that of $a$. The energy of $a$ light is greater than that of $b$ light, so (4) is correct; When $a , b$ light passes through an obstacle, since $a$ is violet and $b$ is red, the wavelength of $a$ light is less than $b$ light, and the wavelength of $b$ light is less than $b$ light. ] is red, then $a$ is smaller than $b$ and $b$ is more prone to diffraction than $a$, so (5) is wrong; According to $$ \Delta x = \frac { L } { d } \lambda $$ If the double-slit interference experiments with $a , b$ light are performed separately, although the wavelengths of $a , b$ light are not equal, the $L , d$ may also be unequal when the double-slit interference experiments are performed separately, then $a , b$ The fringe spacing of the light may be equal, so (6) is correct; Let the width of the glass block be $d$, and let the angle of refraction of light be $r$, which is obtained according to the law of refraction $$ n = \frac { \sin \left( 90 ^ { \circ } - \theta \right) } { \sin r } $$ The speed of propagation of light in a glass block is $$ v = \frac { c } { n } $$ The distance traveled by light in a glass block is $$ s = \frac { d } { \cos r } $$ Then the propagation time of light in the glass block is $$ t = \frac { s } { v } $$ The association gives $$ t = \frac { d \sin \left( 90 ^ { \circ } - \theta \right) } { c \sin r \cos r } = \frac { 2 d \cos 45 ^ { \circ } } { c \sin 2 r } $$ Since the angle of refraction of $a$ light is small, the propagation time of $a$ light in a glass block is greater than the propagation time of $b$ light, so (1) is correct.

Question 36: 38. As shown in the figure, a beam of visible light at an angle of incidence $\theta$ from a glass b...

38. As shown in the figure, a beam of visible light at an angle of incidence $\theta$ from a glass block into the air, after refraction into $a , b$ two beams of monochromatic light. $a , b$ Compared to the two beams of light, the following statements are correct ( ) ![](/images/questions/phys-geometric-optics/image-032.jpg)

  • A. A. The refractive index of glass to $a$ light is less than that to $b$ light.
  • B. B. In glass, $a$ light has a greater $b$ speed of light
  • C. C. Increase the angle of incidence $\theta , b$ Light is first totally reflected.
  • D. D. $a$ Critical angle of total reflection of light is less than $b$ Critical angle of total reflection of light

Answer: D

Solution: A. The angle of deflection of light $a$ is large, and the refractive index of glass to $a$ light is greater than that to $b$ light, so A is wrong; B. According to $v = \frac { c } { n }$, the larger the refractive index, the smaller the speed in the medium, so in the glass, $a$ the refractive index of the light is large, $a$ the speed of the light is smaller than the speed of the light of the $b$, B error; CD. According to $\sin C = \frac { 1 } { n } , ~ a$ the refractive index of light is large, the critical angle is small, increasing the angle of incidence $\theta , ~ a$ the light first undergoes total reflection, C is wrong, D is correct.

Question 37: 39. A cross-section of a piece of prism made of glass is shown, where $A B O D$ is a rectangle, $O C...

39. A cross-section of a piece of prism made of glass is shown, where $A B O D$ is a rectangle, $O C D$ is a quarter-circle arc of radius $R$, and the center of the circle is $O$. A red laser beam is incident from a point on the $A B$ surface at an angle of $\theta = 60 ^ { \circ }$ It enters the prism and then shoots at a critical angle at the $B C$ surface of the $O$ point, and part of the optical path is shown in the figure. The following statement is correct ( ) ![](/images/questions/phys-geometric-optics/image-033.jpg)

  • A. A. It is possible that the red laser is fully reflected at the $C D$ side.
  • B. B. The refractive index of the red laser in the prism is $\frac { \sqrt { 7 } } { 2 }$
  • C. C. Red laser light travels twice as fast in a prism as $\sqrt { 3 }$ in a vacuum.
  • D. D. If the angle of incidence remains the same and the green laser is replaced, total reflection is not possible on the $B C$ side.

Answer: B

Solution: A. The optical path of the red laser in the prism is shown in the figure ![](/images/questions/phys-geometric-optics/image-034.jpg) After total reflection at $O$, the red laser light is directed to the $C D$ side along the radius, and the light is directed along the normal direction of the interface, so the light doesn't change the direction of propagation, and total reflection doesn't occur at the $C D$ side, so A is wrong; B. The refractive index of the red laser is $n$, and the angle of refraction at the $A B$ side is $r$, according to mathematical knowledge, the angle of incidence of light at the $B C$ side is $B C$. $$ i = C = 90 ^ { \circ } - r $$ At the point of incidence, according to the law of refraction $$ n = \frac { \sin 60 ^ { \circ } } { \sin r } $$ the solution is $$ \sin r = \frac { \sqrt { 3 } } { 2 n } $$ Since the red laser light is fully reflected in the $B C$ face, according to the critical angle equation $$ \sin C = \sin \left( 90 ^ { \circ } - r \right) = \frac { 1 } { n } $$ the solution is $$ \cos r = \frac { 1 } { n } $$ Also by the math $$ \sin ^ { 2 } r + \cos ^ { 2 } r = 1 $$ Substituting in the data and solving associatively, we get $$ n = \frac { \sqrt { 7 } } { 2 } $$ ## Therefore, B is correct; C. According to the formula $n = \frac { c } { v }$, we can get $$ \frac { v } { c } = \frac { 1 } { n } = \frac { 2 \sqrt { 7 } } { 7 } $$ Therefore, C is wrong; D. According to the electromagnetic spectrum, the frequency of red light is smaller than the frequency of green light, so the refractive index of red light is smaller than the refractive index of green light, i.e. $$ n _ { \text {红 } } < n _ { \text {绿 } } $$ According to the law of refraction $n = \frac { \sin \theta } { \sin r }$, the angle of refraction of light in the $A B$ surface meets the $$ n _ { \text {红 } } < n _ { \text {绿 } } $$. $$ r _ { \text {红 } } > r _ { \text {绿 } } $$ Then the angle of incidence of light in the $B C$ plane satisfies $$ i _ { \text {红 } } < i _ { \text {绿 } } $$ According to the critical angle formula $\sin C = \frac { 1 } { n }$, the critical angle relationship of light in the $B C$ plane is $$ C _ { \text {红 } } > C _ { \text {绿 } } $$ Since the red light is directed at the point $O$ on the $B C$ surface at the critical angle, if the incident angle is constant, the green laser light will be totally reflected at the $B C$ side, and therefore D is wrong.

Question 38: 40 . As shown in the figure, a cylindrical oil storage drum has a bottom diameter of $A B = 8 d$ and...

40 . As shown in the figure, a cylindrical oil storage drum has a bottom diameter of $A B = 8 d$ and a height of $B C = 6 d$. When there is no oil in the drum, a point $E$ is visible from a point $E$ exactly at a point $A$ at the bottom edge of the drum. When the depth of the oil in the barrel is equal to half of the height of the barrel, still looking along the $E A$ direction, the point on the bottom of the barrel $F , F , A$ is exactly visible from a point $F , F , A$ The two points are separated by a distance of $\frac { 7 } { 4 } d$ , then the refractive index of the oil is ( ) ![](/images/questions/phys-geometric-optics/image-035.jpg) High School Physics Assignment, October 30, 2025

  • A. A. $n = \frac { 4 } { 3 }$
  • B. B. $n = \sqrt { 2 }$
  • C. C. $n = \sqrt { 3 }$
  • D. D. $n = \frac { 5 } { 3 }$

Answer: A

Solution: Let the point of intersection of the light ray and the oil surface be $O$ point, the geometric relationship can be seen $A C$ distance is $10 d$, the angle of incidence, the angle of refraction is $\theta _ { 1 } , \theta _ { 2 }$, so the angle of incidence ${ } ^ { \theta _ { 1 } }$ sine value is ${ } ^ { \theta _ { 1 } }$. FORMULA_9]] sine value is $$ \sin \theta _ { 1 } = \frac { 4 d } { 5 d } = \frac { 4 } { 5 } $$ The geometric relationship shows that the $O F$ distance is $$ x _ { O F } = \sqrt { ( 3 d ) ^ { 2 } + \left( 4 d - \frac { 7 } { 4 } d \right) ^ { 2 } } = \frac { 15 } { 4 } d $$ Therefore, the sine of the refraction angle is $$ \sin \theta _ { 2 } = \frac { 4 d - \frac { 7 } { 4 } d } { x _ { O F } } = \frac { 3 } { 5 } $$ Therefore the refractive index $$ n = \frac { \sin \theta _ { 1 } } { \sin \theta _ { 2 } } $$ Combining the above solutions yields $$ n = \frac { 4 } { 3 } $$
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Geometric Optics

几何光学

38 Practice Questions

Practice with Chinese questions to prepare for the CSCA exam. You can toggle translations while practicing.

Topic Overview

Geometrical optics studies the laws of light propagation in straight lines through a homogeneous medium, focusing on the phenomena of reflection, refraction and total reflection of light and their applications. In the CSCA Physics exam, this section is often examined in conjunction with light path diagrams, requiring candidates to be able to analyze the behavior of light at the interfaces of different media and to compare the propagation properties of different colors of light. Questions usually involve calculations and judgments on concepts such as the index of refraction, critical angle, and speed of light.

Questions:38

Key Points

  • 1Application of the Laws of Reflection and Refraction of Light (Snell's Law)
  • 2Conditions for Total Reflection and Calculation of Critical Angle
  • 3Comparison of refractive indices of different colors of light (e.g., red and violet) and analysis of light paths
  • 4Path and rate changes in light propagation through multilayer media

Study Tips

Become proficient in the refractive index formula n = c/v and Snell's law n₁sinθ₁ = n₂sinθ₂, and practice more on questions that analyze the behavior of light in conjunction with light path diagrams.

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