Skip to main content

Physical Optics - Practice Questions (38)

Question 1: 1. A student used the experimental setup shown in Figure A to study the phenomenon of interference a...

1. A student used the experimental setup shown in Figure A to study the phenomenon of interference and diffraction of light, the width of the slit $S _ { 1 } , S _ { 2 }$ $d$ can be adjusted, the slit to the screen distance is $L$. The same monochromatic light irradiated perpendicular to the slit, the experiment was obtained on the screen Figure B, Figure C shown in the pattern. The following description is correct ( ) ![](/images/questions/phys-physical-optics/image-001.jpg) A ![](/images/questions/phys-physical-optics/image-002.jpg) B ![](/images/questions/phys-physical-optics/image-003.jpg) C

  • A. A. Figure B shows the diffraction pattern of light
  • B. B. When irradiating two slits, increasing the width of the slits, all else being equal, increases the distance between the centers of adjacent dark stripes in Figure B.
  • C. C. When illuminating both slits, increasing the distance $L$, other things being equal, the distance between the centers of the adjacent dark stripes in Fig. B increases
  • D. D. When two slits are illuminated, if the difference in the distance traveled by light from the slit ${ } ^ { S _ { 1 } , S _ { 2 } }$ to the $P$ point on the screen is an odd multiple of the half-wavelength, there must be a bright stripe at the $P$ point.

Answer: C

Solution: A. The stripes in Figure B are equally spaced, so it is a double-slit interference stripe pattern, A is wrong; BC. according to the interference formula $$ \lambda = \frac { L \lambda } { d } $$ When increasing the slit width $d$ other unchanged, the center spacing of interference fringes decreases, when increasing the distance $L$ other unchanged, the center spacing of interference fringes increases, B is wrong, C is correct; D. When the distance difference of the light is an odd multiple of the half wavelength, it is the mutual weakening zone, i.e., the dark fringe, so the P point must be the dark fringe, D is wrong.

Question 2: 2. The following phenomena are diffraction of light ( )

2. The following phenomena are diffraction of light ( )

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

Answer: B

Solution: A. Figure A is an application of total reflection, so A is wrong; B. B is the diffraction of light, so B is correct; C. C is the phenomenon of light refraction, so C is wrong; D. D figure for the interference phenomenon of light, so D error;

Question 3: 3. Under the light of an incandescent lamp, colored streaks can be seen on the surface of two glass ...

3. Under the light of an incandescent lamp, colored streaks can be seen on the surface of two glass plates pressed together. This is the light )

  • A. A. double-slit interference phenomenon
  • B. B. thin-film interference phenomenon
  • C. C. single-slit diffraction phenomenon (physics)
  • D. D. dispersion

Answer: B

Solution: Light striking an air film between two glass plates reflects off the upper and lower surfaces, creating superposition and forming interference fringes, which are thin-film interference. The interference phenomenon shows that light has volatility. Therefore, B is correct, ACD is wrong.

Question 4: 4. Children blowing soap bubbles, hot vegetable soup surface of the oil splashes, the oil film on th...

4. Children blowing soap bubbles, hot vegetable soup surface of the oil splashes, the oil film on the surface of the road water, all have colored patterns. Which of the following is the cause of this

  • A. A. dispersion of light
  • B. B. interference of light
  • C. C. polarization of light
  • D. D. diffraction of light

Answer: B

Solution: Children blowing out of the soap bubble bubbles, hot vegetable soup surface of the oil flower, the road on the surface of the oil film of water, all have color patterns. This is due to light irradiation on the water surface of the oil film on the light in the oil film on the upper and lower surfaces of the two surfaces, respectively, the two columns of reflected light in the oil film on the surface of the film interference occurs in the formation of colored interference fringes.

Question 5: 5. Check whether the upper surface of the glass plate below is flat as shown in the figure, monochro...

5. Check whether the upper surface of the glass plate below is flat as shown in the figure, monochromatic light from above will produce streaks, this is the use of light ( ) ![](/images/questions/phys-physical-optics/image-004.jpg)

  • A. A. circular hole diffraction
  • B. B. single-slit diffraction (physics)
  • C. C. double-slit interference
  • D. D. thin-film interference (physics)

Answer: D

Solution: To check whether the upper surface of the glass plate is flat, the interference fringe is formed by the interference of reflected light from the lower surface of the a plate and the upper surface of the b plate. Specifically: when the distance difference between the two reflected light (i.e., 2 times the film thickness) is an even number of times the half-wavelength, there is a bright streak, is an odd number of times the half-wavelength, there is a dark streak, if the streak is straight then the plate is flat, if the streak is curved, then the glass plate is not flat, so the check below the surface of the glass plate whether the surface is flat is to utilize the thin-film interference of light, so the ABC is wrong, D is correct. Therefore, D is correct. [Eyesight] thin-film interference stripes formed by the interference of reflected light on the upper and lower surfaces of the film; when the difference in distance between the two reflected light (i.e., two times the thickness of the film) is an even multiple of the half-wavelength, there are bright stripes; when the difference in distance between the two reflected light is an odd multiple of the half-wavelength, there are dark stripes.

Question 6: 6. Of the following four electromagnetic waves, the most likely to exhibit diffraction is

6. Of the following four electromagnetic waves, the most likely to exhibit diffraction is

  • A. A. infrared ray
  • B. B. ultraviolet ray
  • C. C. wireless electric wave
  • D. D. $\gamma$ ray

Answer: C

Solution: Among the above electromagnetic waves, radio waves have the largest wavelength, so it is easiest to show diffraction phenomenon, so C meets the meaning of the question, ABD does not meet the meaning of the question.

Question 7: 7. In a single-slit diffraction experiment, by reducing only the width of the single slit, the scree...

7. In a single-slit diffraction experiment, by reducing only the width of the single slit, the screen

  • A. A. Wider stripes and increased light intensity
  • B. B. Narrower stripes and increased light intensity
  • C. C. Streaks widen and light intensity decreases
  • D. D. Narrowing of stripes and reduction of light intensity

Answer: C

Solution: In the single-slit diffraction experiment, the smaller the width of the single slit, the more obvious the diffraction phenomenon, the greater the spacing of the stripes; the smaller the width of the slit, the smaller the light flux, the brightness of the field of view will become darker, the weaker the intensity of the light, so only to reduce the width of the single slit, the stripes will become wider, the intensity of the light is weakened, so the ABD is wrong, C is correct.

Question 8: 8. The following phenomena show that light is a transverse wave ![](/images/questions/phys-physical...

8. The following phenomena show that light is a transverse wave ![](/images/questions/phys-physical-optics/image-005.jpg) Figure (a) ![](/images/questions/phys-physical-optics/image-006.jpg) Figure (b) ![](/images/questions/phys-physical-optics/image-007.jpg) Figure (c) ![](/images/questions/phys-physical-optics/image-008.jpg) Figure (d)

  • A. A. Figure (a), X-ray diffraction of a DNA molecule
  • B. B. Figure (b), Color interference in soap film
  • C. C. Figure (c), the bubbles in the water are particularly bright due to total reflection
  • D. D. Figure (d), rotating the polarizer in front of the camera lens to change the direction of polarization, shooting the same scene presents different scenes

Answer: D

Solution: A. X-ray diffraction of DNA molecules, the diffraction of light shows that light is volatile, but it does not show that light is B.The interference of light proves that light is fluctuating, but it does not show that light is a transverse wave; B. The interference of light proves that light has fluctuation, but it cannot show that light is a transverse wave, so B is wrong; C. Total reflection can not show that light has volatility, so C error; D. Polarizer on the incident light has the function of shielding and transmittance, can make the longitudinal light or transverse light through a polarization phenomenon is unique to transverse waves, the direction of light vibration for the propagation of the direction of the asymmetry is called polarization, which is the transverse wave is different from other longitudinal waves, one of the most obvious signs. Therefore, light with polarization indicates that light is a transverse wave. Therefore, D is correct.

Question 9: 9. The figure shows the diffraction image of a water wave observed using a water wave tank, and from...

9. The figure shows the diffraction image of a water wave observed using a water wave tank, and from the image it can be seen that ![](/images/questions/phys-physical-optics/image-009.jpg)

  • A. A. The water wave on the B side is a diffracted wave
  • B. B. A-side wave speed is equal to B-side wave speed.
  • C. C. Decreasing the distance between the baffles, the wavelength of the diffracted wave will decrease
  • D. D. Increasing the distance between the baffles makes diffraction more pronounced

Answer: B

Solution: A. $B$ The side water wave diffracts when it encounters a baffle with a narrow slit in the process of propagation, then $A$ the side water wave is a diffracted wave, so A is wrong; B. Mechanical waves of the same kind propagate at the same speed in the same medium, so B is correct; C. Reduce the distance between the baffle, diffraction phenomenon will be more obvious, but the diffraction of the wave speed is unchanged, the frequency is unchanged, so the wavelength is unchanged, so C is wrong; D. Because only when the distance between the baffle and the wavelength is about the same or more than the wavelength, the diffraction phenomenon is obvious, so when increasing the distance between the baffle, diffraction phenomenon will be less obvious, so D error.

Question 10: 10. Recently, the Chinese Academy of Sciences (CAS) announced that the technology of 5 nm photolitho...

10. Recently, the Chinese Academy of Sciences (CAS) announced that the technology of 5 nm photolithography has advanced by leaps and bounds. Photolithography is the core equipment for the production of large-scale integrated circuits (ICs), and the most central component of its exposure system is the ultraviolet (UV) light source. In the fight against the new crown virus, a large number of infrared thermometers used to measure body temperature. About infrared and ultraviolet light, the following statements are correct ( )

  • A. A. Only hot objects radiate infrared.
  • B. B. Only cold objects radiate UV
  • C. C. Infrared rays are more diffractive than ultraviolet rays.
  • D. D. Infrared photons have more energy than UV light

Answer: C

Solution: AB . All the temperature higher than the absolute 0 degrees (0 K) of the object are constantly radiating infrared radiation, so AB is wrong; C. The wavelength of infrared light is greater than ultraviolet light, so infrared light diffraction ability than ultraviolet light is stronger, so C is correct; D. The frequency of infrared light is smaller than that of ultraviolet light, so the energy of infrared photons is smaller than that of ultraviolet light according to $E = h v$, so D is wrong.

Question 11: 13. Optics has a wide range of applications in life, the following statements are correct

13. Optics has a wide range of applications in life, the following statements are correct

  • A. A. Transmission Enhancement Films on optical lenses are applied to the polarization of light
  • B. B. The glasses worn to watch stereoscopic movies are an application of light interference
  • C. C. Soap film showing colored streaks in sunlight is an interference phenomenon of light
  • D. D. In order to reduce the interference of reflected light from the water surface when photographing swimming fish in a pool, a filter is installed in front of the lens to apply the diffraction of light.

Answer: C

Solution: A. Optical lens on the transmittance film is the use of light interference phenomenon, so A error; B.The glasses worn when watching stereoscopic movies are applied to the polarization of light, so B is wrong; C. Soap film in the sunlight showing colored stripes is the interference phenomenon of light, so C is correct; D. Reflected light is polarized light, the photographer in the pool of swimming fish is installed in front of the camera polarizing filter, the application of light polarization phenomenon attenuates the impact of reflected light on the water surface, so D error.

Question 12: 14. A rainbow is formed when sunlight enters a water droplet, is refracted once, then reflected at t...

14. A rainbow is formed when sunlight enters a water droplet, is refracted once, then reflected at the back of the droplet, and finally refracted once more when it leaves the droplet. The following figure for the formation of the rainbow diagram, a beam of white light L from the left into the droplet, $a , b$ is the white light into the droplet after a reflection and two refraction after two monochromatic light lines. Then. ![](/images/questions/phys-physical-optics/image-010.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 $\sin C = \frac { 1 } { n }$, we can know that the critical angle of ${ } _ { a }$ light is smaller than that of ${ } _ { b }$ light, i.e. ${ } _ { a }$ is easier to cause total reflection than ${ } _ { b }$. Total reflection occurs, so C is correct; 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 13: 15. with the same double-slit interference experimental device in a vacuum to do red and violet ligh...

15. with the same double-slit interference experimental device in a vacuum to do red and violet light double-slit interference experiments, to obtain A, B two kinds of interference fringes, as shown in the figure, then the following statements are correct ![](/images/questions/phys-physical-optics/image-011.jpg) ## A \section*{| | | | | | | | | :--- | :--- | :--- | :--- | :--- | :--- | | | | | | | | |}

  • A. A. A is violet, B is red
  • B. B. The rate of propagation of light A in water is greater than the rate of propagation of light B in water
  • C. C. In the same medium, the refractive index of light A is large and that of light B is small
  • D. D. In the same medium, the critical angle of total reflection of light A is small, and the critical angle of total reflection of light B is large.

Answer: B

Solution: A. The formula for the spacing of the double-slit interference fringes is $$ \Delta x = \frac { L } { d } \lambda $$ The larger spacing of the fringes in figure A and the larger wavelength of light show that A is red light and B is violet light, A is wrong; $B C$. A light (red light) has a smaller refractive index in water, by the $$ v = \frac { c } { n } $$ It can be seen that the propagation rate of A light in water is greater than the propagation rate of Z light in water, B is correct, C is wrong; D. By $$ \sin C = \frac { 1 } { n } $$ can be known, in the same medium, the critical angle of total reflection of A light is large, the critical angle of total reflection of B light is small, D error.

Question 14: 16. Active braking system in the car has been widely used, people in the front of the car to install...

16. Active braking system in the car has been widely used, people in the front of the car to install millimeter-wave radar, the use of radar waves to monitor the driver is difficult to observe the blind spot, can be prompted by voice signals to the driver to pay attention to the safety of driving, or the car's active braking. About millimeter wave (microwave), the following statements are correct

  • A. A. Millimeter waves can't travel in a vacuum
  • B. B. Millimeter waves have shorter wavelengths than UV
  • C. C. Millimeter waves propagate in a straight line in a homogeneous medium
  • D. D. Millimeter waves cannot diffract

Answer: C

Solution: A. Millimeter waves are radio waves, which can propagate in both medium and vacuum; B. Millimeter waves have longer wavelengths than ultraviolet light; C. Millimeter wave wavelength is short, the form of propagation with the same light, in the same kind of uniform medium is along the straight line propagation, C is correct; D. Diffraction is the phenomenon of all Porter, just millimeter wave wavelength is short, encountered obstacles when the diffraction phenomenon is not obvious, D error.

Question 15: 17. The following statements are incorrect

17. The following statements are incorrect

  • A. A. The reason why you hear the sound before you see the person is because sound waves have a large wavelength and are prone to diffraction.
  • B. B. The speed of propagation of a mechanical wave in a medium is independent of the frequency of the wave
  • C. C. In a double-slit interference experiment, the experiment with violet light gives narrower stripes than the experiment with red light under the same conditions
  • D. D. The greater the mass of the pendulum ball, the greater the period of simple harmonic vibration of the pendulum at the same location when the pendulum length is constant.

Answer: D

Solution: A. The end of the person first heard the sound, because the sound wave wavelength is larger, easy to diffraction phenomenon. Therefore, A is correct and does not meet the meaning of the question; B.The propagation speed of mechanical waves in a medium is only related to the medium itself. Therefore, B is correct, does not meet the meaning of the question; C. In the double-slit interference experiment, the spacing of the stripes is proportional to the wavelength of the incident light, so the violet stripes are narrower, C is correct, does not meet the meaning of the question; D. According to the formula for the period of a single pendulum, the period is only related to the length of the pendulum and the size of the acceleration of gravity, so D is wrong, in line with the meaning of the question.

Question 16: 18. Regarding the optical lens permeation enhancement film, the following statements are incorrect (

18. Regarding the optical lens permeation enhancement film, the following statements are incorrect (

  • A. A. Transmission Enhancement Films are designed to reduce light reflection losses and increase the intensity of transmitted light
  • B. B. The thickness of the permeation-enhancing film is equal to one-fourth of the wavelength of the incident light in the film.
  • C. C. Transmission Enhancement Membrane utilizes the principle of interference of light
  • D. D. Lenses coated with a transmittance-enhancing film, the reflected light of various color light all cancel each other, so the imaging effect of such lenses is better.

Answer: D

Solution: ABC. Transmission enhancement film is made using the principle of interference of light, when the optical range difference is half the wavelength of light in the film, there is a reflection of light weakened to reduce the loss of light reflection, thereby increasing the transmission of light, the thickness of the transmission enhancement film should be the wavelength of the incident light in the transmission enhancement film $1 / 4$, so the ABC is correct, does not meet the meaning of the question; D. In the choice of transmittance enhancement film, generally to make the human eye sensitive to the green light in the vertical incidence of each other, red, violet light reflection can not be completely offset, so the lens coated with a transmittance enhancement film is mauve, so D is wrong, in line with the meaning of the question.

Question 17: 19. The soft headphone cable contains fine copper wires of uniform thickness. The processing of this...

19. The soft headphone cable contains fine copper wires of uniform thickness. The processing of this fine copper wire is very demanding. In the factory, the diagram is utilized. The device shown in the factory, with a laser irradiation of fine copper wire, the implementation of automatic control of the extraction process, the image on the light screen may be ( ) ![](/images/questions/phys-physical-optics/image-012.jpg)

  • A. A. ![](/images/questions/phys-physical-optics/image-005.jpg)
  • B. B. |II|
  • C. C. $\square | \mid$
  • D. D. |||||

Answer: D

Solution: When the size of the obstacle is comparable to the wavelength of the wave, or smaller than the wavelength of the wave, a significant diffraction phenomenon occurs. The device utilizes the diffraction phenomenon of light, which is the same as the diffraction fringes seen by a laser through a single slit, i.e., it produces unequally spaced fringes, a particularly wide central fringe, narrower fringes on both sides, and symmetrically distributed fringes.

Question 18: 20. In a Young's double-slit interference experiment, if a monochromatic light source $S$ is moved a...

20. In a Young's double-slit interference experiment, if a monochromatic light source $S$ is moved a small distance upward from the center axis position shown in the figure in a direction perpendicular to $S O$. In what direction do the center interference fringes move? How does the spacing of neighboring bright fringes change? ( ) $\stackrel { \bullet } { S }$ ![](/images/questions/phys-physical-optics/image-013.jpg)

  • A. A. The spacing of adjacent bright stripes remains constant and the center interfering bright stripe moves upward
  • B. B. The spacing between adjacent bright fringes becomes larger and the center interfering bright fringe moves downward
  • C. C. The spacing of adjacent bright fringes remains constant and the center interfering bright fringe moves downward
  • D. D. The spacing between adjacent bright fringes becomes smaller and the center interfering bright fringe moves upward

Answer: C

Solution: After the light source $S$ is moved upward for a small distance, the light emitted by $S$ arrives at a point below the $O$ point and the optical range difference is zero, so the center interfering fringe moves downward, as shown in the figure. ![](/images/questions/phys-physical-optics/image-014.jpg) According to the formula for the distance between neighboring bright fringes $\Delta y = \frac { l } { d } \lambda$, since $l , d , \lambda$ does not change, the spacing of the fringes remains unchanged.

Question 19: 21. Using the device shown in the figure to study the phenomenon of double-slit interference and mea...

21. Using the device shown in the figure to study the phenomenon of double-slit interference and measure the wavelength of light, the following statements are incorrect ![](/images/questions/phys-physical-optics/image-015.jpg)

  • A. A. The (1) (2) (3) elements of the experimental setup are filters, single slit, double slit
  • B. B. By replacing the green filter with a red filter, the interference fringe spacing becomes larger
  • C. C. After moving the single slit to the double slit by a small distance, the spacing of the interference fringes remains unchanged
  • D. D. Decreasing the double slit spacing in the experiment increases the number of stripes in the eyepiece

Answer: D

Solution: A. The light from the source is filtered into monochromatic light, which illuminates the single slit. The single slit is equivalent to a line light source, which in turn illuminates the double slit. Light from the double slit interferes in the space to the right of the double slit. One end of the light-blocking cylinder is equipped with a woolen glass screen to observe the interference fringes, so the (1) (2) (3) components in the experimental setup are the filter, single slit, and double slit, respectively, so A is correct; B. By replacing the green filter with a red filter, the wavelength of the light increases, and according to $\Delta x = \frac { l \lambda } { d }$, the interference fringes become more widely spaced, so B is correct; C. According to $\Delta x = \frac { l \lambda } { d }$, it can be known that the interference fringe spacing has nothing to do with the distance between the single slit and the double slit, so after moving the single slit to the double slit by a small distance, the interference fringe spacing remains unchanged, so C is correct; D. According to $\Delta x = \frac { l \lambda } { d }$, it can be known that by decreasing the distance between the double slits in the experiment, the interference fringe spacing increases and the number of fringes in the eyepiece decreases, so D is wrong. The question is wrong, so choose D.

Question 20: 22. A beam of monochromatic parallel light passing through a double slit gives interference fringes ...

22. A beam of monochromatic parallel light passing through a double slit gives interference fringes on a screen, the following statements are correct (

  • A. A. The spacing between two adjacent light stripes is larger than the spacing between two adjacent dark stripes
  • B. B. The spacing between the bright or dark stripes obtained in an experiment with red light is smaller than in an experiment with violet light
  • C. C. Stripes cannot be formed by doing this experiment with white light
  • D. D. When this experiment is done with white light, colored streaks appear on the screen

Answer: D

Solution: The widths of the bright and dark interference fringes are equal, A is wrong; the spacing of the fringes is proportional to the wavelength, so the experiment with red light is greater than the experiment with violet light, the spacing between the bright fringes or dark fringes is greater, B is wrong; using white light to do this experiment, the color interference fringes appear on the screen, C is wrong, D is correct.

Question 21: 23. The following statements about the phenomenon of light are correct

23. The following statements about the phenomenon of light are correct

  • A. A. The rainbow seen after rain is a total reflection phenomenon
  • B. B. Tail lights on cars use the dispersion of light.
  • C. C. The colored streaks on the dewdrops belong to the phenomenon of dispersion of light
  • D. D. The bright spots seen in the shade are refractions of light

Answer: C

Solution: A. After the rain in the air there are a number of small droplets of water, sunlight through the small droplets can be dispersed into seven colors of light, which produces the rainbow phenomenon, so A error; B. car taillights for red, because red light than other visible light wavelength, easy diffraction, so B error; C. Dewdrops on the colored stripes is the sunlight through the droplets can be dispersed into seven colors that belongs to the phenomenon of dispersion, so C is correct; D. light in the shade of dense trees on the ground in the round bright spot is a small hole imaging, so D error.

Question 22: 24. In order to enhance the shopping experience, a shopping mall has introduced a 5G indoor navigati...

24. In order to enhance the shopping experience, a shopping mall has introduced a 5G indoor navigation system, which transmits electromagnetic waves of a specific frequency to accurately locate the position of customers. It is known that the system uses electromagnetic waves at a higher frequency than the 4G network. Regarding these electromagnetic waves, the following statements are correct

  • A. A. The electromagnetic waves used for 5G indoor navigation are more susceptible to diffraction than those used in 4G networks
  • B. B. These electromagnetic waves can propagate in a vacuum and the speed of propagation increases with frequency
  • C. C. Electromagnetic waves used for 5G indoor navigation can be polarized
  • D. D. Maxwell predicted and confirmed the existence of electromagnetic waves

Answer: C

Solution: A. According to ${ } ^ { \lambda = \frac { c } { f } }$, 5G indoor navigation uses electromagnetic waves at a higher frequency compared to the electromagnetic waves of the 4G network, and the According to ${ } ^ { \lambda = \frac { c } { f } }$, the electromagnetic waves used in 5G indoor navigation have a higher frequency than those used in 4G networks, and the wavelength is shorter; since the longer the wavelength, the easier the diffraction phenomenon occurs, the electromagnetic waves used in 5G indoor navigation are less prone to the diffraction phenomenon; B. The propagation speed of all electromagnetic waves in vacuum is the speed of light $c = 3 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s }$, independent of the frequency, so B is wrong. C. Polarization is a characteristic of transverse waves, electromagnetic waves are transverse waves, so no matter how high or low the frequency of polarization phenomenon can occur, so C is correct. D. Maxwell predicted the existence of electromagnetic waves through theory, but experiments confirmed by Hertz completed, so D error. Therefore, C.

Question 23: 25. The following statements about are correct

25. The following statements about are correct

  • A. A. When a uranium nucleus undergoes $\alpha$ decay, it releases $\alpha$ particles and a certain amount of energy, and it is this spontaneous release of energy that is currently utilized in nuclear power plants.
  • B. B. If the photoelectric effect can occur by irradiating a metal with violet light, the photoelectric effect must not occur by using red light instead.
  • C. C. Hydrogen atoms absorb photons of a certain frequency when they jump from a higher energy level to a lower energy level
  • D. D. Both mechanical and electromagnetic waves have interference and diffraction properties

Answer: D

Solution: A, $\alpha$ decay of the energy released is relatively small, the current nuclear power plant is utilized by the fission of uranium nuclei release of energy, so A error. B, irradiation of a certain metal with violet light can be photoelectric effect can be known that the frequency of violet light is greater than the metal's limiting frequency, the frequency of red light is less than the frequency of violet light, red irradiation does not necessarily produce the photoelectric effect, so B error; C, the hydrogen atom outside the nuclear electron from a higher energy level jump to a lower energy level, the energy decreases to release a certain frequency of photons, so C error. error; C, the hydrogen atom's electrons outside the nucleus from the higher energy level jump to the lower energy level, the energy decreases, to release a certain frequency of photons, so C error D, interference and diffraction are Porter has the nature of mechanical and electromagnetic waves have interference, diffraction properties, the Therefore, D is correct. Therefore, D is correct. This question examines the photoelectric effect, Bohr's theory and other basic knowledge, the key to familiarize themselves with the textbook, more accumulation, remember these basic knowledge.

Question 24: 26. As shown in the figure, there is water in the container, $P M$ is the water surface, and a beam ...

26. As shown in the figure, there is water in the container, $P M$ is the water surface, and a beam of white light is emitted from the point $A$ to the point $O$ on the surface of the water, and after it is emitted to the point $O$ on the surface of the water. The refracted light is dispersed and shines between $a , b$ on the vessel wall, corresponding to $a , b$ two colors of monochromatic light, then ![](/images/questions/phys-physical-optics/image-016.jpg)

  • A. A. If the luminous point $A$ is unchanged and the incident point $O$ is shifted to the left, then the light from $b$ may be totally reflected.
  • B. B. If $a$ light irradiates a metal and the photoelectric effect occurs, $b$ light irradiates the metal and the photoelectric effect does not necessarily occur.
  • C. C. Experiments were done with $a$ light and $b$ light on the same double-slit interference experimental setup, and the fringes of the $a$ light were wider spaced out
  • D. D. The photon energy of $a$ light is greater than the photon energy of $b$ light.

Answer: C

Solution: A. If the luminous point $A$ is unchanged and the incident point $O$ is shifted to the right, total reflection of $b$ light is possible, so A is wrong; B. From the meaning of the question, it can be seen that $b$ has a higher refractive index, then $b$ has a higher frequency, and if $a$ light irradiated by a certain metal can occur photoelectric effect, then $b$ light irradiated by the metal must also occur photoelectric effect, so B is wrong. Therefore, B is wrong; C. From the meaning of the question, it can be seen that $b$ light has a shorter wavelength, so in the same set of double-slit interference experimental device, the interference fringe spacing of $b$ light is smaller than $a$ light, so C is correct; D. By the photon energy formula $\varepsilon = h v$, it can be seen that $b$ light has a higher photon energy, so D is wrong.

Question 25: 27 . The following statements about the phenomenon of light are correct a thin slice of light ![](...

27 . The following statements about the phenomenon of light are correct a thin slice of light ![](/images/questions/phys-physical-optics/image-017.jpg) A ![](/images/questions/phys-physical-optics/image-018.jpg) B ![](/images/questions/phys-physical-optics/image-019.jpg) C ![](/images/questions/phys-physical-optics/image-020.jpg) D

  • A. A. Figure A shows the interference pattern obtained when the flatness of the surface of a workpiece is examined by interferometric methods, the curved interference fringes indicate that the plane being examined is raised here
  • B. B. The stereoscopic movie in Figure B is based on the principle of interference of light.
  • C. C. Figure C shows the interference pattern obtained by passing a monochromatic light through a double slit
  • D. D. Fig. Din shows the diffraction pattern formed behind the disk when light hits the disk

Answer: D

Solution: A. Figure A, the interference fringe bending position is the air film layer thickness of the region in advance of the emergence of the air film layer thickness of the thicker region of the fringe can be known, the curved interference fringe shows that the detected plane here is concave, A error; B. Figure B shows the principle of stereoscopic film is the use of light polarization, B error; C. Interference fringe is equal width, equal brightness, diffraction fringe is the central fringe is the brightest, widest, to the sides of the darker and narrower, C is a monochromatic light through the single slit after the diffraction fringe, C error; D. Figure D for the diffraction pattern obtained by light to the disk, if the light irradiation opaque disk, behind the center of a bright spot will appear, which is known as Poisson bright spot, D is correct;

Question 26: 28. Light in science and technology, production and life has a wide range of applications, the follo...

28. Light in science and technology, production and life has a wide range of applications, the following statement is correct

  • A. A. Soap bubbles floating in the air often look colored, which is a diffraction phenomenon of light
  • B. B. On a clear summer day, at noon, you see the pavement in the distance especially bright, as if there is water, which is the phenomenon of diffraction of light
  • C. C. Checking the flatness of an optical plane with a standard plane sample, which utilizes the interference of light
  • D. D. When photographing swimming fish in a pool, a polarizing filter in front of the camera can filter out light reflected from the water surface, which indicates that the light wave is a longitudinal wave.

Answer: C

Solution: A. Soap bubbles floating in the air often look colored, which is the interference phenomenon of light, option A error; B. Sunny summer, at noon to see the far side of the road is particularly bright, as if there is water, which is the phenomenon of total reflection of light, option B error; C. Check the flatness of the optical plane with a standard plane sample, which is the use of light interference, option C is correct; D. Photographing swimming fish in the pool in front of the camera with a polarizing filter can filter out the reflected light from the water surface, which indicates that the light wave is transverse to the surface of the water. waves, option D is wrong.

Question 27: 29. China's scientific and technological test satellite for solar exploration, Xixihe, has realized ...

29. China's scientific and technological test satellite for solar exploration, Xixihe, has realized for the first time in the international arena the spectral imaging of the solar ${ } ^ { \mathrm { H } }$ band. Space observation. The light emitted when a hydrogen atom leaps from the $n = 3 , 4 , 5 , 6$ energy level to the $n = 2$ energy level corresponds to a spectral line in the visible region. The corresponding spectral lines in the visible region are the four spectral lines of $\mathrm { H } _ { \alpha } , \mathrm { H } _ { \beta } , \mathrm { H } _ { \gamma } , \mathrm { H } _ { \delta }$, as shown in the figure. The following statement is correct () | $n$ | | | | | $E / \mathrm { eV }$ | | :--- | :--- | :--- | :--- | :--- | :--- | | 4 | | | | | | - 0.85 | | 3 | | | | | | | | $\mathrm { H } _ { \alpha }$ | $\mathrm { H } _ { \beta }$ | $\mathrm { H } _ { \gamma }$ | $\mathrm { H } _ { \delta }$ | - 1.51 | | 2 | | | | | | | 1 -13.6

  • A. A. $\mathrm { H } _ { \alpha } , \mathrm { H } _ { \beta } , \mathrm { H } _ { \gamma } , \mathrm { H } _ { \delta }$ The wavelengths of the four lights get progressively larger
  • B. B. $\mathrm { H } _ { \alpha } , \mathrm { H } _ { \beta } , \mathrm { H } _ { \gamma } , \mathrm { H } _ { \delta }$ Four types of light propagate progressively faster in a vacuum
  • C. C. When the same double-slit interference setup is illuminated with $\mathrm { H } _ { \alpha , } \mathrm { H } _ { \beta , } \mathrm { H } _ { \gamma } , \mathrm { H } _ { \delta }$ of four types of light, the neighboring fringes corresponding to $\mathrm { H } _ { \delta }$ are less spaced apart.
  • D. D. If the photoelectric effect occurs when $\mathrm { H } ^ { \gamma }$ irradiates a metal, then the photoelectric effect must occur when $\mathrm { H } _ { \beta }$ irradiates the metal.

Answer: C

Solution: A. Of the four, the light that jumps from a higher energy level to the $n = 2$ energy level has more energetic photons. According to $\varepsilon = h v = \frac { h c } { \lambda }$ INLINE_FORMULA_2]], the energies of the four types of light gradually increase, so the wavelengths gradually become smaller, so A is wrong; B. The propagation speed of electromagnetic waves in vacuum is the speed of light $c$, so B is wrong; C. According to the double-slit interference fringe spacing formula $\Delta x = \frac { l } { d } \lambda$ C. According to the formula $\Delta x = \frac { l } { d } \lambda$ of double-slit interference, the smaller the wavelength, the smaller the spacing between neighboring fringes, in which the wavelength of ${ } ^ { \mathrm { H } }$ is the smallest, so C is correct; D. The frequency of $\mathrm { H } _ { { } ^ { \gamma } \text { 能让金属发生光电效应,说明其频率超过材料的截止频率. } } \mathrm { H } _ { \beta }$ is lower than that of $\mathrm { H } _ { { } ^ { \gamma } }$, so it cannot be determined. D. The frequency of $\mathrm { H } _ { \beta }$ is lower than that of $\mathrm { H } _ { { } ^ { \gamma } }$, so it is not possible to determine whether the frequency of $\mathrm { H } _ { \beta }$ can be greater than the cutoff frequency of the material.

Question 28: 30. The following statements about the phenomenon of light are correct ( )

30. The following statements about the phenomenon of light are correct ( )

  • A. A. Rainbows in the sky after rain are an interference phenomenon.
  • B. B. The apparent depth of the water is shallower than the actual depth, which is a phenomenon of total reflection of light.
  • C. C. The blurred edges of the shadow of a shaving blade are diffraction of light
  • D. D. The appearance of colored patterns in the oil film on the water surface under sunlight is a dispersion phenomenon of light

Answer: C

Solution: A: The rainbow in the sky after rain is the phenomenon of dispersion of light, so A is wrong; B: The apparent depth of water is shallower than the actual depth is the phenomenon of light refraction, so B is wrong; C: The blurring of the edge of the shadow of the razor blade is the diffraction of light, so C is correct; D: in the sunlight irradiation, the oil film on the surface of the water appeared color pattern is the interference phenomenon of light, so D error.

Question 29: 31. As shown in the figure for an experimental setup to show the phenomenon of thin film interferenc...

31. As shown in the figure for an experimental setup to show the phenomenon of thin film interference, P is attached to the soap film of the wire ring placed straight, due to gravity, the formation of thin on the bottom of the thick wedge, S is a lit alcohol lamp, to the flame sprinkled with some salt, in the soap film on the observation of interference, if the experiment will be the metal coil in the plane in which it is slowly rotated, the phenomenon observed ( ![](/images/questions/phys-physical-optics/image-021.jpg) ![](/images/questions/phys-physical-optics/image-022.jpg) ![](/images/questions/phys-physical-optics/image-023.jpg)

  • A. A. Interference fringes remain constant
  • B. B. When the metal coil rotates $30 ^ { \circ }$, the interference fringes rotate in the same direction $30 ^ { \circ }$.
  • C. C. When the metal coil rotates $45 ^ { \circ }$, the interference fringes rotate in the same direction $90 ^ { \circ }$.
  • D. D. When the metal coil rotates $60 ^ { \circ }$, the interference fringes rotate in the same direction $30 ^ { \circ }$.

Answer: A

Solution: In the process of turning the metal wire circle, because the thickness of the two layers of film is unchanged, so the optical range difference formed by the reflected light will not change, so the resulting interference fringes remain unchanged from the original state.

Question 30: 32. Among the following natural phenomena that can be explained by the theory of interference of lig...

32. Among the following natural phenomena that can be explained by the theory of interference of light are

  • A. A. Rainbow in the sky
  • B. B. Sunlight creates circular patches of light on the forest floor
  • C. C. Colorful patterns on soap bubbles
  • D. D. Very bright air bubbles in the water

Answer: C

Solution: A. The rainbow in the sky is formed because of the reflection and refraction of light; B.The spots formed by sunlight on the forest floor are due to the formation of light traveling in a straight line, B is wrong; C. Soap bubbles on the color pattern, is due to the formation of light interference, C is correct; D. The very bright air bubbles in the water are formed due to total reflection, C error.

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

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

  • A. A. Only visible light can be polarized, invisible light cannot be polarized
  • B. B. When photographing objects in the water, a polarizer in front of the camera lens can attenuate the effect of reflected light from the water surface.
  • C. C. Holograms are taken using the principle of diffraction of light
  • D. D. Red light enters water from air, the wavelength becomes longer and the color remains the same

Answer: B

Solution: A. Polarization is a characteristic of transverse waves, all transverse waves (including visible and invisible light) can be polarized. Invisible light (such as infrared, ultraviolet) with electromagnetic waves, belonging to the transverse wave, so A error; B. The light reflected from the water surface is partially polarized light, and the polarizer can filter the polarized light in a specific direction, thus reducing the impact of reflected light, so B is correct; C. Holograms use light interference (object light and reference light superimposed on the formation of interference fringes) to record information, rather than diffraction, so C error; D. Red light into the water when the frequency remains unchanged (color unchanged), but the speed of the wave decreases, by $v = \lambda f$ can be shortened by the wavelength $\lambda$, so D error.

Question 32: 34. September 23, 2023 to October 8, 2023 will be held in Hangzhou, Zhejiang Province, the 19th Asia...

34. September 23, 2023 to October 8, 2023 will be held in Hangzhou, Zhejiang Province, the 19th Asian Games, is located in the southwest of the Asian Games Village and about 36 kilometers away from the Fuyang District Water Sports Center for the Asian Games rowing, canoeing (including static water, slalom) project competitions and training venues, and at the same time, through the laying of different lights, projections, water curtains, etc., to create a set of large-scale water shows, light shows, Naked eye 3D light and shadow show and other scenes as one of the night tour light and shadow version of the "Dwelling in the Fuchun Mountains", presenting "50 miles of Spring River Flower and Moon Night" scene, night tour of the water sports center is a good choice for tourists. The following statements about the phenomenon of light are correct ![](/images/questions/phys-physical-optics/image-024.jpg)

  • A. A. Interference may occur in the superimposed region when blue light and red light are dimmed to the same brightness
  • B. B. Blue light and red light travel at different speeds in water, with blue light traveling a bit faster
  • C. C. Blue light is directed obliquely into the water at the same angle of incidence as red light, and the red light is deflected at a greater angle from its original direction
  • D. D. When watching a 3D movie in a movie theater, the glasses worn by the viewer are a pair of polarizers with perpendicular transmittance directions.

Answer: D

Solution: A. Blue light and red light of different frequencies, in the superposition region can not occur interference phenomena, so A error; B. Blue light is higher than the frequency of red light, the speed of propagation in water is different, blue light speed is smaller, so B error; C. Blue light is higher than the frequency of red light, the refractive index is larger, blue light and red light with the same angle of incidence into the water obliquely, red light deviates from the original direction of the angle is smaller, so C error; D. Cinema to watch 3D movies, the audience wears glasses is a pair of perpendicular to the direction of transmittance polarizer, so D is correct.

Question 33: 35. As shown in the figure is a schematic diagram to study the two-slit interference of light, there...

35. As shown in the figure is a schematic diagram to study the two-slit interference of light, there are two slits $S _ { 1 } , S _ { 2 }$ on the baffle plate, the two beams of light emitted by the ${ } ^ { S _ { 1 } }$ and ${ } ^ { S _ { 2 } }$ reach the screen when the interference fringes will be produced, it is known that the wavelength of incident light is $\lambda$, and the points ${ } ^ { P }$ on the screen to the two slits ${ } ^ { S }$ and ${ } ^ { P }$ will produce interference fringes. FORMULA_3]], and the distance from the ${ } ^ { P }$ point on the screen to the two slits ${ } ^ { S }$ and ${ } ^ { S }$ is equal, and if the bright fringe at ${ } ^ { P }$ is recorded as bright stripe No. 0, then the interference fringe from ${ } ^ { S }$ will be generated by ${ } ^ { S }$ and ${ } ^ { S }$. INLINE_FORMULA_8]] upwards, the bright stripe at ${ } ^ { P }$ is exactly the 5th bright stripe. If the length of the line ${ } ^ { S } P _ { 1 }$ is $y _ { 1 } , S _ { 2 } P _ { 1 }$ and the length of the line $y _ { 2 }$ is $y _ { 2 } - y _ { 1 }$, then $y _ { 2 } - y _ { 1 }$ is equal to () ![](/images/questions/phys-physical-optics/image-025.jpg)

  • A. A. $5 \lambda$
  • B. B. $7.5 \lambda$
  • C. C. $10 \lambda$
  • D. D. $15 \lambda$

Answer: A

Solution: Based on $\Delta s = y _ { 2 } - y _ { 1 } = n \lambda ( n = 012 , \ldots )$ Since the bright pattern at ${ } ^ { P }$ is exactly the 5th bright pattern, it can be seen that $n = 5$, i.e. $y _ { 2 } - y _ { 1 }$ is equal to ${ } ^ { 5 \lambda }$.

Question 34: 36. In 1801, Thomas Young with a double-slit interference experiment to study the nature of light wa...

36. In 1801, Thomas Young with a double-slit interference experiment to study the nature of light waves. 1834, the use of single-sided mirrors Loeb also obtained the results of Young's interference (called Loeb mirror experiment). The basic device of the Loet's mirror experiment is shown in the figure, an experimental group used the experimental device to investigate the interference phenomenon of light. $S$ is a monochromatic light source, $S$ emits light directly on the light screen, at the same time $S$ emitted light is also through the plane mirror $M$ reflected in the light screen from the plane mirror $M$. FORMULA_4]], the light reflected from the plane mirror $S$ is equivalent to that emitted by the virtual image of $S$ in the plane mirror, so that the formation of two of the same coherent light source, the formation of light and dark stripes on the light screen. Let the wavelength of monochromatic light is $\lambda$, the distance from the light source S to the plane mirror is $a$, the horizontal distance to the left endpoint of the plane mirror is $b$, the horizontal distance from the left endpoint of the plane mirror to the light screen is $C$. Then the following judgment is correct ( ) ![](/images/questions/phys-physical-optics/image-026.jpg)

  • A. A. The distance between two adjacent bright stripes is $\frac { c } { 2 a } \lambda$
  • B. B. The distance between two adjacent bright stripes is $\frac { b + c } { a } \lambda$
  • C. C. If the plane mirror M is panned upward a little, the central distance between two adjacent bright stripes becomes larger
  • D. D. If the plane mirror M is shifted slightly to the right, the central spacing of the two adjacent bright stripes becomes smaller

Answer: C

Solution: AB . The central spacing of two adjacent bright stripes is the interference fringe spacing $\Delta x = \frac { L \lambda } { d }$ The meaning of the question is known $d = 2 a , L = b + c$ The distance between two neighboring bright stripes is $\Delta x = \frac { ( b + c ) \lambda } { 2 a }$ Therefore, AB is wrong; CD . If the plane mirror M is moved upward a little, that is, the double-slit spacing $d$ is reduced, the central distance between the two adjacent bright stripes becomes larger; if the plane mirror M is moved to the right a little, the double-slit spacing $d$ and the distance from the double-slit to the light screen both remain unchanged, and the central distance between the two adjacent bright stripes remains unchanged. Therefore, C is correct and D is incorrect.

Question 35: 37. As shown in the figure, with the interference method to check the flatness of the plane, with a ...

37. As shown in the figure, with the interference method to check the flatness of the plane, with a monochromatic light from above, from the top of the sample plate to the downward observation can be seen interference fringes. Then ( ) ![](/images/questions/phys-physical-optics/image-027.jpg) ![](/images/questions/phys-physical-optics/image-028.jpg) ![](/images/questions/phys-physical-optics/image-029.jpg)

  • A. A. To make the stripes thinner, you can move the sheet to the right
  • B. B. If the interference fringe is curved as shown in Figure B, it indicates that the corresponding position of the measured surface is concave downward.
  • C. C. Move the sample plate up parallel, with the stripes moving toward the split tip
  • D. D. The spacing of the interference fringes formed by irradiating with yellow light is smaller than when irradiating with green light

Answer: C

Solution: A. The difference in thickness of the air film between neighboring bright stripes is equal to half a wavelength, when the sheet is moved a little to the right, the split angle of the air film becomes larger, and the distance between the positions where the difference in thickness of the air film is equal to half a wavelength becomes smaller, so the spacing between neighboring bright stripes becomes smaller, and the stripes become denser, so A is wrong; B. Air thin layer interference is equal thickness interference, that is, the thickness of the air film at the bright stripes is the same, from the curved stripes can be seen, the thickness of the air film at the left side of the inspection plane is the same as the thickness of the air film at the back of the curved, so the bend is convex, so B is wrong; C. According to the position of the stripes and the thickness of the air film is corresponding, when the sample plate parallel upward, the same thickness of the air film to the tip of the cleavage to move, so the stripes toward the tip of the cleavage to move, so C is correct; D. Stripe spacing is directly proportional to the wavelength of light, due to the long wavelength of yellow light, so the yellow light stripe spacing is large, green light stripe spacing is small, so D is wrong.

Question 36: 38. In the double-slit interference experiment, the light source emits orange light with a wavelengt...

38. In the double-slit interference experiment, the light source emits orange light with a wavelength of $6.0 \times 10 ^ { - 7 } \mathrm {~m}$, the orange interference fringes are obtained on the light screen, and the $A$ point on the light screen is exactly the second bright stripe from the center stripe. Other conditions remain unchanged, now change to other colors of visible light for the experiment, the $A$ point on the light screen is the location of the dark fringe, the frequency range of visible light is $3.9 \times 10 ^ { 14 } \sim 7.5 \times 10 ^ { 14 } \mathrm {~Hz}$, then the wavelength of the incident light may be ()

  • A. A. $8.0 \times 10 ^ { - 7 } \mathrm {~m}$
  • B. B. $4.8 \times 10 ^ { - 7 } \mathrm {~m}$
  • C. C. $4.0 \times 10 ^ { - 7 } \mathrm {~m}$
  • D. D. $3.4 \times 10 ^ { - 7 } \mathrm {~m}$

Answer: B

Solution: AD. The frequency range of visible light is $3.9 \times 10 ^ { 14 } \sim 7.5 \times 10 ^ { 14 } \mathrm {~Hz}$, based on the equation $$ c = \lambda f $$ The wavelength range is $4.0 \times 10 ^ { - 7 } \sim 7.69 \times 10 ^ { - 7 } \mathrm {~m}$. AD is wrong; B. According to the formula $$ \Delta r = \frac { n \lambda } { 2 } \quad ( n \text { 为整数 } ) $$ It can be seen that bright stripes appear where $n$ is even, and dark stripes appear where $n$ is odd, so the second bright stripe from the center stripe appears at $n = 4$, so the distance from the point $A$ to the point $A$ is about 1.2 mm. Difference in the optical range of the two slits $$ \Delta r = 4 \times \frac { 6 \times 10 ^ { - 7 } } { 2 } \mathrm {~m} = 1.2 \times 10 ^ { - 6 } \mathrm {~m} $$ For the dark stripe to appear, $n$ needs to be an odd number in order to appear, and when the wavelength of the incident light is $4.8 \times 10 ^ { - 7 } \mathrm {~m}$, it is necessary to take an odd number. $$ 1.2 \times 10 ^ { - 6 } \mathrm {~m} = n \frac { 4.8 \times 10 ^ { - 7 } } { 2 } \mathrm {~m} $$ $n = 5$ is odd, so dark stripes appear at $A$, option B is correct. C. When the wavelength of incident light is $4.0 \times 10 ^ { - 7 } \mathrm {~m}$ $$ 1.2 \times 10 ^ { - 6 } \mathrm {~m} = n \times \frac { 4 \times 10 ^ { - 7 } } { 2 } \mathrm {~m} $$ $n = 6$ is even, so bright streaks appear at $A$, and C is wrong.

Question 37: 39. A student used the device shown below to measure the wavelength of a monochromatic light. During...

39. A student used the device shown below to measure the wavelength of a monochromatic light. During the experiment, the power supply to make the light source normal light; adjust the optical path, so that from the eyepiece can be observed in the interference fringes. If you want to increase the number of fringes observed from the eyepiece, the student can ![](/images/questions/phys-physical-optics/image-030.jpg)

  • A. A. Move the single seam closer to the double seam
  • B. B. Move the screen closer to the double slit
  • C. C. Move the screen away from the double slit
  • D. D. Use of double seams with smaller spacing

Answer: B

Solution: If you want to increase the number of stripes observed from the eyepiece, you need to decrease the stripe spacing, which is given by the equation $$ \Delta x = \frac { L } { d } \lambda $$ The distance from the double slit to the screen $L$ or the distance between the double slits $d$ needs to be decreased, so B is correct, A is wrong, C is wrong, and D is wrong.

Question 38: 40. As shown in the figure, two beams of monochromatic light of different colors $a , b$ parallel to...

40. As shown in the figure, two beams of monochromatic light of different colors $a , b$ parallel to the bottom edge of the triangular mirror $B C$ from the $A B$ side of the injection, the refraction by the triangular mirror intersects at the point $P$, the following statements are correct is ( ) ![](/images/questions/phys-physical-optics/image-031.jpg) High School Physics Assignment, October 30, 2025

  • A. A. The refractive index of a prism for $a$ light is less than that for $b$ light.
  • B. B. In a prism $a$ the speed of propagation of light is greater than the speed of propagation of $b$ light
  • C. C. The critical angle of total reflection for $a$ light from the same medium is greater than the critical angle of total reflection for $b$ light.
  • D. D. In diffraction experiments using $a$ light and $b$ light, the experimental phenomenon of $b$ light is more obvious.

Answer: D

Solution:
Back to Topics

Physical Optics

物理光学

38 Practice Questions

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

Topic Overview

Physical optics focuses on the study of the fluctuations of light, including phenomena such as interference, diffraction and polarization. In the CSCA exam, this section is often examined through diagrams of experimental set-ups, identification of phenomena and principle analysis questions, which require candidates to be able to distinguish between different optical phenomena and understand the conditions under which they arise.

Questions:38

Key Points

  • 1Characterization of interference conditions and fringes of light
  • 2Recognition and Comparison of Single-Slit/Circular-Aperture Diffraction Patterns
  • 3Polarized light generation and application scenarios
  • 4Optical phenomena in practical experiments to distinguish judgment

Study Tips

It is recommended to memorize the core features of interference and diffraction by comparing and contrasting them, and to understand the experimental setup and phenomenon correspondences by combining more schematic diagrams.

Practicing topics ≠ Passing the exam

Full mock exam based on official syllabus, with mixed topics like the real test