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Optics - Practice Questions (28)

Question 1: A beam of light from the air into a medium, the incident light and reflected light between the angle...

A beam of light from the air into a medium, the incident light and reflected light between the angle of $90 ^ { \circ }$ , refracted light and reflected light between the angle of $105 ^ { \circ }$ , then the refractive index of the medium ()

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

Answer: B

Solution: [Knowledge Points] The law of refraction of light [Explanation] Because the angle between the incident ray and the reflected ray is $90 ^ { \circ }$, according to the law of reflection of light, the angle of incidence is $90 ^ { \circ }$. $$ i = 45 ^ { \circ } $$ The angle of refraction is $105 ^ { \circ }$, and the angle of refraction is $$ i = 45 ^ { \circ } $$. $$ r = 30 ^ { \circ } $$ According to the law of refraction of light $$ n = \frac { \sin i } { \sin r } = \frac { \sin 45 ^ { \circ } } { \sin 30 ^ { \circ } } = \sqrt { 2 } $$ ACD is wrong, B is correct.

Question 2: A ray of light is directed from air into a medium with refractive index $\sqrt { 2 }$, and the angle...

A ray of light is directed from air into a medium with refractive index $\sqrt { 2 }$, and the angle of incidence at the interface is $45 ^ { \circ }$ . Some of the incident light is reflected and some is refracted. The angle between the reflected and refracted rays is ()

  • A. A. $.75 ^ { \circ }$
  • B. B. $90 ^ { \circ }$
  • C. C. $105 ^ { \circ }$
  • D. D. $120 ^ { \circ }$

Answer: C

Solution: Knowledge point] The law of refraction of light [Detailed Explanation]As shown in the figure, [IMAGE_4]. ![](/images/questions/phys-optics/bbdf5a480c78.jpg) According to the law of refraction $$ \frac { \sin i } { \sin r } = n $$ then $\sin r = \frac { \sin i } { n } = \frac { \sin 45 ^ { \circ } } { \sqrt { 2 } } = \frac { 1 } { 2 }$ $$ r = 30 ^ { \circ } $$ The angle between the reflected and refracted rays $$ \theta = 180 ^ { \circ } - 45 ^ { \circ } - 30 ^ { \circ } = 105 ^ { \circ } $$

Question 3: Light to $30 ^ { \circ }$ angle of incidence from the glass to the interface between the glass and t...

Light to $30 ^ { \circ }$ angle of incidence from the glass to the interface between the glass and the air, it reflects the light and refracts the light at an angle of $90 ^ { \circ }$, then the refractive index of the glass should be ( )

  • A. A. 0.866
  • B. B. 1.732
  • C. C. 1.414
  • D. D. 1.500

Answer: B

Solution: [Knowledge Points] The law of refraction of light [Detailed Explanation]Because the angle between the reflected light and the refracted light is $90 ^ { \circ }$, we can see from the figure that the angle of refraction is $r = 60 ^ { \circ }$ ![](/images/questions/phys-optics/530ea8d75b2d.jpg) Therefore the refractive index $$ n = \frac { \sin 60 ^ { \circ } } { \sin 30 ^ { \circ } } = \sqrt { 3 } = 1.732 $$

Question 4: As shown in the figure, a fine beam of monochromatic light from the semicircular glass block on the ...

As shown in the figure, a fine beam of monochromatic light from the semicircular glass block on the $P$ point along the radius direction to the center of the circle $O$, just in the bottom surface of the glass block after the total reflection from the glass block, monochromatic light in the glass block under the surface of the angle of reflection for the $53 ^ { \circ }$, take $\sin 53 ^ { \circ } = \frac { 4 } { 5 }$ ,the refractive index of the glass block to the monochromatic light is ( ) ![](/images/questions/phys-optics/238b858a191d.jpg)

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

Answer: C

Solution: Knowledge Points] Conditions for total reflection, critical angle [Detailed explanation] According to the refractive index and the critical conditions of total reflection can be seen $$ n = \frac { 1 } { \sin C } = \frac { 1 } { \sin 53 ^ { \circ } } = \frac { 5 } { 4 } $$

Question 5: As shown in Figure 1, a beam of light is directed from glass to air at an angle of incidence of $45 ...

As shown in Figure 1, a beam of light is directed from glass to air at an angle of incidence of $45 ^ { \circ }$ . Known that the beam of light from the glass to the air when the critical angle of $42 ^ { \circ }$ , the following four light path diagrams in the correct ( )

  • A. A. ![](/images/questions/phys-optics/fa486867190e.jpg)
  • B. B. ![](/images/questions/phys-optics/fa0c2c561dec.jpg)
  • C. C. ![](/images/questions/phys-optics/07eecc9d5d01.jpg)
  • D. D. ![](/images/questions/phys-optics/ef114552b9ff.jpg)

Answer: A

Solution: Knowledge Points] Conditions for Total Reflection, Critical Angle The critical angle of glass is $\mathrm { C } = 42 ^ { \circ }$. The angle of incidence is $\mathrm { i } = 45 ^ { \circ } > \mathrm { C }$ when the light rays from the glass are directed into the air, so the light rays will undergo total reflection at the interface between the glass and the air, and all the light rays will be reflected back to the glass. A. The diagram is consistent with the conclusion, option A is correct; B. The diagram is not consistent with the conclusion, option B is wrong; C. The diagram is not consistent with the conclusion, option C is wrong; D. The graph does not match the conclusion, option D is wrong;

Question 6: As shown in the figure, a beam of monochromatic light from a rectangular glass block on the upper su...

As shown in the figure, a beam of monochromatic light from a rectangular glass block on the upper surface of the $M$ point at a certain angle into the glass block, now slowly change the angle of incidence $\alpha$, when $\alpha = 45 ^ { \circ }$ when the right side of the block can not be seen in the light out. The refractive index of this glass block is ( ) ![](/images/questions/phys-optics/267581446668.jpg)

  • A. A. $\frac { 3 } { 2 }$
  • B. B. $\frac { \sqrt { 6 } } { 2 }$
  • C. C. $\frac { \sqrt { 5 } } { 2 }$
  • D. D. $\frac { \sqrt { 7 } } { 2 }$

Answer: B

Solution: Knowledge]The law of refraction of light, the conditions of total reflection, the critical angle [Detailed explanation] according to the question, draw the light path, as shown in the figure ![](/images/questions/phys-optics/c990116da952.jpg) According to the law of refraction, we have $n = \frac { \sin \alpha } { \sin \left( 90 ^ { \circ } - C \right) }$ and $n = \frac { 1 } { \sin C }$ Substituting the data solves for $n = \frac { \sqrt { 6 } } { 2 }$

Question 7: As shown in the figure, the dotted line represents the interface of the two media and its normal, th...

As shown in the figure, the dotted line represents the interface of the two media and its normal, the solid line represents a beam of light directed to the interface after the occurrence of reflection and refraction of the light, the following statements are correct ( ) ![](/images/questions/phys-optics/9ee8d0d358d3.jpg)

  • A. A. $O b$ may be the incident line of light
  • B. B. $a O$ must be the incident line of light
  • C. C. $O c$ may be the incident line of light
  • D. D. $O b$ must not be a reflected ray.

Answer: B

Solution: Knowledge point] The law of refraction of light [Detailed explanation] by the law of reflection can be known, the angle of reflection is equal to the angle of incidence, the incident light and the reflected ray of light on both sides of the normal; by the law of refraction can be known, the angle of refraction and the angle of incidence is not equal, and divided into the normal on both sides. In summary, $P Q$ is the normal, $M N$ is the interface, $a O$ is the incident ray, $O b$ is the reflected ray, $O c$ is the refracted ray. $O c$ is the refracted ray.

Question 8: A beam of complex color light from the air to the glass, refraction and divided into $a , b$ two bea...

A beam of complex color light from the air to the glass, refraction and divided into $a , b$ two beams of monochromatic light, the direction of propagation is shown in the figure. Let the refractive index of glass to $a , b$ for ${ } ^ { n _ { a } }$ and ${ } ^ { n _ { b } } , ~ a , b$ in the glass propagation speed of ${ } ^ { v _ { a } }$ and ${ } ^ { v _ { b } }$, then ([IMAGE_6]]). ![](/images/questions/phys-optics/cdb2f565c441.jpg)

  • A. A. $n _ { a } = n _ { b }$
  • B. B. $n _ { a } < n _ { b }$
  • C. C. $v _ { a } > v _ { b }$
  • D. D. $v _ { a } < v _ { b }$

Answer: D

Solution: Knowledge Points] The law of refraction of light, the wavelength of the refractive index expression and speed expression [Detailed explanation] AB. According to the law of refraction $$ n = \frac { \sin i } { \sin r } $$ According to the law of refraction $$ n = \frac { \sin i } { \sin r } $$, the angle of incidence of two beams of monochromatic light is the same as the angle of incidence of $a , b$, and the angle of refraction of $a$ is less than the angle of refraction of $b$, the refractive index of $a$ is greater than the refractive index of $b$ light. _7]] is greater than the refractive index of $b$, and option AB is incorrect; CD. $$ v = \frac { c } { n } $$ Since the refractive index of $a$ light is greater than the refractive index of $b$ light, then there are $$ v _ { a } < v _ { b } $$ Option C is wrong, D is correct.

Question 9: A beam of light from the air to a liquid, the angle of incidence is $45 ^ { \circ }$ , the angle bet...

A beam of light from the air to a liquid, the angle of incidence is $45 ^ { \circ }$ , the angle between the reflected and refracted light is $105 ^ { \circ }$ , then the refractive index of light in the liquid is

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

Answer: C

Solution: [Knowledge Points] The law of refraction of light [Explanation] By the law of reflection, the angle of reflection is $45 ^ { \circ }$, from the geometric relationship, the angle of refraction is $30 ^ { \circ }$, so the refractive index of light in the liquid is $$ n = \frac { \sin 45 ^ { \circ } } { \sin 30 ^ { \circ } } = \sqrt { 2 } $$ C is correct.

Question 10: The speed of propagation of light in a certain glass is $\sqrt { 3 } \times 10 ^ { 8 } \mathrm {~m} ...

The speed of propagation of light in a certain glass is $\sqrt { 3 } \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s }$ ,To make the refracted and reflected rays of light from the glass into the air into $90 ^ { \circ }$ ,then the angle of incidence should be

  • A. A. $30 ^ { \circ }$
  • B. B. $60 ^ { \circ }$
  • C. C. $45 ^ { \circ }$
  • D. D. $90 ^ { \circ }$

Answer: A

Solution: Knowledge points] The law of refraction of light [Detailed explanation] According to the question to make the light path diagram, as shown below ![](/images/questions/phys-optics/c4e40bc2ad22.jpg) The refractive index is $$ n = \frac { c } { v } = \sqrt { 3 } $$ Let the angle of incidence be ${ } ^ { \theta }$ ,then we have $$ n = \frac { \sin \theta _ { 2 } } { \sin \theta _ { 1 } } = \frac { \cos \theta _ { 1 } } { \sin \theta _ { 1 } } $$ The solution is given by $$ n = \frac { \sin \theta _ { 2 } } { \sin \theta _ { 1 } } = \frac { \cos \theta _ { 1 } } { \sin \theta _ { 1 } } $$ $$ \theta _ { 1 } = 30 ^ { \circ } $$

Question 11: As shown in the figure is a beam of monochromatic light from a medium into the air medium occurs in ...

As shown in the figure is a beam of monochromatic light from a medium into the air medium occurs in the phenomenon, known as the speed of light in the air is c , then the monochromatic light in the medium of the wave speed is ![](/images/questions/phys-optics/b360ce83a1e2.jpg)

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

Answer: D

Solution: [Knowledge Points] The law of refraction of light [Detailed Explanation]From the figure, we can see that the angle of incidence $i = 30 ^ { \circ }$ , the angle of refraction is $r = 60 ^ { \circ }$ , then the refractive index of the medium is : $n = \frac { \sin \gamma } { \sin i } = \frac { \sin 60 ^ { \circ } } { \sin 30 ^ { \circ } } = \sqrt { 3 }$ ,the monochromatic light in the medium of the wave speed is: $v = \frac { c } { n } = \frac { c } { \sqrt { 3 } } = \frac { \sqrt { 3 } } { 3 } c$ ,so D is correct,ABC is wrong.

Question 12: A beam of monochromatic light from the air to the surface of a medium, the light path as shown in th...

A beam of monochromatic light from the air to the surface of a medium, the light path as shown in the figure, then the refractive index of the medium is about () ) ![](/images/questions/phys-optics/5ba2fcec2b60.jpg)

  • A. A. 1.50
  • B. B. 1.41
  • C. C. 0.71
  • D. D. 0.67

Answer: B

Solution: [Knowledge Points] The law of refraction of light [Detailed Explanation]From the question of monochromatic light from the air to the medium, the angle of incidence $\mathrm { i } = 45 ^ { \circ }$ , the angle of refraction $\mathrm { r } = 30 ^ { \circ }$, the medium of refraction The refractive index of the medium is $n = \frac { \sin i } { \sin r } = \sqrt { 2 } = 1.41$ , the

Question 13: As shown in the figure, a glass prism has an isosceles triangle with $a$ at $30 ^ { \circ }$. A ray ...

As shown in the figure, a glass prism has an isosceles triangle with $a$ at $30 ^ { \circ }$. A ray of light enters the prism perpendicular to the $a b$ plane and exits through the $a c$ plane. The angle between the outgoing and incoming rays is $30 ^ { \circ }$, then the refractive index of the prism material is ( ) ![](/images/questions/phys-optics/8df1e2d467d6.jpg)

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

Answer: D

Solution: [Knowledge Point]Law of Refraction of Light [Detailed Explanation]Rays of light are refracted in the $a c$ interface, the angle of incidence is $30 ^ { \circ }$, the angle of refraction is $60 ^ { \circ }$, according to the law of refraction of light. $$ n = \frac { \sin 60 ^ { \circ } } { \sin 30 ^ { \circ } } = \sqrt { 3 } $$

Question 14: As shown in the figure, the cross-section of a transparent medium for the equilateral triangle, a be...

As shown in the figure, the cross-section of a transparent medium for the equilateral triangle, a beam of monochromatic light from the $A B$ side of a point into the medium, the angle of incidence is $60 ^ { \circ }$, light from the $A C$ side of the medium. If the angle between the incident light and the incident light is $60 ^ { \circ }$, then the refractive index of the transparent medium for this monochromatic light is ( ) ![](/images/questions/phys-optics/974e6da2bafe.jpg)

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

Answer: B

Solution: [Knowledge Points] The law of refraction of light [Detailed Explanation]The light path is shown in the figure ![](/images/questions/phys-optics/a04cb715e347.jpg) The angle of incidence of light at $D$ point $\theta _ { 1 } = 60 ^ { \circ }$ ,the angle between the incident ray and the incident ray is $60 ^ { \circ }$ ,then the incident ray is parallel to the normal line through the $D$ point, according to the geometric relationship can be known, the angle of refraction $\theta _ { 3 } = 60 ^ { \circ }$ ,monochromatic light at $\theta _ { 3 } = 60 ^ { \circ }$ point ,the light path is as shown in the figure ![](/images/questions/phys-optics/a04cb715e347.jpg) $\theta _ { 3 } = 60 ^ { \circ }$, the angle of incidence of monochromatic light on the $A C$ side of $\theta _ { 2 } = 30 ^ { \circ }$, then transparent The refractive index of this monochromatic light $n = \frac { \sin \theta _ { 3 } } { \sin \theta _ { 2 } } = \sqrt { 3 }$ is the refractive index of the transparent medium.

Question 15: As shown in the figure, a beam of monochromatic light from the air to the surface of a medium, the a...

As shown in the figure, a beam of monochromatic light from the air to the surface of a medium, the angle of incidence of light $\theta _ { 1 } = 60 ^ { \circ }$, if the reflected ray perpendicular to the refracted ray, then the refractive index of the medium is ![](/images/questions/phys-optics/4d4a7e902327.jpg)

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

Answer: C

Solution: [Knowledge Points] The law of refraction of light [Detailed explanation] monochromatic light from the air to the medium, the angle of incidence $\theta _ { 1 } = 60 ^ { \circ }$, set the angle of refraction for $\theta _ { 2 }$, from the geometry of knowledge can be obtained $\theta _ { 2 } = 90 ^ { \circ } - \theta _ { 1 }$ Then the refractive index of the medium $n = \frac { \sin \theta _ { 1 } } { \sin \theta _ { 2 } } = \frac { \sin \theta _ { 1 } } { \sin \left( 90 ^ { \circ } - \theta _ { 1 } \right) } = \sqrt { 3 }$

Question 16: As shown in the problem, a beam of light with vacuum wavelength $\lambda$ is injected into a double ...

As shown in the problem, a beam of light with vacuum wavelength $\lambda$ is injected into a double slit, and the spacing of the double slits is $d$; a light screen $( l > d )$ is placed in parallel far away from the double slits $l$, and a medium with refractive index $n$ is filled with the light screen. The distance between the double slit and the light screen is filled with a medium of refractive index $n$, then the spacing of neighboring bright stripes on the light screen is ![](/images/questions/phys-optics/0c5d80656212.jpg)

  • A. A. $\frac { n \lambda l } { d }$
  • B. B. $\frac { \lambda d } { n l }$
  • C. C. $\frac { n \lambda d } { l }$
  • D. D. $\frac { \lambda l } { n d }$

Answer: D

Solution: [Knowledge point] $\Delta \mathrm { x } = \mathrm { L } \lambda / \mathrm { d }$ formula simple calculation [Details] The wavelength of light in a medium is $\lambda ^ { \prime } = \frac { v } { f } = \frac { c } { n f } = \frac { \lambda } { n }$ According to the expression of stripe spacing $\Delta x = \frac { l } { d } \lambda ^ { \prime }$ It can be obtained as $\Delta x = \frac { \lambda l } { n d }$

Question 17: In the experiment "Measuring the wavelength of light by double-slit interference", the spacing of th...

In the experiment "Measuring the wavelength of light by double-slit interference", the spacing of the double slits is $d$, the distance from the double slits to the screen is $l , 6$, and the distance between the bright fringes is $a$, then the light's wavelength is

  • A. A. $\frac { d a } { 6 l }$
  • B. B. $\frac { d a } { 5 l }$
  • C. C. $\frac { l a } { 6 d }$
  • D. D. $\frac { l a } { 5 d }$

Answer: B

Solution: [Knowledge] $\Delta \mathrm { x } = \mathrm { L } \lambda / \mathrm { d }$ formula simple calculation [Detailed explanation] The question shows that the stripe spacing $\Delta x = \frac { a } { 6 - 1 } = \frac { a } { 5 }$ Because $\Delta x = \frac { l } { d } \lambda$ The solution is $\lambda = \frac { d a } { 5 l }$

Question 18: Using parallel monochromatic light $a$ to carry out Young's double-slit interference experiment, the...

Using parallel monochromatic light $a$ to carry out Young's double-slit interference experiment, the distance between the centers of two adjacent bright fringes is ${ } ^ { X _ { a } }$; if other conditions remain unchanged, only the monochromatic light $a$ is replaced by monochromatic light [[INLINE_FORMULA_3 The distance between the centers of two adjacent bright stripes is ${ } ^ { X _ { b } }$. If $x _ { a } : x _ { b } = 2 : 3$ is known, the monochromatic $a , b$

  • A. A. The ratio of wavelengths is $3 : 2$
  • B. B. The ratio of wavelengths is $4 : 9$
  • C. C. The ratio of frequencies is $3 : 2$
  • D. D. The ratio of frequencies is $9 : 4$

Answer: C

Solution: Knowledge Points] Relationship between wavelength, frequency and wave speed, $\Delta \mathrm { x } = \mathrm { L } \lambda / \mathrm { d }$ formula for simple calculations [Detailed Explanation]AB . According to the principle of Young's double-slit interference experiment $\Delta x = \frac { l } { d } \lambda$ other conditions do not become $\frac { \lambda _ { a } } { \lambda _ { b } } = \frac { x _ { a } } { x _ { b } } = \frac { 2 } { 3 }$, so AB is wrong. AB Wrong ; According to $c = \lambda f$, we can know $\frac { f _ { a } } { f _ { b } } = \frac { \lambda _ { b } } { \lambda _ { a } } = \frac { 3 } { 2 }$, so C is correct, D is wrong.

Question 19: A student through two fingers side by side to form a slit, through this slit to look at the light-em...

A student through two fingers side by side to form a slit, through this slit to look at the light-emitting fluorescent lamps parallel to it, and observed colored stripes, this phenomenon belongs to the

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

Answer: C

Solution: [Knowledge Points] Single-slit diffraction and small hole diffraction patterns [Detailed explanation] a student through the two fingers side by side to form a slit, through this slit to look at its parallel to the light-emitting fluorescent lamps, observed the color stripes, this phenomenon belongs to the single-slit diffraction phenomenon.

Question 20: Will be two pencils side by side together, leaving a slit in the middle, through this slit to look a...

Will be two pencils side by side together, leaving a slit in the middle, through this slit to look at the fluorescent lamps parallel to it, you can observe the color stripes, which is the phenomenon of light

  • A. A. dispersion
  • B. B. reflect (in the figurative sense: to show the nature of)
  • C. C. intervene
  • D. D. diffraction

Answer: D

Solution: [Knowledge points] single-slit diffraction and small hole diffraction pattern [Detailed explanation] will be two pencils side by side together, leaving a slit in the middle, through this slit to look at its parallel to the fluorescent lamp, you can observe the color stripes, which is the phenomenon of diffraction of light.

Question 21: In a single-slit diffraction experiment, by decreasing the width of the single slit only, the screen

In a single-slit diffraction experiment, by decreasing the width of the single slit only, the screen

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

Answer: C

Solution: Knowledge point] Diffraction conditions [Details] 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 throughput, the field of view of the brightness 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 will be weakened, so the ABD is wrong, C is correct.

Question 22: When doing diffraction experiments with a water wave tank, if the vibration frequency of the oscilla...

When doing diffraction experiments with a water wave tank, if the vibration frequency of the oscillator striking the water surface is 5 Hz and the propagation speed of the water wave in the tank is $0.05 \mathrm {~m} / \mathrm { s }$, in order to observe the obvious diffraction phenomenon, the diameter of the small hole $d$ should be

  • A. A. 10 cm
  • B. B. 50 cm
  • C. C. $d > 10 \mathrm {~cm}$
  • D. D. $d < 1 \mathrm {~cm}$

Answer: D

Solution: Knowledge] Diffraction conditions, wavelength, frequency and speed of the relationship between the wave [Detailed explanation] The wavelength of water wave is $$ \lambda = \frac { v } { f } = 0.01 \mathrm {~m} = 1 \mathrm {~cm} $$ In order to have obvious diffraction phenomenon, the size of the obstacle or empty should be similar to or smaller than the wavelength, D is correct.

Question 23: The following phenomena are diffraction of light

The following phenomena are diffraction of light

  • A. A. Rainbow in the sky after rain
  • B. B. Colored streaks observed in fluorescent lamps viewed through a slit
  • C. C. Soap film appears colored when exposed to daylight
  • D. D. The bubbles in the water look especially bright

Answer: B

Solution: [Knowledge points] the conditions of total reflection, critical angle, thin film interference phenomena and principles, single-slit diffraction and small hole diffraction pattern [Detailed explanation] A. Rainbow after the rain is sunlight in the raindrops in the refraction, reflection and dispersion formation, belong to the dispersion phenomenon of light, so A does not meet the question; B. Through the slit observation fluorescent lamps appear colorful stripes, is the light around the edge of the slit diffraction phenomenon, so B meets the meaning of the question; C. Soap film colored stripes are light in the film before and after the surface of the reflection of the interference, belonging to the thin film interference, so C does not meet the meaning of the question; D. Bubbles in the water is bright because the light from the water into the bubble when the total reflection, so that more light into the human eye, so D does not meet the meaning of the question.

Question 24: As shown in the figure, a mixed beam of $a , b$ is directed from a medium to a vacuum, and it is fou...

As shown in the figure, a mixed beam of $a , b$ is directed from a medium to a vacuum, and it is found that only $b$ light is emitted from the interface, and the reflected beam is denoted by $c$. ![](/images/questions/phys-optics/748f5444e277.jpg)

  • A. A. $c$ is $a$ light.
  • B. B. In a vacuum, the speed of light in $b$ is greater than the speed of light in $a$.
  • C. C. In a vacuum, the wavelength of $b$ light is longer than that of $a$ light.
  • D. D. $a$ Light is more likely to diffract significantly

Answer: C

Solution: Knowledge Points] Conditions for total reflection, critical angle, diffraction conditions A.lightatinterfaceareaonallwilloccursreflection,$c$ is$a , b$mixedcombinedlight,therefore A wrong; B. In a vacuum, the speed of light is certain, i.e., in a vacuum, the speed of $b$ light is equal to the speed of $a$ light, so B is wrong; C. When the angle of incidence is the same, $a$ light undergoes total reflection and $b$ light does not undergo total reflection, then the critical angle of $a$ light is smaller than that of $b$ light, and according to $n = \frac { 1 } { \sin C } , b$ light is not the same as $a$ light, so $b$ light is the same as $a , b$ light, so B is wrong. The refractive index of $n = \frac { 1 } { \sin C } , b$ light is smaller than that of $a$ light, and since the smaller the refractive index is, the smaller the frequency is, and the longer the wavelength is, the wavelength of $b$ light is longer than that of $a$ light, and so C is correct; D. In a vacuum, the wavelength of $b$ light is longer than that of $a$ light, and diffraction is more likely to occur, so D is wrong.

Question 25: The four scenes shown in the figure belong to the phenomenon of interference of light is ![](/image...

The four scenes shown in the figure belong to the phenomenon of interference of light is ![](/images/questions/phys-optics/dc526cb6fd25.jpg) A ![](/images/questions/phys-optics/56cb48b7fb08.jpg) B ![](/images/questions/phys-optics/2b67f8f22b94.jpg) C ![](/images/questions/phys-optics/29d8c8f75484.jpg) D

  • A. A. Poisson Brightening
  • B. B. Colorful Soap Bubbles
  • C. C. rainbow
  • D. D. three-dimensional movie

Answer: B

Solution: [Knowledge points] light refraction phenomenon, thin film interference phenomena and principles, Poisson bright spot, polarization phenomenon and its interpretation [Detailed explanation] A. Figure A is the principle of light diffraction, A error; B. Figure B is the principle of light interference, B correct; C. The principle of Figure C is the refraction of light, C error; D. The principle of figure D is the polarization of light,D error.

Question 26: Related to the principle of polarization of light is

Related to the principle of polarization of light is

  • A. A. Poisson Brightening
  • B. B. 3D movie
  • C. C. small hole imaging
  • D. D. mirage (lit. or fig.)

Answer: B

Solution: [Knowledge points] light refraction phenomenon, single-slit diffraction and small hole diffraction pattern, Poisson bright spot, the application of polarization [Details] A. Poisson bright spot is the diffraction of light, so A error; B.3D movie playback and viewing the use of light polarization phenomenon, so B correct; C. small hole imaging is the use of light propagation along a straight line, so C error; D. mirage is due to the refraction of light in an inhomogeneous medium, so D error.

Question 27: As shown in the figure, a beam of red light from the air into a certain medium, the interface on bot...

As shown in the figure, a beam of red light from the air into a certain medium, the interface on both sides of the light and the interface angle were $45 ^ { \circ }$ and $60 ^ { \circ }$ . Then the ratio of the sine of the angle of incidence to the sine of the angle of refraction is ![](/images/questions/phys-optics/90c744ece3f5.jpg)

  • A. A. $\frac { \sqrt { 2 } } { 2 }$
  • B. B. $\frac { \sqrt { 6 } } { 3 }$
  • C. C. $\frac { \sqrt { 6 } } { 2 }$
  • D. D. $\sqrt { 2 }$ ## Reference answer | Question No. | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | Answers | B | C | B | C | D | A | B | B | D | C | | Question No. | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | | Answers | C | A | D | B | D | B | C | Question No. | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | Question No. | Question No. | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 | Answer | C | 30 | Question No. | Answers | C | D | C | D | C | C | B | C | B | B | D |

Answer: D

Solution: [Knowledge Point]Law of Refraction of Light [Details] The ratio of the sine of the angle of incidence to the sine of the angle of refraction is $n = \frac { \sin 45 ^ { \circ } } { \sin 30 ^ { \circ } } = \sqrt { 2 }$

Question 28: A Knowledge]Conditions for total reflection,Critical angle The critical angle of glass is $\mathrm {...

A Knowledge]Conditions for total reflection,Critical angle The critical angle of glass is $\mathrm { C } = 42 ^ { \circ }$. The angle of incidence is $\mathrm { i } = 45 ^ { \circ } > \mathrm { C }$ when the light rays from the glass enter the air, so the light rays will be totally reflected at the interface between the glass and the air, and all the light rays will be reflected back to the glass.

  • A. A. The graph is consistent with the conclusion, and option A is correct;
  • B. B. This figure is not consistent with the conclusion, and option B is incorrect;
  • C. C. This figure is not consistent with the conclusion, and option C is incorrect;
  • D. D. The graph is not consistent with the conclusion, and option D is wrong; Therefore, choose A.

Answer: A

Solution: Knowledge Points] Conditions for Total Reflection, Critical Angle [Explanation] The critical angle of glass is $\mathrm { C } = 42 ^ { \circ }$. The angle of incidence is $\mathrm { i } = 45 ^ { \circ } > \mathrm { C }$ when the light rays from the glass are directed into the air, so the light rays will undergo total reflection at the interface between the glass and the air, and all the light rays will be reflected back to the glass. A. The diagram is consistent with the conclusion, option A is correct; B. The diagram is not consistent with the conclusion, option B is wrong; C. The diagram is not consistent with the conclusion, option C is wrong; D. The graph does not match the conclusion, option D is wrong;
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