Wave Optics Class 12 MCQs Questions with Answers
Wave Optics MCQ Physics Chapter 10 Class 12 Question 1.
Wavefront generated from a line source is
(A) cylindrical wavefront
(B) spherical wavefront
(C) plane wavefront
(D) either (A) or (B)
Answer:
(A) cylindrical wavefront
MCQ On Wave Optics Class 12 Chapter 10 Question 2.
Phase difference between any two points of a wavefront is
(A) π
(B) π/2
(C) 0
(D) π/4
Answer:
(C) 0
Explanation:
Wavefront is the locus of all points those are in same phase.
Wave Optics Class 12 MCQ Chapter 10 Question 3.
In Huygens theory, light waves
(A) are transverse waves and require a medium to travel.
(B) are longitudinal waves and require a medium to travel.
(C) are transverse waves and require no medium to travel.
(D) are longitudinal waves and require no medium to travel.
Answer:
(B) are longitudinal waves and require a medium to travel.
Explanation:
According to Huygens, light waves are longitudinal waves and require a material medium to travel. For this reason Huygens assumed the existence of a hypothetical medium called luminiferous ether.
MCQ Of Wave Optics Class 12 Chapter 10 Question 4.
Huygens theory could not explain
(A) photoelectric effect.
(B) reflection of light.
(C) diffraction of light.
(D) interference of light.
Answer:
(A) photoelectric effect.
Explanation:
Wave nature of light cannot explain the photoelectric effect. Particle nature of light can only explain it.
Wave Optics Class 12 MCQ Pdf Chapter 10 Question 5.
Which of the following statement is true?
(A) According to both Maxwell’s electromagnetic theory and Huygens wave theory light is treated as a wave in nature and require medium to travel.
(B) According to both Maxwell’s electromagnetic theory and Huygens wave theory light is treated as a particle in nature and require medium to travel.
(C) According to both Maxwell’s electromagnetic theory and Huygens wave theory light is treated as a wave in nature and does not require medium to travel.
(D) According to Maxwell’s electromagnetic theory light is treated as a wave in nature and require no medium to travel. According to Huygens theory light is treated as a wave in nature and require medium to travel.
Answer:
(D) According to Maxwell’s electromagnetic theory light is treated as a wave in nature and require no medium to travel. According to Huygens theory light is treated as a wave in nature and require medium to travel.
MCQ Wave Optics Class 12 Chapter 10 Question 6.
In a Young’s double-slit experiment the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case,
(A) there shall be alternate interference patterns of red and blue.
(B) there shall be an interference pattern for red distinct from that for blue.
(C) there shall be no interference fringes.
(D) there shall be an interference pattern for red mixing with one for blue.
Answer:
(C) there shall be no interference fringes.
Wave Optics MCQ Chapter 10 Class 12 Question 7.
Figure shows a standard two-slit arrangement with slits S1, S2. P1, P2 are the two minima points on either side of P shows in Figure. At P2 on the screen, there is a hole and behind P2 is a second screen, 2-slit arrangement with slits S3 and S4 and a second screen behind them.
(A) There would be no interference pattern on the second screen but it would be lighted.
(B) The second screen would be totally dark.
(C) There would be a single bright point on the second screen.
(D) There would be a regular two slit pattern on the second screen.
Answer:
(D) There would be a regular two slit pattern on the second screen.
Explanation:
At P2 is minima due to two wave fronts in opposite phase coming from, two slits S1 and S2, but there is wave fronts from S1, S2. So P2 will act as a source of secondary wavelets. Wave front starting from P2 reaches at S3 and S4 slits which will again act as two monochromatic or coherent sources and will form pattern on second screen.
Wave Optics MCQ Class 12 Chapter 10 Question 8.
In Young’s double slit experiment, the distance between the slits is reduced to half and the distance between the slits and the screen is doubled. The fringe width
(A) will be double.
(B) will be half.
(C) will remain same.
(D) will be four times.
Answer:
(D) will be four times.
Explanation:
Fringe width = β = λD/d
Initially, β = λD/d
Finally, β’ = \(\frac{\lambda \times 2 \mathrm{D}}{d / 2}\) = 4 x \(\frac{\lambda \mathrm{D}}{d}\) = 4β
Class 12 Wave Optics MCQ Chapter 10 Question 9.
A Young’s double slit experiment is performed with blue (wavelength 460 nm) and green light (wavelength 550 nm) respectively. If y is the distance of 4th maximum from the central fringe then
(A) yB = yG
(B) yB > yG
(C) yG > yB
(D) yB/yG = 550/460
Answer:
(C) yG > yB
Explantion:
yn = nλD/d
So, yn ∝λ
Since λG > λB
∴ø G > yB
MCQ On Wave Optics Chapter 10 Question 10.
A Young/s Double slit experiment is performed in air and in water. Which of the following relationship is true regarding fringe width (P)?
(A) βAIR > βWATER
(B) βWATER > βAIR
(C) βAIR = βWATER
(D) βWATER = 0
Answer:
(A) βAIR > βWATER
Explantion:
β ∝λ and λ.∝1/µ
So, β ∝1/µ
Since µWATER > µAIR
βAIR > βWATER
Physics MCQ Class 12 Chapter 10 Question 11.
The penetration of light into the region of geometrical shadow is known as
(A) interference of light.
(B) diffraction of light.
(C) refraction of light.
(D) polarisation of light.
Answer:
(B) diffraction of light.
Physics Class 12 MCQ Chapter 10 Question 12.
Angular width of central maxima of a single slit diffraction pattern is independent of
(A) slit width
(B) frequency of the light used
(C) wavelength of the light used
(D) distance between slit and screen
Answer:
(D) distance between slit and screen
Explanation:
Angular width = 2 sin-1 λ/d So, it is independent of D (distance between slit and screen).
Physics Class 12 MCQ Questions Question 13.
When a monochromatic light is passed around a file wire a diffraction pattern is observed. How the fringe width will change by increasing the diameter?
(A) Fringe width has no relation with the diameter of wire
(B) Increases
(C) Decreases
(D) Fringe width changes with change of wavelength only
Answer:
(C) Decreases
Explanation:
β = λD/d, where d is the diameter of the wire. So, if the diameter increases, fringe with decreases.
Question 14.
The main condition for diffraction to be observed is
(A) size of obstacle should be comparable to the wavelength of the wave
(B) size of obstacle should be much larger than the wavelength of the wave
(C) size of obstacle should be much smaller than the wavelength of the wave
(D) for any size of obstacle
Answer:
(A) size of obstacle should be comparable to the wavelength of the wave
Assertion And Reason Based MCQs (1 Mark each)
Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is true
Question 1.
Assertion (A): According to Huygens theory no back-ward wavefront is possible.
Reason (R): Amplitude of secondary wavelets is proportional to (1+ cos 0), where 0 is the angle between the ray at the point of consideration and direction of secondary wavelet.
Answer:
(A) Both A and R are true and R is the correct explanation of A
Explanation:
According to Huygens theory each and every point on a wavefront is the source of secondary wavelets. Secondary wavelets do not proceed backward. So the assertion is true. Kirchhoff’s explained that amplitude of secondary wavelets is proportional to (1 + cos 0), where 0 is the angle between the ray at the point of consideration and direction of secondary wavelets. In the backward direction 0 = 180°; so 1 + cos 0 = 0; so there secondary wavelets do not proceed backward. Hence assertion and both are true and the reason properly explains the assertion.
Question 2.
Assertion (A): Wavefront emitted by a point source of light in an isotropic medium is spherical.
Reason (R): Isotropic medium has same refractive index in all directions.
Answer:
(A) Both A and R are true and R is the correct explanation of A
Explanation:
If a medium has same refractive index at every point in all directions, then the wavefrom obtained from a point source in such a medium is spherical since wave travels in all direction with same speed. Such a medium is known as isotropic medium. So, the assertion and reasoji both are true and the reason explain the assertion properly.
Question 3.
Assertion (A): When a light wave travels from rarer to denser medium, its speed decreases. Due to this reduction of speed the energy carried by the light wave reduces.
Reason (R): Energy of wave is proportional to the frequency.
Answer:
(D) A is false and R is true
Explanation:
When a light wave travels from rarer to denser medium, its speed decreases. But this reduction of speed does not imply the loss of energy carried by the light wave. So the assertion is false. Energy of wave is proportional to the frequency of the wave which remains same in very medium. Hence there is no loss of energy. So, the reason is true.
Question 4.
Assertion (A): No interference pattern is detected when two coherent sources are too close to each other.
Reason (R): The fringe width is inversely proportional to the distance between the two slits.
Answer:
(A) Both A and R are true and R is the correct explanation of A
Explanation:
No interference pattern is detected when two coherent sources are too close to each other. The assertion is true.
Fringe width is proportional to 1/d. When d becomes too small, the fringe width becomes too large. So no pattern will be visible. So, the reason is also true. Reason also explains the assertion.
Question 5.
Assertion (A): For best contrast between maxima and minima in the interference pattern of Young’s double slit experiment, the amplitudes of light waves emerging from the two sources should be equal.
Reason (R): For interference, the sources must be coherent.
Answer:
(B) Both A and R are true but R is NOT the correct explanation of A
Explanation:
For destructive interference, a = ~ a2. When aj = ay only the minima will be completely dark. This will create the best contrast. So the assertion is true.
For interference, the sources must be coherent. Reason is also true. But the reason does not explain the assertion.
Question 6.
Assertion (A): Fringes of interference pattern produced by blue light is narrower than that produced by red light.
Reason (R): In Young’s double slit experiment, fringe width = XD/d
Answer:
(C) A is true but R is false
Explanation:
Fringes of interference pattern produced by blue light is narrower than that produced by red light. The assertion is true. Fringe width = XD/d. Since blue light has wavelength smaller than that of red light, blue light produces narrower fringes. So, reason is also true and explains the assertion.
Question 7.
Assertion (A): Diffraction takes place with all types of waves.
Reason (R): Diffraction is perceptible when the wavelength of the wave is comparable to the dimension of the diffracting device.
Answer:
(B) Both A and R are true but R is NOT the correct explanation of A
Explanation:
Diffraction is spreading of waves around obstacle. It takes place with all types of waves (mechanical, non-mechanical, transverse, longitudinal) and with very small moving particles (atom, neutron, electron etc.) which show wave like property. So, the assertion is true. Diffraction is perceptible when the wavelength of the wave is comparable to the dimension of the diffracting device. The reason is also true. But it does not explain the assertion.
Case-Based MCQS’
Attempt any 4 sub-parts out of 5. Each sub-part carries 1 mark.
I. Read the following text and answer the following questions on the basis of the same:
In one of his experiments on interference, August Jean Fresnel used a biprism to induce interference between two beams. He split a diverging beam of light into two parts by using the biprism to refract them. This resulted in two split beams which acted as if they were from two coherent sources and which therefore interfered with each other. A Fresnel Biprism is a thin double prism placed base to base and have very small refracting angle (0.5°). This is equivalent to a single prism with one of its angle nearly 179° and other two of 0.5° each.
In Young’s double Slits experiment, a single source is split in two coherent sources. For the Young’s slits experiment, we must approximate that the slits act as point sources. This however is not the case, since the slits have finite width. In this way, it gives rise to unwanted diffraction effects that causes errors. The Fresnel biprism experiment overcomes this problem.
A Fresnel biprism is a variation of Young’s Slits experiment. When monochromatic light through a narrow slit falls on biprism that divides it into two components. One of these component is refracted from upper portion of biprism and the other one refracted through lower portion. Two virtual coherent sources formed from the original source. In this case, two virtual coherent sources are point sources and replace slits in Young’s experiment.
Question 1.
The Fresnel biprism is:
(A) a combination of two prisms with their bases in contact.
(B) a combination of two prisms with their refracting surfaces in contact.
(C) single prism
(D) not a prism actually.
Answer:
(A) a combination of two prisms with their bases in contact.
Explanation:
A Fresnel Biprism is a thin double I prism placed base to base.
Question 2.
Base angles of Fresnel biprism are:
(A) 179°
(B) 90°
(C) 0.50°
(D) None of these
Answer:
(C) 0.50°
Explanation:
A Fresnel Biprism is a thin double I prism placed base to base and have very small I refracting angle (0.5°). |
Question 3.
Fresnel biprism produces:
(A) two real coherent sources.
(B) two virtual coherent sources.
(C) a number of real coherent sources.
(D) a number of virtual coherent sources.
Answer:
(A) two real coherent sources.
Explanation:
When monochromatic light through a narrow slit falls on Fresnel biprism that divides it into two components. One of these component is refracted from upper portion of biprism and the other one refracted through lower portion. Thus, two virtual coherent sources formed from the original source.
Question 4.
What is the difference between the coherent sources produced by Youngf s double slit arrangement and Fresnel biprism?
(A) Young’s double slit arrangement produces virtual coherent sources whereas Fresnel biprism produces real coherent sources
(B) Young’s double slit arrangement produces coherent point sources whereas Fresnel biprism produces coherent sources which are not point sources
(C) Both Young’s double slit arrangement and Fresnel biprism produce similar coherent sources
(D) Fresnel birism produces virtual coherent point sources whereas Young’s double slit arrangement produces real coherent sources which are not point sources.
Answer:
(D) Fresnel birism produces virtual coherent point sources whereas Young’s double slit arrangement produces real coherent sources which are not point sources.
Explanation:
In Young’s double Slits experiment, a single source is split in two coherent sources. Both are real Both the slits have finite width. Fresnel biprism divides the beam of monochromatic light incident on it into two components. One of these component
is refracted from upper portion of biprism and the other one refracted through lower portion. Thus two virtual coherent sources are formed from the original source.
Question 5.
Which problem of Young’s double slit experiment is overcome by Fresnel biprism?
(A) Young’s double slit arrangement gives rise to irregular interference fringe pattern which is overcome by Fresnel biprism which produces coherent sources by refraction in a prism
(B) Finite width of slits in Young’s double slit experiment gives rise to unwanted diffraction effects that causes errors. This is overcome by Fresnel biprism by producing virtual coherent point sources.
(C) Young’s double slit arrangement produces interference fringe pattern of low intensity which is overcome by Fresnel biprism.
(D) All of the above
Answer:
(B) Finite width of slits in Young’s double slit experiment gives rise to unwanted diffraction effects that causes errors. This is overcome by Fresnel biprism by producing virtual coherent point sources.
Explanation:
In Young’s double Slits experiment, a single source is split in two coherent sources. For the Young’s slits experiment, we must approximate that the slits act as point sources. This however is not the case, since the slits have finite width. In this way, it gives rise to unwanted diffraction effects that causes errors. The Fresnel biprism experiment overcomes this problem.
When monochromatic light through a narrow slit falls on biprism that divides it into two components. One of these component is refracted from upper portion of biprism and the other one refracted through lower portion. Two virtual coherent point sources are formed from the original source.
II. Read the following text and answer the following questions on the basis of the same:
Diffraction in a hall:
A and B went to purchase a ticket of a music programme. But unfortunately only one ticket was left. They purchased the single ticket and decided that A would be in the hall during the 1st half and B during the 2nd half. Both of them reached the hall together. A entered the hall and found that the seat was behind a pillar which creates an obstacle. He was disappointed. He thought that he would not be able to hear the programme properly. B was waiting outside the closed door. The door was not fully closed. There was a little opening. But surprisingly, A could hear the music programme.
This happened due to diffraction of sound. The fact we hear sounds around corners and around barriers involves both diffraction and reflection of sound. Diffraction in such cases helps the sound to ‘bend around” the obstacles. In fact, diffraction is more pronounced with longer wavelengths implies that we can hear low frequencies around obstacles better than high frequencies.
B was outside the door. He could also hear the programme. But he noticed that when the door opening is comparatively less he could hear the programme even being little away from the door. This is because when the width of the opening is larger than the wavelength of the wave passing through the gap then it does not spread out much on the other side. But when the opening is smaller than the wavelength more diffraction occurs and the waves spread out greatly – with semicircular wavefront. The opening in this case functions as a localized source of sound.
Question 1.
A and B could hear the music programme due to phenomenon named
(A) interference.
(B) scattering.
(C) diffraction.
(D) dispersion.
Answer:
(C) diffraction.
Explanation:
The fact we hear sounds around corners and around barriers involves both diffraction and reflection of sound.
Question 2.
Diffraction is more pronounced with ………… wavelengths.
(A) Longer
(B) Shorter
(C) fluctuating
(D) all
Answer:
(A) Longer
Explanation:
In fact, diffraction is more pronounced with longer wavelengths
Question 3.
The minimum and maximum frequencies in the musical programme were 550 Hz and 10 kHz. Which frequency was better audible around the pillar obstacle?
(A) 10 kHz
(B) 550 kHz
(C) Mid frequency
(D) The complete frequency range
Answer:
(A) 10 kHz
Explanation:
In fact, diffraction is more pronounced with longer wavelengths implies that you can hear low frequencies around obstacles better than high frequencies.
Question 4.
Diffraction of sound takes place more when :
(A) sound is diffracted through an opening having width equal to the wavelength of the sound.
(B) sound is diffracted through an opening having width more than the wavelength of the sound.
(C) sound is diffracted through an opening having width less than the wavelength of the sound.
(D) diffraction of sound does not depend on the width of the opening.
Answer:
(C) sound is diffracted through an opening having width less than the wavelength of the sound.
Explanation:
When the width of opening is comparatively less than the wavelength of sound wave, the sound spread out much better i.e. better diffraction occurs. When the width of the opening is larger than the wavelength, the wave passing through the opening does not spread out much on the other side.
Question 5.
How the waveform will look like outside the door of the hall?
(A) Sound repeater
(B) Sound reflector
(C) Localized sound source
(D) None of the above
Answer:
(C) Localized sound source
Explanation:
spreads out well through a gap whose width is slightly smaller than the wavelength of the sound wave as if it is a localised source of sound.