Chapter Exam

Chapter-10 Wave Optics — Online MCQ Test

Q: According to Huygens' principle, what is a wavefront?

Answer: C. Both A and B — Huygens' principle defines a wavefront as a surface of constant phase, and it can also be understood as the envelope of all secondary wavelets generated from previous wavefront positions.

Q: What type of wavefront is produced by a point source?

Answer: B. Spherical wavefront — A point source emits light equally in all directions, creating wavefronts that are spherical surfaces centered at the source.

Q: In Young's double slit experiment, what is the condition for constructive interference?

Answer: B. Path difference = nλ, where n = 0, 1, 2... — Constructive interference occurs when the path difference is an integral multiple of wavelength (nλ), causing the waves to reinforce each other.

Q: When red light and blue light are used in Young's double slit experiment, which light produces wider fringes?

Answer: A. Red light (larger wavelength) — Fringe width β = λD/d increases with wavelength. Red light has larger wavelength than blue light, so it produces wider fringes.

Q: In Young's double slit experiment, if the distance between slits is doubled while keeping other parameters constant, the fringe width will:

Answer: A. Become half — Since fringe width β = λD/d, doubling d (slit separation) results in β becoming β/2, i.e., half of the original value.

Q: In Young's double slit experiment, if the screen is moved closer to the slits, which parameter changes?

Answer: B. Fringe width decreases — Since β = λD/d, decreasing D (distance to screen) directly decreases fringe width β proportionally.

Q: In Young's double slit experiment, the condition for the mth dark fringe is path difference = (2m+1)λ/2. For m = 2, the path difference is:

Answer: A. 5λ/2 — Substituting m = 2 in (2m+1)λ/2 gives (2×2+1)λ/2 = 5λ/2, which corresponds to the second dark fringe from the central bright fringe.

Q: In a double slit interference pattern, the intensity of light at a point is I_0. If the intensity from each slit alone is I, what is the relationship between I_0 and I when the path difference is zero?

Answer: B. I_0 = 4I — Intensity in interference is I_total = I₁ + I₂ + 2√(I₁I₂)cosδ. For equal slits (I₁=I₂=I) and zero path difference (δ=0), I₀ = I + I + 2I = 4I.

PHYSICS · Grade 12 · CBSE(NCERT)

Practice Chapter-10 Wave Optics with a free chapter-wise online MCQ test for CBSE(NCERT) Grade 12 PHYSICS. This chapter covers: Huygens principle - wavefront - interference - Young's double slit experiment - diffraction. AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter-10 Wave Optics — Important Questions & Answers (FAQ)

Frequently asked questions from CBSE(NCERT) Grade 12 PHYSICS — Chapter-10 Wave Optics, with answers and explanations. These are sample questions; the exam has its own separate question set.

According to Huygens' principle, what is a wavefront?

A. A surface joining all points that vibrate in the same phase at a given instant
B. The envelope of all secondary wavelets at a given time
C. Both A and B ✓
D. The path traced by a single particle of the medium

Answer: C. Both A and B
Huygens' principle defines a wavefront as a surface of constant phase, and it can also be understood as the envelope of all secondary wavelets generated from previous wavefront positions.

What type of wavefront is produced by a point source?

A. Plane wavefront
B. Spherical wavefront ✓
C. Cylindrical wavefront
D. Elliptical wavefront

Answer: B. Spherical wavefront
A point source emits light equally in all directions, creating wavefronts that are spherical surfaces centered at the source.

In Young's double slit experiment, what is the condition for constructive interference?

A. Path difference = (2n+1)λ/2, where n = 0, 1, 2...
B. Path difference = nλ, where n = 0, 1, 2... ✓
C. Path difference = nλ/2, where n = 0, 1, 2...
D. Path difference = (n+1)λ, where n = 0, 1, 2...

Answer: B. Path difference = nλ, where n = 0, 1, 2...
Constructive interference occurs when the path difference is an integral multiple of wavelength (nλ), causing the waves to reinforce each other.

What is the condition for destructive interference in Young's double slit experiment?

A. Path difference = nλ
B. Path difference = (2n+1)λ/2, where n = 0, 1, 2... ✓
C. Path difference = (n+1)λ
D. Path difference = 2nλ

Answer: B. Path difference = (2n+1)λ/2, where n = 0, 1, 2...
Destructive interference occurs when waves are out of phase by odd multiples of λ/2, resulting in cancellation.

When red light and blue light are used in Young's double slit experiment, which light produces wider fringes?

A. Red light (larger wavelength) ✓
B. Blue light (smaller wavelength)
C. Both produce equally wide fringes
D. It depends on the slit separation

Answer: A. Red light (larger wavelength)
Fringe width β = λD/d increases with wavelength. Red light has larger wavelength than blue light, so it produces wider fringes.

In Young's double slit experiment, if the distance between slits is doubled while keeping other parameters constant, the fringe width will:

A. Become half ✓
B. Become double
C. Remain unchanged
D. Become four times

Answer: A. Become half
Since fringe width β = λD/d, doubling d (slit separation) results in β becoming β/2, i.e., half of the original value.

Which statement about the fringe visibility in Young's double slit experiment is correct?

A. It depends only on the wavelength
B. It is highest when the two slits have equal intensity ✓
C. It decreases with increasing slit width
D. It is independent of light intensity

Answer: B. It is highest when the two slits have equal intensity
Fringe visibility is defined as (I_max - I_min)/(I_max + I_min) and is maximum (= 1) when the two slits have equal intensity; unequal intensities reduce visibility.

In Young's double slit experiment, if the screen is moved closer to the slits, which parameter changes?

A. Wavelength of light
B. Fringe width decreases ✓
C. Slit separation
D. Wavelength becomes smaller

Answer: B. Fringe width decreases
Since β = λD/d, decreasing D (distance to screen) directly decreases fringe width β proportionally.

In Young's double slit experiment, the condition for the mth dark fringe is path difference = (2m+1)λ/2. For m = 2, the path difference is:

A. 5λ/2 ✓
B. 3λ/2
C. 7λ/2
D. 9λ/2

Answer: A. 5λ/2
Substituting m = 2 in (2m+1)λ/2 gives (2×2+1)λ/2 = 5λ/2, which corresponds to the second dark fringe from the central bright fringe.

In a double slit interference pattern, the intensity of light at a point is I_0. If the intensity from each slit alone is I, what is the relationship between I_0 and I when the path difference is zero?

A. I_0 = 2I
B. I_0 = 4I ✓
C. I_0 = I
D. I_0 = √(2I)

Answer: B. I_0 = 4I
Intensity in interference is I_total = I₁ + I₂ + 2√(I₁I₂)cosδ. For equal slits (I₁=I₂=I) and zero path difference (δ=0), I₀ = I + I + 2I = 4I.

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