JEE Physics Wave Optics MCQs Set 13

Answer: (b) A

Explanation:
1. \( A_R=2A\cos\frac\phi2=2A\cos\frac\pi3=2A\times\frac12=A \).

Teacher's Note:
1. Resultant amplitude of two equal waves is 2A cos(φ/2).
2. cos 60° = 1/2.

Answer: (a) 2 : 1

Explanation:
1. The intensity from a slit is proportional to its width, \( I\propto w \).
2. Intensity is proportional to the square of the amplitude, so \( \frac{A_1}{A_2}=\sqrt{\frac{w_1}{w_2}}=\sqrt4=2 \).
3. The amplitudes are in the ratio 2 : 1.

Teacher's Note:
1. Amplitude varies as the square root of intensity.
2. Light energy through a slit is proportional to its width.

Answer: (d) \( a_1^2+a_2^2 \)

Explanation:
1. \( y_2=a_2\cos\left(\frac\pi2-\omega t\right)=a_2\sin\omega t \), which differs in phase from \( y_1 \) by \( \frac\pi2 \).
2. For a phase difference of 90° the resultant intensity \( I\propto a_1^2+a_2^2+2a_1a_2\cos90^\circ=a_1^2+a_2^2 \).

Teacher's Note:
1. Rewrite y2 as a sine to see the 90° phase difference.
2. The cross term vanishes at a phase difference of π/2.

Answer: (c) \( \frac{d_1}{d_2}=\frac{\lambda_1}{\lambda_2} \)

Explanation:
1. Position of the nth maximum \( y_n=\frac{n\lambda D}{d} \), so \( y\propto\lambda \) for fixed n, D and d.
2. \( \frac{d_1}{d_2}=\frac{\lambda_1}{\lambda_2} \).

Teacher's Note:
1. Fringe positions are proportional to the wavelength.
2. Same order n and same geometry.

Answer: (b) \( \frac\beta2 \)

Explanation:
1. Doubling the frequency halves the wavelength: \( \lambda'=\frac\lambda2 \).
2. Fringe width \( \beta=\frac{\lambda D}{d}\propto\lambda \), so the new width is \( \frac\beta2 \).

Teacher's Note:
1. c = fλ, so λ ∝ 1/f.
2. Fringe width is proportional to λ.

Answer: (a) \( \frac{2\pi}3 \)

Explanation:
1. Resultant amplitude \( 2A\cos\frac\phi2=A\Rightarrow\cos\frac\phi2=\frac12 \).
2. \( \frac\phi2=\frac\pi3\Rightarrow\phi=\frac{2\pi}{3} \).

Teacher's Note:
1. Set the resultant amplitude equal to A.
2. cos 60° = 1/2.

Answer: (d) 0.30 mm

Explanation:
1. In water the wavelength becomes \( \lambda'=\frac\lambda\mu \).
2. Fringe width \( \beta'=\frac\beta\mu=\frac{0.4}{4/3}=0.30\ \text{mm} \).

Teacher's Note:
1. The wavelength decreases in a medium, but the frequency does not change.
2. β is proportional to λ.

Answer: (c) shift to one side without changing the fringe width

Explanation:
1. The slab adds an extra optical path \( (\mu-1)t \) to one beam.
2. The point of zero path difference moves towards the covered slit, so the whole pattern shifts sideways by \( \frac{(\mu-1)tD}{d} \).
3. The fringe width \( \frac{\lambda D}{d} \) is unchanged.

Teacher's Note:
1. Extra optical path shifts the pattern but does not alter λD/d.
2. The shift is towards the side of the slit covered with the slab.

Answer: (b) \( \frac{9\lambda}{2} \)

Explanation:
1. The nth minimum has path difference \( (2n-1)\frac\lambda2 \).
2. For n = 5: \( 9\times\frac\lambda2=\frac{9\lambda}{2} \).

Teacher's Note:
1. The first minimum is at λ/2.
2. The fifth minimum is at (2 × 5 − 1)λ/2.

Answer: (a) coordinated waves of exactly the same wavelength

Explanation:
1. Coherence requires a constant phase relation between waves.
2. A laser emits waves of the same wavelength in step with each other (stimulated emission), so it is coherent.

Teacher's Note:
1. Coherent sources have a constant phase difference.
2. Divergence is unrelated to coherence.

Answer: (d) \( \frac{d^2}{2\lambda} \)

Explanation:
1. Width of the central maximum \( =\frac{2\lambda D}{d} \).
2. Set equal to the slit width: \( \frac{2\lambda D}{d}=d\Rightarrow D=\frac{d^2}{2\lambda} \).

Teacher's Note:
1. The central maximum extends from the first minimum on one side to the other.
2. First minima at d sin θ = λ.