JEE Physics Wave Optics MCQs Set 16

Answer: (b) 2 : 1

Explanation:
1. At P the phase difference is 0: \( I_P=4I_0 \).
2. At Q the phase difference is \( \pi/2 \): \( I_Q=4I_0\cos^2\frac\pi4=2I_0 \).
3. Ratio \( 2:1 \).

Teacher's Note:
1. Intensity at a point is Imax cos squared of half the phase difference.

Answer: (c) \( \frac{2D\lambda}{a} \)

Explanation:
1. First minima at \( y=\pm\frac{D\lambda}{a} \).
2. Width of the central maximum \( =\frac{2D\lambda}a \).

Teacher's Note:
1. The central maximum extends between the first minima on either side.

Answer: (c) 0.2 mm

Explanation:
1. Central single-slit width \( =\frac{2\lambda D}{a} \); fringe width \( \beta=\frac{\lambda D}{d} \).
2. Ten fringes inside: \( \frac{2\lambda D}a=10\frac{\lambda D}d\Rightarrow a=\frac d5=0.2\ \text{mm} \)

Teacher's Note:
1. Equate the central diffraction width to ten interference fringe widths.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Reflection from the glass plate (air to glass) gives a phase change of pi, while reflection at the lens-air boundary gives none, so A is true.
2. At the centre the film thickness is zero, so the net path difference is lambda/2: destructive interference, a dark centre. R follows from A.

Teacher's Note:
1. The extra half-wavelength shift at one reflection makes the zero-thickness point dark.

Answer: (c) \( 4\pi \)

Explanation:
1. Second minimum: path difference \( a\sin\theta=2\lambda \).
2. Phase difference \( =\frac{2\pi}\lambda\times2\lambda=4\pi \)

Teacher's Note:
1. The nth minimum has a path difference of n wavelengths between the slit edges.

Answer: (d) \( \frac34 \)

Explanation:
1. Phase difference \( \phi=\frac{2\pi}\lambda\cdot\frac\lambda6=\frac\pi3 \).
2. \( \frac I{I_0}=\cos^2\frac\phi2=\cos^230^\circ=\frac34 \)

Teacher's Note:
1. I over Imax equals cos squared of half the phase difference.

Answer: (b) \( 4I_0 \)

Explanation:
1. The amplitude at the central maximum is proportional to the slit width, since all the wavelets add in phase.
2. Doubling the width doubles the amplitude, so the intensity becomes \( 4I_0 \).

Teacher's Note:
1. Central intensity varies as the square of the slit width, for the same light intensity per unit width.

Answer: (d) Concentric circles

Explanation:
1. The path difference at a screen point depends only on the angle from the axis (S1S2 line), \( \Delta=d\cos\theta \).
2. Points with the same path difference lie on a circle about the axis, so fringes are concentric circles.

Teacher's Note:
1. Rotational symmetry about the line joining the sources gives circular fringes.

Answer: (d) \( \tan^{-1}(n) \)

Explanation:
1. Brewster's law: \( \tan i_B=n \).

Teacher's Note:
1. At the polarizing angle the reflected and refracted rays are at right angles.

Answer: (a) 0.30 mm

Explanation:
1. \( \beta'=\frac\beta\mu=\frac{0.4}{4/3}=0.30\ \text{mm} \)

Teacher's Note:
1. Wavelength shrinks by mu in the medium, so the fringe width does too.

Answer: (d) 5 : 4

Explanation:
1. Resolving power is proportional to \( 1/\lambda \).
2. \( \frac{R_1}{R_2}=\frac{\lambda_2}{\lambda_1}=\frac54 \)

Teacher's Note:
1. Shorter wavelengths resolve better.

Answer: (c) \( a=\sqrt{\lambda L} \) and \( b_{min}=\sqrt{4\lambda L} \)

Explanation:
1. Spot size \( b=a+\frac{\lambda L}a \).
2. Minimum when \( a=\frac{\lambda L}a\Rightarrow a=\sqrt{\lambda L} \), giving \( b_{min}=2\sqrt{\lambda L}=\sqrt{4\lambda L} \)

Teacher's Note:
1. Minimise the sum of geometrical and diffraction spreads.

Answer: (a) \( 10^8 \) km

Explanation:
1. \( \theta=\frac{1.22\lambda}D=\frac{1.22\times6\times10^{-7}}{0.3}=2.44\times10^{-6}\ \text{rad} \)
2. \( s=\theta\times10\times9.46\times10^{15}\approx2.3\times10^{11}\ \text{m}\approx2.3\times10^8\ \text{km} \)

Teacher's Note:
1. Minimum resolvable separation is the angular limit times the distance.

Answer: (d) \( 30\ \mu m \)

Explanation:
1. \( \theta=\frac{1.22\lambda}D=\frac{1.22\times5\times10^{-7}}{5\times10^{-3}}=1.22\times10^{-4}\ \text{rad} \)
2. \( s=\theta\times25\ \text{cm}\approx3\times10^{-5}\ \text{m}=30\ \mu m \)

Teacher's Note:
1. Use the pupil diameter (twice the radius) in Rayleigh's criterion.

Answer: (b) \( \frac\beta4 \)

Explanation:
1. \( I=I_0\cos^2\frac\phi2=\frac{I_0}2\Rightarrow\phi=\frac\pi2 \), a path difference of \( \lambda/4 \).
2. \( y=\frac{D}{d}\cdot\frac\lambda4=\frac\beta4 \)

Teacher's Note:
1. Half intensity corresponds to a path difference of a quarter wavelength.

Answer: (c) 16

Explanation:
1. \( \frac{I_{max}}{I_{min}}=\frac{(5+3)^2}{(5-3)^2}=\frac{64}{4}=16 \)

Teacher's Note:
1. Amplitudes add for maximum and subtract for minimum.

Answer: (b) 1.0

Explanation:
1. \( \lambda'=\frac{700}{4/3}=525\ \text{nm} \).
2. \( \beta=\frac{\lambda'D}{d}=\frac{525\times10^{-9}\times0.48}{0.25\times10^{-3}}\approx1.0\times10^{-3}\ \text{m}=1.0\ \text{mm} \)

Teacher's Note:
1. Use the wavelength in water in the fringe width formula.