JEE Physics Wave Optics MCQs Set 04

Answer: (d) Same frequency having constant phase difference

Explanation:
1. Coherent sources have the same frequency and a constant phase difference.
2. Intensities need not be equal.

Teacher's Note:
1. Same frequency and fixed phase.

Answer: (b) Light is made up of waves

Explanation:
1. Interference fringes can only be explained by the wave theory.
2. Fringe width does depend on the slit spacing.

Teacher's Note:
1. Young's experiment confirmed wave nature.

Answer: (c) Both interference and diffraction occur

Explanation:
1. Each slit diffracts light and the two diffracted beams interfere.
2. So both occur.

Teacher's Note:
1. Interference pattern is modulated by diffraction.

Answer: (a) Violet

Explanation:
1. Fringe width is proportional to \( \lambda \).
2. Violet has the smallest wavelength, so its first maximum is closest to the centre.

Teacher's Note:
1. Violet is nearest, red is farthest.

Answer: (d) \( 2I_0(1+\cos\phi) \)

Explanation:
1. \( I=I_0+I_0+2I_0\cos\phi=2I_0(1+\cos\phi) \).
2. This equals \( 4I_0\cos^2(\phi/2) \).

Teacher's Note:
1. Standard two-beam intensity formula.

Answer: (c) Equal widths and equal intensities

Explanation:
1. The fringe width \( \lambda D/d \) is constant and every maximum has intensity \( 4I_0 \).
2. So all bright fringes are alike.

Teacher's Note:
1. Contrast with diffraction, where intensities fall off.

Answer: (d) \( w'=w \)

Explanation:
1. Fringe width \( \lambda D/d \) is unchanged by the mica sheet.
2. Only the whole pattern shifts.

Teacher's Note:
1. Introducing a sheet shifts but does not widen the fringes.

Answer: (b) 2 : 1

Explanation:
1. Coherent: \( I=(2A)^2=4A^2 \) at the central maximum.
2. Incoherent: \( I=A^2+A^2=2A^2 \).
3. Ratio \( 2:1 \).

Teacher's Note:
1. Incoherent intensities add, coherent amplitudes add.

Answer: (c) 25 : 1

Explanation:
1. Intensity is proportional to slit width, so \( I_1:I_2=4:9 \) and amplitudes are 2 : 3.
2. \( I_{max}/I_{min}=(3+2)^2/(3-2)^2=25 \).

Teacher's Note:
1. Slit width is proportional to intensity.

Answer: (b) \( \dfrac{\lambda D}{d} \)

Explanation:
1. The fringe width is \( \beta=\lambda D/d \).
2. It is the spacing between consecutive maxima or minima.

Teacher's Note:
1. D is the slit-screen distance and d the slit separation.

Answer: (b) Will decrease

Explanation:
1. In water \( \lambda'=\lambda/\mu \) is smaller.
2. Hence \( \beta \) decreases.

Teacher's Note:
1. Wavelength shrinks in a medium.

Answer: (d) Will not be formed

Explanation:
1. The two beams then have different frequencies and are not coherent.
2. So no steady interference pattern forms.

Teacher's Note:
1. Different colours are incoherent.

Answer: (b) \( \dfrac Dd(\mu-1)t \)

Explanation:
1. Extra path is \( (\mu-1)t \).
2. Shift \( =\dfrac{D}{d}(\mu-1)t \).

Teacher's Note:
1. Shift is towards the side of the sheet.

Answer: (a) 0.5 mm

Explanation:
1. \( \beta'=\beta/\mu=0.8/1.6 \).
2. \( \beta'=0.5 \) mm.

Teacher's Note:
1. Everything else is unchanged.

Answer: (b) 2.5 cm

Explanation:
1. Shift \( =\dfrac Dd(\mu-1)t=\dfrac{1}{0.5\times10^{-3}}\times0.5\times2.5\times10^{-5} \).
2. \( =2.5\times10^{-2} \) m = 2.5 cm.

Teacher's Note:
1. Use SI units throughout.

Answer: (a) 4.8 μm

Explanation:
1. \( (\mu-1)t=4\lambda \).
2. \( t=4\times6\times10^{-7}/0.5=4.8\ \mu m \).

Teacher's Note:
1. Shift by n fringes means extra path n lambda.

Answer: (a) 0.4 μm

Explanation:
1. The central fringe stays when \( (\mu_1-1)t_1=(\mu_2-1)t_2 \).
2. \( 0.5\times1.2=1.5\,t \) gives \( t=0.4\ \mu m \).

Teacher's Note:
1. Equal extra optical paths keep the centre fixed.

Answer: (a) 0.20 mm

Explanation:
1. \( \beta\propto\lambda/d \).
2. Both change by the same factor, so \( \beta \) stays 0.20 mm.

Teacher's Note:
1. Ratios cancel.

Answer: (b) Black

Explanation:
1. Reflection at the top surface gives a \( \pi \) phase change and none at the bottom.
2. For \( t\to0 \) the path difference is \( \lambda/2 \), so reflection is destructive for all colours and the film looks black.

Teacher's Note:
1. Very thin films look dark in reflection.

Answer: (c) Only the central fringe is white and all other fringes are coloured

Explanation:
1. At the centre all colours have zero path difference and add to white.
2. Other fringes are coloured because their positions depend on wavelength.

Teacher's Note:
1. Central maximum is white.

Answer: (a) Wavelength increases

Explanation:
1. \( \beta=\lambda D/d \) increases with \( \lambda \).
2. Larger d, smaller D or higher frequency (smaller \( \lambda \)) decrease it.

Teacher's Note:
1. Fringe width proportional to lambda.

Answer: (c) 0.5 mm

Explanation:
1. \( \beta=\lambda D/d=550\times10^{-9}\times1/(1.1\times10^{-3}) \).
2. \( \beta=5\times10^{-4} \) m = 0.5 mm.

Teacher's Note:
1. Convert nm and mm carefully.

Answer: (c) Blue

Explanation:
1. \( \beta\propto\lambda \) and blue has the smallest wavelength among these.
2. So the blue fringes are the narrowest.

Teacher's Note:
1. Violet is the narrowest in the visible range.