JEE Physics Wave Optics MCQs Set 17

Answer: (a) \( 9I,\ 5I,\ I \)

Explanation:
1. \( I=I_1+I_2+2\sqrt{I_1I_2}\cos\phi=5I+4I\cos\phi \)
2. For \( \phi=0 \): 9I. For \( \phi=\pi/2 \): 5I. For \( \phi=\pi \): I.

Teacher's Note:
1. The extreme values are \( (\sqrt{I_1}\pm\sqrt{I_2})^2 \).
2. At quadrature the cross term vanishes and the intensities simply add.

Answer: (c) slightly increasing the widths of the two slits

Explanation:
1. \( \beta=\frac{\lambda D}{d} \) depends on wavelength, screen distance and slit separation only.
2. Widening the slits only raises the intensity and sharpens the fringes.

Teacher's Note:
1. The angular fringe width \( \lambda/d \) is independent of D, but the linear width grows with D.
2. Many wave-optics questions reduce to one idea: find the phase or path difference and compare it with a whole or half number of wavelengths.

Answer: (b) 1.625 mm

Explanation:
1. \( \beta=\frac{\lambda D}{d}=\frac{6.5\times10^{-7}\times1}{10^{-3}}=0.65\ \text{mm} \)
2. Third dark fringe: \( 2.5\beta=1.625\ \text{mm} \). Fifth bright fringe: \( 5\beta=3.25\ \text{mm} \).
3. Separation \( =3.25-1.625=1.625\ \text{mm} \).

Teacher's Note:
1. The n-th dark fringe is at \( (n-\tfrac12)\beta \).
2. The n-th bright fringe is at \( n\beta \).

Answer: (a) \( 2\times10^{-4}\ \text{rad}\ (0.011^\circ) \)

Explanation:
1. \( \theta=\frac{\lambda}{d}=\frac{6\times10^{-7}}{3\times10^{-3}}=2\times10^{-4}\ \text{rad} \) = 0.011 degree.

Teacher's Note:
1. Angular fringe width does not depend on the screen distance.
2. Eliminate options that violate a limiting case (for example phi = 0 or phi = pi) first.

Answer: (d) 0.9 mm

Explanation:
1. \( y_1=\frac{\lambda_1D}{d}=\frac{580\times10^{-9}\times1.5}{2\times10^{-4}}=4.35\ \text{mm} \), \( y_2=5.25\ \text{mm} \).
2. Separation \( 5.25-4.35=0.9\ \text{mm} \).

Teacher's Note:
1. The first maximum of each wavelength is at one fringe width from the centre.
2. Draw the geometry (or the phasor picture) before substituting numbers.

Answer: (c) 0.13 mm (one third of a fringe width)

Explanation:
1. Extra optical path \( \Delta=(1.62-1.52)\times0.40\ \text{mm}=4\times10^{-5}\ \text{m} \).
2. The central maximum shifts by \( \frac{\Delta D}{d}=\frac{4\times10^{-5}}{1.2\times10^{-3}}=33.33\ \text{mm} \), which is \( 83\tfrac13 \) fringe widths.
3. So bright fringes occur at distances of \( \tfrac13\beta \) and \( \tfrac23\beta \) from the geometrical centre; the nearest is \( \tfrac13\times0.4=0.13\ \text{mm} \).

Teacher's Note:
1. The fractional part of the shift in fringe widths decides where the nearest bright fringe falls.
2. A shift of exactly an integer number of fringes would put a maximum at the centre.

Answer: (b) \( \frac{\lambda}{4(\mu-1)} \)

Explanation:
1. \( (\mu-1)t=\frac{\lambda}{4}\Rightarrow t=\frac{\lambda}{4(\mu-1)} \)

Teacher's Note:
1. The slab adds \( (\mu-1)t \) to the optical path compared with air.
2. Intensity problems are easiest with the formula I = I1 + I2 + 2 sqrt(I1 I2) cos(phi).

Answer: (d) 1.45

Explanation:
1. For minimum transmitted intensity the path difference \( 2\mu t=\left(n+\tfrac12\right)\lambda \).
2. With \( t=0.3\ \mu\text{m} \): \( \mu=\frac{(n+\frac12)\,580}{600} \). For n = 0 this is 0.48, for n = 1 it is \( \frac{1.5\times580}{600}=1.45 \), for n = 2 it is 2.42.
3. Only n = 1 gives a value between 1.3 and 1.5, so \( \mu=1.45 \).

Teacher's Note:
1. Try successive values of the integer n and keep the result in the allowed range.
2. In transmitted light the condition for darkness is the same as the condition for brightness in reflected light.

Answer: (d) 100 nm

Explanation:
1. Weak transmission means strong reflection, which occurs when \( 2\mu t=\left(n+\tfrac12\right)\lambda \) (one of the two reflections gives a phase change of \( \pi \)).
2. For the minimum thickness n = 0: \( t=\frac{\lambda}{4\mu}=\frac{560}{4\times1.4}=100\ \text{nm} \).

Teacher's Note:
1. Use the film conditions carefully: the extra half wavelength comes from the phase change on reflection.
2. Draw the geometry (or the phasor picture) before substituting numbers.

Answer: (b) \( 1.8\ \mu\text{m} \)

Explanation:
1. \( a\sin\theta=\frac{3\lambda}{2}\Rightarrow a=\frac{3\times6000\times10^{-10}}{2\times0.5}=1.8\times10^{-6}\ \text{m} \)

Teacher's Note:
1. The first secondary maximum lies roughly midway between the first and second minima.
2. Many wave-optics questions reduce to one idea: find the phase or path difference and compare it with a whole or half number of wavelengths.

Answer: (a) 2 : 1

Explanation:
1. Intensity is proportional to the square of the amplitude, so the amplitude ratio is \( \sqrt4:\sqrt1=2:1 \).

Teacher's Note:
1. Intensity goes as amplitude squared.
2. Intensity problems are easiest with the formula I = I1 + I2 + 2 sqrt(I1 I2) cos(phi).

Answer: (d) the intensity ratio of the sources

Explanation:
1. Visibility is \( \frac{I_{max}-I_{min}}{I_{max}+I_{min}}=\frac{2\sqrt{I_1I_2}}{I_1+I_2} \), which depends only on the intensity ratio.

Teacher's Note:
1. It is maximum (1) when the two intensities are equal.
2. Eliminate options that violate a limiting case (for example phi = 0 or phi = pi) first.

Answer: (c) the fringe width is slightly increased

Explanation:
1. The refractive index of air (about 1.0003) is slightly greater than that of vacuum, so the wavelength in air is slightly shorter.
2. In vacuum the wavelength is a little longer, so \( \beta=\lambda D/d \) increases slightly.

Teacher's Note:
1. The effect is tiny but real: β increases a little.
2. Draw the geometry (or the phasor picture) before substituting numbers.

Answer: (b) 0.577 m

Explanation:
1. \( d\sin\theta=n\lambda\Rightarrow\sin\theta=\frac{1000\times5\times10^{-7}}{10^{-3}}=0.5 \), so \( \theta=30^\circ \).
2. \( y=D\tan30^\circ=0.577\ \text{m} \).

Teacher's Note:
1. The small-angle formula \( y=n\beta=0.5\ \text{m} \) is not valid here because the angle is large.
2. Many wave-optics questions reduce to one idea: find the phase or path difference and compare it with a whole or half number of wavelengths.

Answer: (a) 0.20 mm

Explanation:
1. \( I=I_{max}\cos^2\frac{\phi}{2}=0.75I_{max}\Rightarrow\frac{\phi}{2}=\pm30^\circ \), so \( \phi=\pm60^\circ=\pm\frac{\pi}{3} \).
2. A phase difference of \( 2\pi \) corresponds to one fringe width \( \beta=0.6\ \text{mm} \), so \( \phi=\pi/3 \) corresponds to \( \beta/6=0.1\ \text{mm} \).
3. The nearest two points with 75% intensity are on either side of the centre: separation \( 2\times0.1=0.2\ \text{mm} \).

Teacher's Note:
1. Equal slit intensities give \( I=I_{max}\cos^2(\phi/2) \).
2. Intensity problems are easiest with the formula I = I1 + I2 + 2 sqrt(I1 I2) cos(phi).

Answer: (c) \( X_{blue}\lt X_{green} \)

Explanation:
1. The position of the nth maximum is \( n\lambda D/d \), proportional to wavelength. Blue has the shorter wavelength.
2. So the ratio is \( \frac{4360}{5460} \), not its inverse.

Teacher's Note:
1. X ∝ λ, so X(blue) < X(green).
2. Eliminate options that violate a limiting case (for example phi = 0 or phi = pi) first.

Answer: (c) 1.2

Explanation:
1. \( \text{shift}=\frac{(\mu-1)tD}{d}\Rightarrow5\times10^{-3}=\frac{(\mu-1)(0.5\times10^{-3})(0.1)}{2\times10^{-3}} \).
2. \( \mu-1=\frac{5\times10^{-3}\times2\times10^{-3}}{0.5\times10^{-3}\times0.1}=0.2 \), so \( \mu=1.2 \).

Teacher's Note:
1. Check that the refractive index is greater than 1; this rules out 0.6.
2. Draw the geometry (or the phasor picture) before substituting numbers.

Answer: (a) 600 nm

Explanation:
1. Strong reflection requires \( 2\mu t=\left(m+\tfrac12\right)\lambda \) because one reflection has a phase change of \( \pi \).
2. \( \lambda=\frac{2\times1.5\times500}{m+1/2}=\frac{1500}{m+1/2}\ \text{nm} \): 1000, 600, 428.6, 333 nm for m = 1, 2, 3, 4.
3. Only 600 nm among the options lies in the visible range with this condition.

Teacher's Note:
1. Check each option by computing \( 2\mu t/\lambda \) and seeing whether it is a half-integer.
2. Many wave-optics questions reduce to one idea: find the phase or path difference and compare it with a whole or half number of wavelengths.

Answer: (d) 0.14 cm

Explanation:
1. \( \text{width}=\frac{2\lambda D}{a}=\frac{2\times5\times10^{-7}\times2.1}{1.5\times10^{-3}}=1.4\times10^{-3}\ \text{m}=0.14\ \text{cm} \)

Teacher's Note:
1. Convert all quantities to metres before substituting.
2. Intensity problems are easiest with the formula I = I1 + I2 + 2 sqrt(I1 I2) cos(phi).

Answer: (c) \( \frac12I_0 \)

Explanation:
1. An ideal polaroid transmits half of unpolarised light, so \( I_0/2 \) is transmitted and \( I_0/2 \) is not.

Teacher's Note:
1. The transmitted light is plane polarised.
2. Eliminate options that violate a limiting case (for example phi = 0 or phi = pi) first.

Answer: (b) the diffraction fringes become narrower and crowded together

Explanation:
1. Fringe widths are proportional to the wavelength; blue has the shorter wavelength.
2. Narrower fringes are more crowded.

Teacher's Note:
1. Smaller λ gives narrower fringes.
2. Draw the geometry (or the phasor picture) before substituting numbers.

Answer: (a) 5 m

Explanation:
1. \( \theta=\frac{1.22\lambda}{D}=\frac{1.22\times500\times10^{-9}}{3\times10^{-3}}=2.0\times10^{-4}\ \text{rad} \)
2. \( L=\frac{1\ \text{mm}}{2.0\times10^{-4}}\approx5\ \text{m} \)

Teacher's Note:
1. The dots can be resolved if they subtend at least the angle \( 1.22\lambda/D \) at the eye.
2. Many wave-optics questions reduce to one idea: find the phase or path difference and compare it with a whole or half number of wavelengths.

Answer: (d) \( \lt 0.1\ \text{mm} \)

Explanation:
1. The disc radius is proportional to the wavelength, and X-rays have a much shorter wavelength than visible light.
2. It is smaller but not zero.

Teacher's Note:
1. Radius ∝ λ and λ(X-ray) < λ(visible).
2. Intensity problems are easiest with the formula I = I1 + I2 + 2 sqrt(I1 I2) cos(phi).

Answer: (b) the wavelength of the light illuminating the object

Explanation:
1. For a microscope the limit of resolution is \( \frac{1.22\lambda}{2\mu\sin\beta} \), so the resolving power depends on the wavelength, on the medium and on the objective aperture.
2. Of the given options only the wavelength of the illuminating light appears.

Teacher's Note:
1. Shorter wavelength gives higher resolving power.
2. Eliminate options that violate a limiting case (for example phi = 0 or phi = pi) first.

Answer: (b) \( \frac45 \)

Explanation:
1. \( \sqrt{I_1/I_2}=2\Rightarrow I_{max}\propto(2+1)^2=9,\ I_{min}\propto(2-1)^2=1 \).
2. \( V=\frac{9-1}{9+1}=\frac45 \)

Teacher's Note:
1. Visibility can never exceed 1, which rules out options 4 and 9.
2. Draw the geometry (or the phasor picture) before substituting numbers.

Answer: (d) \( 2\ \text{cm},\ \frac23\ \text{cm},\ \frac25\ \text{cm} \)

Explanation:
1. For a point opposite one slit, the path difference is \( \sqrt{12^2+5^2}-12=13-12=1\ \text{cm} \).
2. For a minimum \( 1=\left(n+\tfrac12\right)\lambda\Rightarrow\lambda=\frac{2}{2n+1}\ \text{cm} \).
3. This gives 2 cm, 2/3 cm, 2/5 cm, and so on.

Teacher's Note:
1. Use Pythagoras for the exact path difference since the screen is close.
2. Many wave-optics questions reduce to one idea: find the phase or path difference and compare it with a whole or half number of wavelengths.

Answer: (c) 0.15 mm

Explanation:
1. The slit and its image are 2 mm apart: d = 2 mm, D = 1 m.
2. Reflection from the mirror adds a phase change of \( \pi \), so the point at the mirror level is dark and the first maximum is at \( \frac{\beta}{2}=\frac{\lambda D}{2d}=\frac{600\times10^{-9}\times1}{2\times2\times10^{-3}}=0.15\ \text{mm} \).

Teacher's Note:
1. With the phase change, bright and dark fringes are interchanged compared with an ordinary double slit.
2. Intensity problems are easiest with the formula I = I1 + I2 + 2 sqrt(I1 I2) cos(phi).

Answer: (b) \( 2\lambda \)

Explanation:
1. \( a\sin\theta=\lambda\Rightarrow a=\frac{\lambda}{\sin30^\circ}=2\lambda \)

Teacher's Note:
1. First minimum: \( a\sin\theta=\lambda \).
2. Eliminate options that violate a limiting case (for example phi = 0 or phi = pi) first.

Answer: (a) 49

Explanation:
1. Beam reflected from the first plate: \( I_1=0.25I \).
2. Beam transmitted by the first plate (0.75), reflected by the second (0.25) and transmitted again by the first (0.75): \( I_2=0.75\times0.25\times0.75\,I=0.1406I \).
3. Amplitudes in the ratio \( \sqrt{I_1}:\sqrt{I_2}=0.5:0.375=4:3 \).
4. \( \frac{I_{min}}{I_{max}}=\left(\frac{4-3}{4+3}\right)^2=\frac{1}{49} \), so n = 49.

Teacher's Note:
1. Square roots of intensities give the amplitudes; subtract and add them for the extremes.
2. Draw the geometry (or the phasor picture) before substituting numbers.