JEE Physics Wave Optics MCQs Set 18

Answer: (d) (ii) and (iv) only

Explanation:
1. \( \frac{I_{max}}{I_{min}}=\left(\frac{a_1+a_2}{a_1-a_2}\right)^2=9\Rightarrow\frac{a_1+a_2}{a_1-a_2}=3\Rightarrow\frac{a_1}{a_2}=2 \).
2. So the intensities are in the ratio 4 : 1 (statements ii and iv).

Teacher's Note:
1. Take the square root of the intensity ratio first.
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) (ii), (iii) and (iv) only

Explanation:
1. All colours have their maximum at the centre, so the central maximum is white.
2. The first maximum of violet is nearest to the centre because its wavelength is smallest, so the maximum next to the centre is violet, not red.
3. Different colours have dark fringes at different places, so no point is dark for all colours.

Teacher's Note:
1. Fringe positions scale with wavelength.
2. Intensity problems are easiest with the formula I = I1 + I2 + 2 sqrt(I1 I2) cos(phi).

Answer: (a) (i) and (iii) only

Explanation:
1. The path difference at the point in front of a slit is \( \sqrt{d^2+b^2}-d\approx\frac{b^2}{2d} \).
2. A wavelength is missing if \( \frac{b^2}{2d}=\left(n+\tfrac12\right)\lambda\Rightarrow\lambda=\frac{b^2}{(2n+1)d} \).
3. This gives \( \frac{b^2}{d},\frac{b^2}{3d},\frac{b^2}{5d},\dots \)

Teacher's Note:
1. Use the binomial approximation for the path difference.
2. Eliminate options that violate a limiting case (for example phi = 0 or phi = pi) first.

Answer: (a) (i) and (ii) only

Explanation:
1. Maxima need \( d\sin\theta=m\lambda \) with \( |\sin\theta|\le1 \). For d = \( \lambda \) the first order would be at 90° (grazing), so only the central maximum is seen. For \( \lambda\lt d\lt2\lambda \) the first order occurs at a finite angle, so statement (ii) is true.
2. Let the intensities be 4 and 1: \( I_{max}=9,\ I_{min}=1 \). If slit 1 is reduced to 1: \( I_{max}=4,\ I_{min}=0 \), both decrease, so (iii) is false. If slit 2 is raised to 4: \( I_{max}=16,\ I_{min}=0 \), the dark fringe becomes darker, so (iv) is false.

Teacher's Note:
1. Evaluate the extremes with \( (\sqrt{I_1}\pm\sqrt{I_2})^2 \).
2. Draw the geometry (or the phasor picture) before substituting numbers.

Answer: (c) \( (2n+1)\frac{\lambda}{4} \)

Explanation:
1. \( I=I_0\cos^2\frac{\phi}{2}=\frac{I_0}{2}\Rightarrow\frac{\phi}{2}=(2n+1)\frac{\pi}{4} \), so \( \phi=(2n+1)\frac{\pi}{2} \).
2. Path difference \( =\frac{\phi\lambda}{2\pi}=(2n+1)\frac{\lambda}{4} \).

Teacher's Note:
1. Half intensity means cos²(φ/2) = 1/2.
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) the fractional error in d decreases

Explanation:
1. \( d=\frac{\lambda}{2\sin\theta}\Rightarrow\frac{\Delta d}{d}=\cot\theta\,\Delta\theta \).
2. As \( \theta \) increases, \( \cot\theta \) decreases, so the fractional error decreases. (The absolute error \( \frac{\lambda\cos\theta}{2\sin^2\theta}\Delta\theta \) also decreases, so it is not constant and does not increase.)

Teacher's Note:
1. Differentiate d with respect to \( \theta \).
2. Intensity problems are easiest with the formula I = I1 + I2 + 2 sqrt(I1 I2) cos(phi).

Answer: (a) (i) and (iii) only

Explanation:
1. \( d/\lambda=\frac{1.8\times10^{-3}}{6\times10^{-7}}=3000 \).
2. At P1 the path difference is 0 and at P2 it is d = 3000 wavelengths, so there are about 3000 fringes between them and P2 is a bright spot of the maximum order (3000). So (i) and (iii) are true and (ii) is false.
3. With \( \Delta=d\cos\varphi \) (angle measured from the line joining the sources) the spacing of bright spots in angle is \( \frac{\lambda}{d\sin\varphi} \), which is smallest near P1 and increases toward P2, so (iv) is false.

Teacher's Note:
1. Maximum possible path difference equals the separation of the sources.
2. Eliminate options that violate a limiting case (for example phi = 0 or phi = pi) first.

Answer: (d) concentric circles

Explanation:
1. Points on the screen at the same distance from the axis have the same path difference to the two sources, so lines of equal path difference are circles about the axis.
2. Hence bright and dark rings (concentric circles) are seen.

Teacher's Note:
1. If the screen were parallel to the line joining the sources, hyperbolic fringes would result.
2. Draw the geometry (or the phasor picture) before substituting numbers.

Answer: (c) planar

Explanation:
1. A parallel (collimated) beam has plane wavefronts perpendicular to the direction of propagation.
2. The wavefront bends only after the beam has travelled in the medium.

Teacher's Note:
1. Parallel beam means plane wavefront.
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) minimum on the axis of the beam

Explanation:
1. \( v=\frac{c}{\mu} \) and \( \mu \) is largest where the intensity is largest, i.e. on the axis.
2. So the speed is minimum on the axis; it is inversely related to the intensity.

Teacher's Note:
1. Higher refractive index means lower speed.
2. Intensity problems are easiest with the formula I = I1 + I2 + 2 sqrt(I1 I2) cos(phi).

Answer: (d) 2

Explanation:
1. Coherent: amplitudes add, so \( I_1\propto(2A)^2=4A^2 \).
2. Incoherent: intensities add, so \( I_2\propto A^2+A^2=2A^2 \).
3. Ratio \( I_1/I_2=2 \).

Teacher's Note:
1. Amplitudes add for coherent sources; intensities add for incoherent ones.
2. Eliminate options that violate a limiting case (for example phi = 0 or phi = pi) first.

Answer: (d) \( \frac13 \)

Explanation:
1. After rotating through 30°, beam A gives \( I_A\cos^230^\circ=\tfrac34I_A \) and beam B gives \( I_B\sin^230^\circ=\tfrac14I_B \).
2. Equal brightness: \( \frac34I_A=\frac14I_B\Rightarrow\frac{I_A}{I_B}=\frac13 \).

Teacher's Note:
1. Use Malus's law for each beam separately.
2. Draw the geometry (or the phasor picture) before substituting numbers.

Answer: (c) bends upwards

Explanation:
1. The lower part of a wavefront moves in lower-index air, i.e. faster, than the upper part.
2. The wavefront tilts so that the beam turns toward the region of higher refractive index, which is upward (this is how a mirage is formed).

Teacher's Note:
1. Light bends toward the slower (higher index) region.
2. Check that your answer has the right units and a sensible order of magnitude.

Answer: (b) \( 30\ \mu\text{m} \)

Explanation:
1. \( \theta=\frac{1.22\lambda}{D}=\frac{1.22\times5\times10^{-7}}{5\times10^{-3}}=1.22\times10^{-4}\ \text{rad} \) (pupil diameter 0.5 cm).
2. \( s=\theta\times25\ \text{cm}=3.05\times10^{-5}\ \text{m}\approx30\ \mu\text{m} \)

Teacher's Note:
1. Use the diameter, not the radius, in the formula.
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) 7.8 mm

Explanation:
1. Bright fringes coincide when \( n_1\times650=n_2\times520\Rightarrow\frac{n_1}{n_2}=\frac{520}{650}=\frac45 \), so the smallest values are \( n_1=4,\ n_2=5 \).
2. \( y=\frac{4\times650\times10^{-9}\times1.5}{0.5\times10^{-3}}=7.8\times10^{-3}\ \text{m} \)

Teacher's Note:
1. Reduce the ratio of the wavelengths to the smallest integers.
2. Intensity problems are easiest with the formula I = I1 + I2 + 2 sqrt(I1 I2) cos(phi).

Answer: (a) 45°

Explanation:
1. After A the intensity is \( I/2 \). With C at angle \( \theta \) to A and B parallel to A, the final intensity is \( \frac I2\cos^2\theta\cos^2\theta=\frac I8 \).
2. \( \cos^4\theta=\frac14\Rightarrow\cos^2\theta=\frac12\Rightarrow\theta=45^\circ \)

Teacher's Note:
1. Apply Malus's law twice.
2. Fringe width depends only on wavelength, screen distance and slit separation; remember how each enters.

Answer: (d) \( 25\ \mu\text{m} \)

Explanation:
1. The central maximum has half-angle 30°: \( a\sin30^\circ=\lambda\Rightarrow\lambda=0.5\ \mu\text{m} \).
2. \( \beta=\frac{\lambda D}{d}\Rightarrow d=\frac{0.5\times10^{-6}\times0.5}{10^{-2}}=2.5\times10^{-5}\ \text{m}=25\ \mu\text{m} \)

Teacher's Note:
1. Find the wavelength from the single-slit condition first.
2. Eliminate options that violate a limiting case (for example phi = 0 or phi = pi) first.

Answer: (c) 428.6 nm and 600 nm

Explanation:
1. Path difference at the hole: \( \Delta=\frac{yd}{D}=\frac{1.5\times10^{-3}\times10^{-3}}{1}=1.5\ \mu\text{m}=1500\ \text{nm} \).
2. Minima: \( \Delta=(m+\tfrac12)\lambda\Rightarrow\lambda=\frac{1500}{m+1/2} \) = 1000, 600, 428.6, 333 nm for m = 1, 2, 3, 4.
3. Only 600 nm and 428.6 nm lie in 400-700 nm.

Teacher's Note:
1. List the wavelengths for successive m and keep those in the visible range.
2. Draw the geometry (or the phasor picture) before substituting numbers.

Answer: (b) 500 nm

Explanation:
1. The path difference at the hole is 1500 nm.
2. Maxima: \( \Delta=m\lambda\Rightarrow\lambda=\frac{1500}{m} \) = 1500, 750, 500, 375 nm for m = 1, 2, 3, 4.
3. Only 500 nm lies in the visible range 400-700 nm.

Teacher's Note:
1. A wavelength can be both a candidate for a maximum and not for a minimum.
2. Check that your answer has the right units and a sensible order of magnitude.

Answer: (b) 1.17 mm

Explanation:
1. \( y_3=\frac{3\lambda D}{d}=\frac{3\times6.5\times10^{-7}\times1.2}{2\times10^{-3}}=1.17\times10^{-3}\ \text{m} \)

Teacher's Note:
1. Convert angstrom to metres: 1 Å = 10⁻¹⁰ m.
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) 1.56 mm

Explanation:
1. Coincidence needs \( n_1\times6500=n_2\times5200\Rightarrow n_1:n_2=4:5 \).
2. \( y=\frac{4\times6.5\times10^{-7}\times1.2}{2\times10^{-3}}=1.56\ \text{mm} \)

Teacher's Note:
1. Reduce the ratio 6500 : 5200 = 5 : 4.
2. Intensity problems are easiest with the formula I = I1 + I2 + 2 sqrt(I1 I2) cos(phi).

Answer: (d) zero

Explanation:
1. Extra path \( (\mu-1)t=0.5\times1.5\ \mu\text{m}=0.75\ \mu\text{m} \) = 1.5 wavelengths (of 0.5 µm).
2. A path difference of an odd number of half wavelengths gives destructive interference, so the intensity is zero.

Teacher's Note:
1. Express the extra path in wavelengths.
2. Fringe width depends only on wavelength, screen distance and slit separation; remember how each enters.

Answer: (c) 1.5 mm

Explanation:
1. \( \text{shift}=\frac{(\mu-1)tD}{d}=\frac{0.75\times10^{-6}\times1}{5\times10^{-4}}=1.5\times10^{-3}\ \text{m} \)

Teacher's Note:
1. The shift is towards the slit that carries the plate.
2. Eliminate options that violate a limiting case (for example phi = 0 or phi = pi) first.

Answer: (c) 600 nm

Explanation:
1. \( \frac{\lambda D}{d}=0.25\ \text{mm},\ \frac{\lambda D}{d+1.2\ \text{mm}}=\frac23\times0.25\ \text{mm} \).
2. Dividing: \( \frac{d+1.2}{d}=\frac32\Rightarrow d=2.4\ \text{mm} \).
3. \( \lambda=\frac{\beta d}{D}=\frac{0.25\times10^{-3}\times2.4\times10^{-3}}{1}=6\times10^{-7}\ \text{m} \)

Teacher's Note:
1. Divide the two fringe-width equations to eliminate the wavelength.
2. Draw the geometry (or the phasor picture) before substituting numbers.

Answer: (b) \( 24\ \mu\text{m} \)

Explanation:
1. Final slit separation \( d'=3.6\ \text{mm} \) and fringe width \( \beta'=\frac23\times0.25=\frac16\ \text{mm} \).
2. Shift = 20β' = 3.33 mm = \( \frac{(\mu-1)tD}{d'} \).
3. \( t=\frac{3.33\times10^{-3}\times3.6\times10^{-3}}{0.5\times1}=2.4\times10^{-5}\ \text{m}=24\ \mu\text{m} \)

Teacher's Note:
1. Use the final slit separation, not the original one.
2. Check that your answer has the right units and a sensible order of magnitude.

Answer: (a) 90 nm

Explanation:
1. The reflection at the top surface (air to 1.8) has a phase change of \( \pi \); the reflection at the lower surface (1.8 to 1.5, into a rarer medium) has none.
2. Constructive interference: \( 2\mu t=\left(n+\tfrac12\right)\lambda \); the least thickness is for n = 0: \( t=\frac{\lambda}{4\mu}=\frac{648}{4\times1.8}=90\ \text{nm} \).

Teacher's Note:
1. Check the phase changes at both surfaces by comparing refractive indices.
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) 4.33 mm towards the covered slit

Explanation:
1. Relative index of glass with respect to the medium: \( \frac{1.5}{4/3}=1.125 \).
2. Extra optical path: \( (1.125-1)\times10.4\ \mu\text{m}=1.3\ \mu\text{m} \).
3. \( \text{shift}=\frac{\Delta D}{d}=\frac{1.3\times10^{-6}\times1.5}{0.45\times10^{-3}}=4.33\ \text{mm} \), toward the covered slit.

Teacher's Note:
1. Wavelengths are quoted in the medium, so use the relative refractive index of the sheet.
2. Intensity problems are easiest with the formula I = I1 + I2 + 2 sqrt(I1 I2) cos(phi).

Answer: (d) \( \frac34 \)

Explanation:
1. Path difference at O is \( 1.3\ \mu\text{m}=\frac{13}{6}\lambda \) (with \( \lambda=0.6\ \mu\text{m} \)), so \( \phi=2\pi\times\frac{13}{6} \).
2. \( \frac{I}{I_{max}}=\cos^2\frac{\phi}{2}=\cos^2\frac{13\pi}{6}=\cos^2\frac{\pi}{6}=\frac34 \)

Teacher's Note:
1. Reduce the phase modulo 2π before taking the cosine.
2. Fringe width depends only on wavelength, screen distance and slit separation; remember how each enters.

Answer: (c) 650 nm and 433.3 nm

Explanation:
1. Path difference at O: \( 1.3\ \mu\text{m}=1300\ \text{nm} \).
2. Maxima: \( \lambda=\frac{1300}{m} \) = 1300, 650, 433.3, 325 nm for m = 1, 2, 3, 4.
3. Only 650 nm and 433.3 nm lie in 400-700 nm.

Teacher's Note:
1. Use the same list-and-select method as for the earlier fixed-hole problem.
2. Eliminate options that violate a limiting case (for example phi = 0 or phi = pi) first.