JEE Physics Wave Optics MCQs Set 11

Answer: (a) 5850 Å

Explanation:
1. Object distance \( u=30\ \text{cm} \), image distance \( v=70\ \text{cm} \), so \( m=\frac{70}{30} \).
2. Slit separation \( d=\frac{0.7}{7/3}=0.3\ \text{cm} \).
3. \( \lambda=\frac{\beta d}{D}=\frac{0.0195\times0.3}{100}=5.85\times10^{-5}\ \text{cm}=5850\ \text{Å} \).

Teacher's Note:
1. The lens magnifies the separation of the two virtual sources.
2. Magnification = v/u.

Answer: (d) \( \frac\lambda4 \)

Explanation:
1. With index between air and glass, both reflections are from a denser medium, so the phase changes cancel.
2. Destructive interference needs \( 2t=\frac\lambda2 \), so \( t=\frac\lambda4 \).

Teacher's Note:
1. Minimum reflection needs a path difference of half a wavelength in the film.
2. λ here is the wavelength inside the coating.

Answer: (c) \( \frac12I_0 \)

Explanation:
1. A polariser transmits half of the unpolarised light: \( I_t=\frac{I_0}{2} \).
2. The part not transmitted is \( I_0-\frac{I_0}{2}=\frac{I_0}{2} \).

Teacher's Note:
1. Unpolarised light has equal components in all directions.
2. Average of cos² is 1/2.

Answer: (a) straight line

Explanation:
1. The fringes are the intersection of the screen with hyperboloids of revolution about the line joining the slits.
2. On a screen parallel to the plane of the slits and far away, the portions near the axis are essentially straight lines parallel to the slits.

Teacher's Note:
1. Fringes are loci of constant path difference.
2. Near the axis the hyperbolic curves are almost straight.

Answer: (c) 442.5 nm

Explanation:
1. Coincidence: \( 3\times590=4\lambda \).
2. \( \lambda=\frac{1770}{4}=442.5\ \text{nm} \).

Teacher's Note:
1. Position of the nth bright fringe is nλD/d.
2. Equate positions.

Answer: (b) 2

Explanation:
1. Coherent: amplitude \( 2A \), so \( I_1\propto4A^2 \).
2. Incoherent: intensities add, \( I_2\propto A^2+A^2=2A^2 \).
3. \( \frac{I_1}{I_2}=2 \).

Teacher's Note:
1. Coherent waves add amplitudes, incoherent waves add intensities.
2. The central point has zero path difference.

Answer: (a) 2 : 1

Explanation:
1. At P, \( \phi=0 \): \( I_P=I_0 \) (maximum). At Q, \( \phi=\frac\pi2 \): \( I_Q=I_0\cos^2\frac\pi4=\frac{I_0}{2} \).
2. The ratio \( I_P:I_Q=2:1 \).

Teacher's Note:
1. I = I0 cos²(φ/2).
2. Path difference λ/4 gives a phase difference π/2.

Answer: (c) \( \frac{I_0}{4} \)

Explanation:
1. After A: \( \frac{I_0}{2} \).
2. After B (Malus' law): \( \frac{I_0}{2}\cos^2 45^\circ=\frac{I_0}{4} \).

Teacher's Note:
1. Unpolarised light loses half at the first polaroid.
2. Then apply Malus' law.

Answer: (a) \( \frac13 \)

Explanation:
1. After rotation: \( I_A\cos^2 30^\circ=\frac34I_A \) and \( I_B\sin^2 30^\circ=\frac14I_B \).
2. Equal brightness: \( \frac34I_A=\frac14I_B \), so \( \frac{I_A}{I_B}=\frac13 \).

Teacher's Note:
1. Beam B is perpendicular to A, so it transmits sin².
2. Apply Malus' law to each beam.

Answer: (d) 30 cm

Explanation:
1. Refractive index \( \mu=\frac{3\times10^8}{2\times10^8}=1.5 \).
2. Radius of curvature from the sag: \( R=\frac{h^2+s^2}{2s}=\frac{30^2+3^2}{6}\ \text{mm}\approx151\ \text{mm}\approx15\ \text{cm} \).
3. \( f=\frac{R}{\mu-1}=\frac{15}{0.5}=30\ \text{cm} \).

Teacher's Note:
1. Find μ from c/v.
2. Use the chord-sagitta relation for R.

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

Explanation:
1. Limit of resolution \( \theta=\frac{1.22\lambda}{D}=\frac{1.22\times500\times10^{-9}}{5\times10^{-3}}=1.22\times10^{-4}\ \text{rad} \).
2. Separation \( =\theta\times0.25\ \text{m}=3.05\times10^{-5}\ \text{m}\approx30\ \mu m \).

Teacher's Note:
1. The aperture diameter is 0.5 cm.
2. Separation = θ × viewing distance.

Answer: (a) bends upwards

Explanation:
1. Near the ground the refractive index is smaller, so light travels faster there.
2. The lower part of each wavefront moves ahead of the upper part, so the wavefront tilts upwards and the beam bends upwards.

Teacher's Note:
1. Speed is c/μ.
2. A beam bends towards the region of higher refractive index.

Answer: (c) 45°

Explanation:
1. Beyond A the intensity is \( \frac I2 \). Since B transmits all of it, B is parallel to A.
2. With C inserted at angle θ to A (and so to B): \( \frac I2\cos^2\theta\cos^2\theta=\frac I8 \), so \( \cos^4\theta=\frac14 \) and \( \cos^2\theta=\frac12 \).
3. Hence \( \theta=45^\circ \).

Teacher's Note:
1. B is parallel to A.
2. Apply Malus' law twice.

Answer: (b) \( 25\ \mu m \)

Explanation:
1. First minimum at 30°: \( a\sin30^\circ=\lambda \), so \( \lambda=0.5\ \mu m \).
2. \( d=\frac{\lambda D}{\beta}=\frac{0.5\times10^{-6}\times0.5}{10^{-2}}=25\ \mu m \).

Teacher's Note:
1. Half of the central width is 30°.
2. Use β = λD/d.

Answer: (a) \( \sin^{-1}\left(\frac{\lambda}{3d}\right) \)

Explanation:
1. \( I\cos^2\frac\phi2=\frac I4 \) gives \( \cos\frac\phi2=\frac12 \), so \( \phi=\frac{2\pi}{3} \).
2. Path difference \( d\sin\theta=\frac\lambda3 \), so \( \theta=\sin^{-1}\frac{\lambda}{3d} \).

Teacher's Note:
1. I = I_max cos²(φ/2).
2. φ = 2π/3 corresponds to a path difference λ/3.

Answer: (c) \( \pi r^2\propto f^2 \)

Explanation:
1. The sun subtends a fixed angle, so the image radius is \( r=f\theta \) and the area \( \pi r^2\propto f^2 \).
2. Covering half the lens does not change the image size, only its brightness, and doubling f increases the image area four times, so the intensity decreases.

Teacher's Note:
1. Image size of a distant object is f × (angle subtended).
2. Blocking part of a lens dims the image.

Answer: (a) I(P2) > I(P1)

Explanation:
1. The sheet adds an optical path λ/2 to S2. The net path difference (S2 side minus S1 side) is \( \frac\lambda2-\Delta x \) where \( \Delta x=S_1P-S_2P \).
2. At P0: \( \frac\lambda2 \), phase π, so intensity is zero. At P1: \( \frac\lambda2-\frac\lambda4=\frac\lambda4 \), phase π/2, so \( I=2I_0 \). At P2: \( \frac\lambda2-\frac\lambda3=\frac\lambda6 \), phase π/3, so \( I=3I_0 \).
3. Thus \( I(P_2)>I(P_1) \); P0 is dark, δ(P0) is not zero, and I(P1) is not zero.

Teacher's Note:
1. The sheet shifts the whole pattern by half a fringe.
2. I = 4I0 cos²(φ/2) for equal slits.

Answer: (d) I(P0) > I(P2) > I(P1) = 0

Explanation:
1. The net path difference at a point is \( \frac{3\lambda}{4}-(S_1P-S_2P) \).
2. At P0: \( \frac{3\lambda}4 \), phase \( \frac{3\pi}2 \), \( I=2I_0 \). At P1: \( \frac{3\lambda}4-\frac\lambda4=\frac\lambda2 \), phase π, so \( I=0 \). At P2: \( \frac{3\lambda}{4}-\frac\lambda3=\frac{5\lambda}{12} \), phase \( \frac{5\pi}{6} \), \( I=4I_0\cos^2\frac{5\pi}{12}\approx0.27I_0 \).
3. So \( I(P_0)>I(P_2)>I(P_1)=0 \).

Teacher's Note:
1. Compute the phase at each point.
2. I = 4I0 cos²(φ/2) for equal slits.

Answer: (c) \( \beta_R>\beta_G>\beta_B \)

Explanation:
1. Fringe width \( \beta=\frac{\lambda D}{d}\propto\lambda \).
2. Wavelengths: red > green > blue, so \( \beta_R>\beta_G>\beta_B \).

Teacher's Note:
1. Fringe width is proportional to wavelength.
2. Red has the longest wavelength.

Answer: (b) \( (2n+1)\frac\lambda4 \)

Explanation:
1. \( \cos^2\frac\phi2=\frac12 \) gives \( \phi=(2n+1)\frac\pi2 \).
2. Path difference \( =\frac{\lambda}{2\pi}\phi=(2n+1)\frac\lambda4 \).

Teacher's Note:
1. Half of the maximum intensity means cos²(φ/2) = 1/2.
2. Convert phase to path with λ/2π.

Answer: (a) decreases

Explanation:
1. \( d=\frac{\lambda}{2\sin\theta} \), so \( \frac{\Delta d}{d}=\cot\theta\ \Delta\theta \).
2. Since cotθ decreases as θ increases from 0° to 90°, the fractional error decreases.

Teacher's Note:
1. Differentiate ln d = ln(λ/2) − ln sinθ.
2. cotθ decreases with θ.

Answer: (a) A, B and C only

Explanation:
1. Fringe width is proportional to λ, so \( \beta_2>\beta_1 \) (A true) and \( m=\frac y\beta \) gives \( m_1>m_2 \) (B true).
2. 3rd maximum of λ2 is at \( 3\times600=1800 \) (in units of nm × D/d) and the 5th minimum of λ1 at \( 4.5\times400=1800 \) (C true).
3. Angular separation is proportional to λ, so it is smaller for λ1 (D false).

Teacher's Note:
1. Maxima at nλ, minima at (n − ½)λ.
2. Angular fringe separation = λ/d.

Answer: (d) B and D only

Explanation:
1. With θ measured from the bisector, bright spots obey \( d\sin\theta=m\lambda \), so \( \Delta\theta=\frac{\lambda}{d\cos\theta} \), which increases towards P2 (A false).
2. At P2, \( \theta=90^\circ \) and \( m=\frac d\lambda=\frac{1.8\times10^{-3}}{600\times10^{-9}}=3000 \), an integer, so P2 is a bright spot of maximum order (B true, C false).
3. From P1 (m = 0) to P2 (m = 3000) there are about 3000 fringes (D true).

Teacher's Note:
1. Path difference is d sinθ from the bisector.
2. d/λ = 3000 is an integer, so P2 is bright.