JEE Physics Wave Optics MCQs Set 15

Answer: (c) shift upward by nearly two fringes

Explanation:
1. Extra path = (μ − 1)t = 0.5 × 2 μm = 1 μm = 2λ.
2. The shift is 2 fringes, towards the side of the slit covered by the film (upward), where the compensating path is zero.

Teacher's Note:
1. The central maximum moves toward the slit with the film.
2. Number of fringes shifted = (μ − 1)t/λ.

Answer: (a) the fringe width remains the same but the pattern shifts

Explanation:
1. The film adds a constant optical path (μ − 1)t to one beam.
2. This shifts the pattern as a whole; β = λD/d is unchanged.

Teacher's Note:
1. Shift of pattern, not change of spacing.
2. Fringe width depends on λ, D and d only.

Answer: (b) 420 nm

Explanation:
1. For the same path difference, nλ is constant: 3 × 700 = 5 × λ′.
2. λ′ = 420 nm.

Teacher's Note:
1. The path difference at that point is fixed.
2. Δx = nλ.

Answer: (a) I0/4

Explanation:
1. With both slits: I0 = (a + a)² ∝ 4a².
2. With one slit: intensity ∝ a² = I0/4.

Teacher's Note:
1. Amplitude halves when one slit is covered.
2. Intensity ∝ amplitude².

Answer: (a) 4.8 μm

Explanation:
1. (μ − 1)t = 4λ.
2. t = 4 × 0.6 μm/0.5 = 4.8 μm.

Teacher's Note:
1. Path introduced by a plate = (μ − 1)t.
2. Shift by n fringes means path difference nλ.

Answer: (d) 0.4 μm

Explanation:
1. The extra optical paths must be equal: (1.5 − 1)(1.2) = (2.5 − 1)t.
2. 0.6 = 1.5t, so t = 0.4 μm.

Teacher's Note:
1. Unaltered central fringe needs equal added path in both beams.
2. Path added = (μ − 1)t.

Answer: (c) interference

Explanation:
1. Light reflected from the top and bottom surfaces of the thin film interferes.
2. Different colours satisfy the constructive condition at different thicknesses.

Teacher's Note:
1. Thin film colours arise from interference.
2. Two reflected beams of path difference ~ λ.

Answer: (b) 440 nm

Explanation:
1. First minimum: a sin θ = λ1.
2. First secondary maximum: a sin θ = 3λ2/2.
3. λ2 = 2λ1/3 = 440 nm.

Teacher's Note:
1. Secondary maxima lie at (n + ½)λ.
2. Equate a sin θ.

Answer: (a) 4.5 × 10⁻³ rad

Explanation:
1. Angular width of the central maximum = 2λ/a.
2. = 2 × 450 × 10⁻⁹/(0.2 × 10⁻³) = 4.5 × 10⁻³ rad.

Teacher's Note:
1. Central maximum spans from −λ/a to +λ/a.
2. Convert mm and nm to metres.

Answer: (d) π radian

Explanation:
1. At the first minimum the path difference between the two edges is λ.
2. The edge and midpoint wavelets differ by λ/2 in path, which is a phase difference of π.

Teacher's Note:
1. Edge to mid-point is half the slit.
2. Path difference λ/2 ↔ phase π.

Answer: (c) 0.4 mm

Explanation:
1. Width of central fringe = 2λD/a.
2. a = 2 × 6.8 × 10⁻⁷ × 2/(6.8 × 10⁻³) = 4 × 10⁻⁴ m = 0.4 mm.

Teacher's Note:
1. Use the formula for the central maximum width.
2. Check all units in metres.

Answer: (b) 0.15 cm

Explanation:
1. First minimum: a sin θ = λ, and the screen is at the focal plane, so y = fλ/a.
2. y = 60 × 5 × 10⁻⁵/0.02 = 0.15 cm.

Teacher's Note:
1. For small angles y = f tan θ ≈ f sin θ.
2. Keep all lengths in cm.

Answer: (a) \( \sin^{-1}\left(\frac1{\sqrt3}\right) \)

Explanation:
1. Brewster's law: μ = tan 60° = √3.
2. Critical angle: sin C = 1/μ = 1/√3.

Teacher's Note:
1. Brewster angle gives μ.
2. sin C = 1/μ.

Answer: (d) tan⁻¹(n)

Explanation:
1. Brewster's law: tan ip = n.

Teacher's Note:
1. Reflected light is fully polarised when the reflected and refracted rays are at 90°.
2. μ = tan ip.

Answer: (c) varies between a maximum and zero

Explanation:
1. By Malus' law I = I0 cos²θ.
2. It falls from I0 at 0° to zero at 90°.

Teacher's Note:
1. Malus' law.
2. Crossed polaroids block plane polarised light completely.

Answer: (b) 12.5%

Explanation:
1. The first Nicol passes half: I1 = I0/2.
2. Second: I2 = I1 cos² 60° = (I0/2)(1/4) = I0/8 = 12.5% of I0.

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

Answer: (a) the reflected ray is completely polarised and the transmitted ray is partially polarised

Explanation:
1. At the Brewster angle the reflected light has only the component perpendicular to the plane of incidence.
2. The refracted light is a mixture and remains partially polarised.

Teacher's Note:
1. Only the reflected beam is fully polarised.
2. The refracted beam loses some of one component only.

Answer: (a) \( \frac{3I_0}{32} \)

Explanation:
1. After P1: I0/2.
2. After P3 (60° to P1): (I0/2) cos² 60° = I0/8.
3. P3 is at 30° to P2 (which is at 90° to P1): I = (I0/8) cos² 30° = 3I0/32.

Teacher's Note:
1. Apply Malus' law at each polaroid.
2. Angle between P3 and P2 is 90° − 60° = 30°.

Answer: (d) The reflected light is polarised with its electric vector perpendicular to the plane of incidence

Explanation:
1. Perpendicular reflected and refracted rays mean i is the Brewster angle, tan i = μ.
2. At this angle the reflected light is plane polarised with E perpendicular to the plane of incidence.

Teacher's Note:
1. tan i = μ, not 1/μ.
2. The component in the plane of incidence is not reflected.

Answer: (c) 7 : 3

Explanation:
1. Resolving power ∝ 1/λ.
2. R1/R2 = λ2/λ1 = 7000/3000 = 7/3.

Teacher's Note:
1. Shorter wavelength resolves better.
2. Invert the wavelength ratio.