JEE Physics Wave Optics MCQs Set 14

Answer: (b) √3 a

Explanation:
1. A = √(a² + a² + 2a² cos 60°) = √(3a²)
2. A = √3 a.

Teacher's Note:
1. Use A² = a1² + a2² + 2a1a2 cos φ.
2. Equal amplitudes: A = 2a cos(φ/2).

Answer: (a) the phase difference between them is constant

Explanation:
1. Coherent sources emit waves of the same frequency with a constant phase difference.
2. Equal amplitude is not essential.

Teacher's Note:
1. A stable pattern needs a steady phase relation.
2. Amplitude only affects contrast.

Answer: (a) is conserved but redistributed

Explanation:
1. Interference only redistributes the energy of the waves.
2. Bright fringes gain exactly what dark fringes lose; the average intensity is I1 + I2.

Teacher's Note:
1. Energy is never created or destroyed.
2. Average over the pattern equals the sum of individual intensities.

Answer: (d) 9 : 4

Explanation:
1. Amplitude ratio = 5 : 1.
2. Imax/Imin = ((5 + 1)/(5 − 1))² = (6/4)² = 9/4.

Teacher's Note:
1. Take the square root of the intensity ratio first.
2. Imax ∝ (a1 + a2)².

Answer: (a) the waves are incoherent

Explanation:
1. Two independent sources do not maintain a fixed phase relationship.
2. Hence they are incoherent and their intensities simply add.

Teacher's Note:
1. Independent sources (two bulbs) cannot be coherent.
2. Only waves derived from a single source can be coherent.

Answer: (c) 16 : 1

Explanation:
1. Imax/Imin = ((a1 + a2)/(a2 − a1))² = ((5 + 3)/(5 − 3))² = 16.

Teacher's Note:
1. Maximum uses the sum, minimum uses the difference.
2. Square the amplitude ratio.

Answer: (b) 9 : 4

Explanation:
1. √(Imax/Imin) = (a1 + a2)/(a1 − a2) = 5, so a1 : a2 = 3 : 2.
2. I1 : I2 = 9 : 4.

Teacher's Note:
1. Solve for the amplitude ratio first.
2. Intensity ratio is the square.

Answer: (a) 2 : 1

Explanation:
1. Amplitude ratio = 3 : 1.
2. Amax : Amin = (3 + 1) : (3 − 1) = 2 : 1.

Teacher's Note:
1. Amplitudes are the square roots of the intensities.
2. Maximum is the sum, minimum is the difference.

Answer: (a) only the central fringe is white and all the other fringes are coloured

Explanation:
1. At the centre the path difference is zero for every colour, so the central fringe is white.
2. Elsewhere each colour has a different fringe width, so the fringes are coloured and overlap further out.

Teacher's Note:
1. β = λD/d depends on colour.
2. Central maximum has zero path difference for all colours.

Answer: (d) Blue

Explanation:
1. β = λD/d, so the fringe width is proportional to λ.
2. Among these colours blue has the smallest wavelength.

Teacher's Note:
1. Violet and blue have short wavelengths.
2. Red has the largest fringe width.

Answer: (c) sin⁻¹(3 × 10⁻³)

Explanation:
1. For the nth maximum d sin θ = nλ, so sin θ = nλ/d.
2. sin θ = 3 × 589 × 10⁻⁹/(0.589 × 10⁻³) = 3 × 10⁻³.

Teacher's Note:
1. Convert nm and mm to metres.
2. The result must be less than 1.

Answer: (b) a path difference of 5λ/2

Explanation:
1. The minima occur at Δx = (2n − 1)λ/2.
2. Third minimum: n = 3, Δx = 5λ/2, which is a phase difference of 5π.

Teacher's Note:
1. Minima have odd multiples of λ/2.
2. Path difference λ corresponds to phase difference 2π.

Answer: (a) the contrast of the fringes decreases

Explanation:
1. Imin = (a1 − a2)² is no longer zero, so the dark fringes are not completely dark.
2. The visibility (Imax − Imin)/(Imax + Imin) is less than 1.

Teacher's Note:
1. Unequal amplitudes give no perfect dark fringes.
2. Fringe width is unaffected.

Answer: (a) the fringe width decreases and the fringes eventually disappear

Explanation:
1. β = λD/d, so increasing d decreases β.
2. When β becomes too small, the fringes cannot be resolved and appear to disappear.

Teacher's Note:
1. β ∝ 1/d.
2. The eye or detector cannot resolve very closely spaced fringes.

Answer: (a) the dark fringes become less dark and the bright fringes become less bright

Explanation:
1. With unequal amplitudes, Imin = (a1 − a2)² > 0 and Imax = (a1 + a2)² is less than for equal slits of the same total intensity.
2. So the contrast decreases.

Teacher's Note:
1. Dark fringes are no longer completely dark.
2. Fringes do not disappear.

Answer: (d) 25 : 1

Explanation:
1. Intensities are proportional to widths: I1 : I2 = 4 : 9, so a1 : a2 = 2 : 3.
2. Imax/Imin = ((3 + 2)/(3 − 2))² = 25.

Teacher's Note:
1. Intensity ∝ slit width; amplitude ∝ √(width).
2. Maximum over minimum = ((a1 + a2)/(a1 − a2))².

Answer: (a) 0.20 mm

Explanation:
1. β = λD/d. Both λ and d increase by the same factor 1.1, so the ratio λ/d is unchanged.
2. β stays 0.20 mm.

Teacher's Note:
1. Compare the factors in numerator and denominator.
2. 1.1/1.1 = 1.

Answer: (c) w′ = w

Explanation:
1. The mica adds an extra constant path (μ − 1)t for that beam.
2. This shifts the whole pattern but does not change λ, D or d, so the fringe width is unchanged.

Teacher's Note:
1. A sheet shifts the fringes, it does not alter β.
2. β = λD/d.

Answer: (b) \( \frac{2\sqrt2}{3} \)

Explanation:
1. Amplitude ratio a1 : a2 = √2 : 1.
2. (Imax − Imin)/(Imax + Imin) = 2a1a2/(a1² + a2²) = 2√2/(2 + 1) = 2√2/3.

Teacher's Note:
1. The visibility is 2a1a2/(a1² + a2²).
2. a1² : a2² = 2 : 1.

Answer: (a) 1 mm

Explanation:
1. Angular fringe width α = λ/d, so α ∝ 1/d.
2. To double α, halve d: d = 1 mm.

Teacher's Note:
1. Angular width depends only on λ and d.
2. Doubling the angle needs half the separation.

Answer: (d) 1.2 mm

Explanation:
1. β = λD/d ∝ λ.
2. β′ = 1 × 6000/5000 = 1.2 mm.

Teacher's Note:
1. Fringe width is directly proportional to wavelength.
2. Same D and d.