Wave Optics MCQs

Huygens' principle states that each point on a wavefront can be considered as a source of new wavelets.

These secondary wavelets propagate isotropically, i.e., in all directions, with the same speed as the original wave (the speed of light for light waves).

The envelope of these wavelets after a short time gives the new position of the wavefront.

Thus the correct choice is “Secondary wavelets that spread out in all directions with the speed of light”.

Constructive interference occurs when the two waves arrive in phase, i.e., their crests coincide.

Phase difference = (2π/λ)·(path difference) must be an integer multiple of 2π.

Thus path difference = n λ where n = 0,1,2,… gives a phase difference of 2πn.

Hence the correct choice is n λ (n = 0, 1, 2, …).

For two waves of equal intensity I0, the electric fields have the same amplitude √I0.

Resultant amplitude = √I0 e^{iωt} + √I0 e^{i(ωt+π)} = √I0 (e^{iωt} – e^{iωt}) = 0.

Resultant intensity ∝ (resultant amplitude)² = 0.

Hence the resultant intensity is Zero.

Path difference between rays from the extreme edges of the slit = a sinθ.

For a minimum the waves from the two halves of the slit must be out of phase by π, i.e. path difference = n λ (n = 1,2,3…).

Thus the condition for minima is a sinθ = n λ.

Diffraction from a circular aperture gives an Airy pattern; the first minimum occurs at angle θ = 1.22 λ/D.

Resolution is defined by Rayleigh’s criterion: two point sources are just resolved when the central maximum of one coincides with the first minimum of the other.

Thus the smallest resolvable angular separation equals the angular radius of the first dark ring, θ = 1.22 λ/D.

Hence the limit of resolution is 1.22 λ/D.

Wavelength in a medium λ = λ₀ / n, where λ₀ is wavelength in vacuum and n is refractive index.

Given λ₀ = 600 nm and n = 1.5, λ = 600 nm ÷ 1.5 = 400 nm.

Thus the wavelength of light in the medium is 400 nm. 400 nm.

For two coherent sources with intensities I1 and I2, the resultant intensity at any point is I = I1 + I2 + 2√(I1 I2) cos δ.

Maximum intensity occurs at δ = 0: Imax = (√I1 + √I2)².

Minimum occurs at δ = π: Imin = (√I1 – √I2)².

With I1 : I2 = 1 : 4, √I1 : √I2 = 1 : 2, so Imax = (1+2)² = 9 and Imin = (2–1)² = 1; thus Imax : Imin = 9 : 1.

Phase difference = (2π/λ) × (path difference).

For a path difference equal to one wavelength (Δx = λ), substitute: (2π/λ)·λ = 2π.

Thus a full‑wave path difference corresponds to a phase shift of 2π radians.

2 π.

For a single‑slit diffraction the first minimum occurs when a·sinθ = λ, where a is the slit width.

Convert a = 0.2 mm = 2×10⁻⁴ m and λ = 600 nm = 6×10⁻⁷ m.

sinθ ≈ λ/a = (6×10⁻⁷)/(2×10⁻⁴) = 3×10⁻³, and for small angles sinθ ≈ θ (in radians).

Thus θ ≈ 3×10⁻³ rad, which matches the given answer.

Brewster’s (polarising) angle θp satisfies tan θp = n (refractive index of the second medium).

For water n = 1.33, so θp = tan⁻¹(1.33).

Evaluating gives θp ≈ 53°.

Hence the polarising angle is approximately 53 degrees.

All points that oscillate together have the same phase of displacement.

A surface joining these points at any instant is called a wavefront.

It is perpendicular to the direction of propagation and differs from a node or antinode which are specific points.

Hence the continuous locus is a wavefront.

Destructive interference occurs when the two waves arrive out of phase by half a wavelength.

The condition for a phase difference of π (180°) is path difference = (2n‑1)·λ/2, where n = 1,2,3…

Substituting n = 1 gives λ/2, n = 2 gives 3λ/2, etc., all giving a half‑integer multiple of λ.

Thus the required path difference is (2n‑1)λ/2, which matches the given answer.

Resultant intensity I = I1 + I2 + 2√(I1I2)cosδ, where δ is the phase difference.

Here I1 = I2 = I0 and δ = 120° → cos120° = –½.

Thus I = I0 + I0 + 2·I0·(–½) = 2I0 – I0 = I0.

Hence the resultant intensity is I0.

For a single slit the first minima occur at a sinθ = ±λ.

The angular position of the first minima on either side of the central maximum is therefore θ ≈ λ/a (small‑angle approximation).

The angular width of the central bright region is the angle between these two minima, i.e. Δθ = 2θ ≈ 2λ/a.

Hence the angular width of the central maximum is 2 λ/a.

Resolving power (angular resolution) θ_min = 1.22 λ/D from Rayleigh criterion.

A larger D makes θ_min smaller, meaning finer detail can be distinguished, so resolving power ∝ 1/θ_min.

Thus resolving power ∝ D/(1.22 λ).

Hence the correct choice is D / (1.22 λ).

A point source emits light equally in all directions in an isotropic medium, so the distance from the source to any point on the wavefront is the same.

Equal distance from a single point defines a sphere.

Thus the wavefront is a spherical surface.

For constructive interference the path difference must be an integer multiple of the wavelength, giving a phase difference of 2π × integer.

Phase difference = (2π/λ)·(path difference) → set path difference = nλ.

Thus Δϕ = 2πn, which is written as 2n π.

2n π.

The central maximum corresponds to zero path difference, so all wavelengths from the white source arrive in phase.

Since every colour interferes constructively at the same point, their superposition reproduces the original white light.

Thus the central fringe is observed as white.

Width of central maximum = 2 λ D / a (single‑slit diffraction).

Substitute λ = 500 nm = 5×10⁻⁷ m, D = 1 m, a = 0.2 mm = 2×10⁻⁴ m.

Width = 2 × 5×10⁻⁷ × 1 / 2×10⁻⁴ = 5×10⁻³ m = 5 mm.

Hence the central maximum is 5 mm wide.

Resolving power ∝ 1/λ (Rayleigh criterion d = 1.22 λ/NA).

A smaller λ reduces the minimum resolvable distance d, giving finer detail.

Thus using light of shorter wavelength improves the microscope’s resolution.