Question 1

What is the formula for fringe width in Young's double slit experiment?

Accepted Answer

Fringe width β = λD/d, where λ is the wavelength of light, D is the distance from the slits to the screen, and d is the separation between the two slits. It is the distance between two consecutive bright (or two consecutive dark) fringes, and is the same for both.

Question 2

What is the condition for bright and dark fringes in YDSE?

Accepted Answer

A bright fringe forms where the path difference Δ = d·sinθ ≈ d·y/D equals a whole number of wavelengths: Δ = nλ (n = 0, 1, 2, …), giving constructive interference. A dark fringe forms where Δ = (n+½)λ, giving destructive interference. The phase difference is φ = 2πΔ/λ.

Question 3

What happens to the fringe width if the apparatus is immersed in water?

Accepted Answer

Inside a medium of refractive index μ the wavelength shrinks to λ/μ while the frequency stays the same. Since β = λD/d, the fringe width also decreases by the same factor μ — the fringes move closer together. For water (μ ≈ 1.33) the fringes are about 1.33 times narrower.

Question 4

Why is the central fringe white in white-light YDSE?

Accepted Answer

At the central point the path difference is zero for every wavelength, so all colours interfere constructively there and combine to give white. Away from the centre each colour has a different fringe width (β ∝ λ), so coloured fringes appear and quickly overlap and wash out at higher orders.

Question 5

How does intensity vary across the interference pattern?

Accepted Answer

For two equal coherent slits the intensity is I = 4I₀cos²(φ/2), where φ = 2πΔ/λ is the phase difference. It is maximum (4I₀) at the bright fringes where φ = 2nπ, and zero at the dark fringes where φ = (2n+1)π, producing the familiar series of evenly spaced bright and dark bands.

Young's Double Slit — Interference & Fringe Width

Practice

Learned it? Now prove it under the clock.