Question

In a Young’s double-slit experiment, two waves each of intensity I superpose each other and produce an interference pattern. Prove that the resultant intensities at maxima and minima are 4 I and zero respectively.

Answer 1

Given: I₁ = I₂ = I

Interference occurs due to the superposition of wave amplitudes. The resultant intensity at a point depends on the phase difference between the interfering waves.

Resultant intensity (I_R) = `I_1 + I_2 + 2 sqrt(I_1 I_2) cos phi`

= 2I + 2I cos Φ

= 2I(1 + cos Φ)

= `2I(2 cos^2 (theta/2))`    ...[1 + cos Φ = 2 cos²(Φ/2)]

= `4I cos^2 (theta/2)`    ...(i)

For Maxima (Constructive Interference), Phase difference (Φ) = 2nπ.

∴ `cos(theta/2)` = ±1

Putting these values in equation (i), we get:

I_max = 4I(1)² = 4I

For Minima (Destructive Interference), Phase difference (Φ) = (2n + 1)π.

∴ `cos(theta/2)` = 0

Putting these values in equation (i), we get:

I_min = 4I(0)² = 0

∴ The intensity at maxima is 4I and at minima is 0, as required.