Question

In biprism experiment the maximum intensity is ‘I₀’. If the path difference between the two interfering waves is ‘λ/4’ then intensity at the point on the screen is ______.

`[sin 45^circ = cos 45^circ = 1/sqrt 2]`

विकल्प

• `I_0/4`

• `I_0/3`

• `I_0/2`

• I₀

Answer 1

In biprism experiment the maximum intensity is ‘I₀’. If the path difference between the two interfering waves is ‘λ/4’ then intensity at the point on the screen is `bbunderline(I_0/2)`.

Explanation:

Given: Path difference (Δx) = `lambda/4`

Phase difference (δ) = `(2 pi)/lambda * lambda/4`

= `pi/2`

Intensity in interference (I) = `I_0 cos^2 (delta/2)`

= `I_0 cos^2 (pi/4)`    ...`[cos 45^circ = 1/sqrt 2]`

= `I_0 (1/sqrt 2)^2`

= `I_0/2`