Question

In Young's double-slit experiment, the intensity at a point where the path difference is `lambda/4` is l. If the maximum intensity is l₀, and then the ratio of `l_0/l` is ______.

`(cos45^circ = 1/sqrt2 = sin45^circ)`

विकल्प

• 4 : 1

• 1 : 2

• 2 : 1

• 1 : 4

Answer 1

In Young's double-slit experiment, the intensity at a point where the path difference is `lambda/4` is l. If the maximum intensity is l₀, and then the ratio of `l_0/l` is 2 : 1.

Explanation:

Since, phase difference, `Φ = (2pi)/lambda` × Path difference

= `(2pi)/lambda xx lambda/4 = pi/2`

∴ Intensity, l = `l_0cos^2(Phi/2)`

⇒ `l_0/l = 1/(cos^2(pi/4)) = 1/((1"/"sqrt2)^2) = 2 : 1`