Question

The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is ______.

विकल्प

• three

• five

• infinite

• zero

Answer 1

The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is five.

Explanation:

For constructive interference path difference (As sin θ ≤ 1)

d sin θ = nλ

Given d = 2λ

∴ 2λ sin θ = nλ ⇒ sin θ = `"n"/2`

n = 0, 1, -1, 2, -2 hence five maxima are possible.