The maximum number of possible interference maxima for slit-seperation equal to twice the wavelength, in young’s double slit experiment is

Options

(a) Infinite
(b) five
(c) three
(d) zero

Correct Answer:

five

Explanation:

For interference maxima, d sin θ = nλ

⇒ 2λ sin θ = nλ ⇒ sin θ = n / 2

sin θ can have values between 0 and ±1.

Hence n can be (-2, -1, 0, +1, +2) or five values.

The possible maxima are five.