In the Young's double slit experiment using a monochromatic light of wavelength $\lambda$, the path difference (in terms of an integer $n$ ) corresponding to any point having half the peak intensity is

(B) $(2 n+1) \frac{\lambda}{4}$

(A) $(2 n+1) \frac{\lambda}{2}$

(C) $(2 n+1) \frac{\lambda}{8}$

(D) $(2 n+1) \frac{\lambda}{16}$

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Wave Optics

In the Young's double slit experiment using a monochromatic light of wavelength

λ

, the path difference (in terms of an integer

n

) corresponding to any point having half the peak intensity is

Answer: (B)

The intensity in Young's double slit experiment is given by:

I = 4 I_{0} cos^{2} \frac{ϕ}{2}

For half the peak intensity:

2 I_{0} = 4 I_{0} cos^{2} \frac{ϕ}{2}

cos \frac{ϕ}{2} = \frac{1}{2}

\frac{ϕ}{2} = \frac{π}{4}

ϕ = \frac{π}{2}

Path difference

Δ x = \frac{λ}{2 π} \times ϕ = \frac{λ}{2 π} \times \frac{π}{2} = \frac{λ}{4}

This corresponds to the general form

(2 n + 1) \frac{λ}{4}

for integer

n

.

Hence the answer is (B).

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Single correct option