First slide

Young's double slit Experiment

Question

In Young's double slit experiment, the intensity at a point is (1 /4) of the maximum intensity. The angular position of this point is

Difficult

A

$\sin^{- 1} (\frac{λ}{d})$
B

$\sin^{- 1} (\frac{λ}{2 d})$
C

$\sin^{- 1} (\frac{λ}{3 d})$
D

$\sin^{- 1} (\frac{λ}{4 d})$

Solution

Let P be the point at angular position $θ$ where the intensity is (1 /4 ) of &e maximum intensity as shown in fig. (a). We know that

$I = I_{max} \cos^{2} (\frac{ϕ}{2}) \frac{I_{max}}{4} = I_{max} \cos^{2} (\frac{ϕ}{2}) or 4 \cos^{2} (\frac{ϕ}{2}) = 1 \cos^{2} (\frac{ϕ}{2}) = \frac{1}{4} or \cos (\frac{ϕ}{2}) = \frac{1}{2} \frac{ϕ}{2} = \cos^{- 1} (\frac{1}{2}) = 60^{\circ} or ϕ = 120^{\circ} = (\frac{2 π}{3}) \frac{2 π}{λ} \times dsin θ = \frac{2 π}{3} or θ = \sin^{- 1} (\frac{λ}{3 d})$

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