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In YDSE, the intensity at a point is $\frac{1}{4}$ of the maximum intensity. Angular position of this point is (d is the saparation between two slits)

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$\sin^{- 1} (\frac{λ}{2 d})$

$\sin^{- 1} (\frac{λ}{d})$

$\sin^{- 1} (\frac{λ}{3 d})$

$\sin^{- 1} (\frac{λ}{4 d})$

answer is C.

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Detailed Solution

$I = I_{M a x} \cos^{2} \frac{ϕ}{2}$

$\frac{I_{M a x}}{4} = I_{m a x} \cos^{2} \frac{ϕ}{2}$

$\cos^{2} \frac{ϕ}{2} = \frac{1}{4}$

$\frac{ϕ}{2} = \frac{π}{3}$

$ϕ = \frac{2 π}{3} Δ x$

$\frac{2 π}{3} = \frac{2 π}{λ} (Δ x)$

$Δ x = \frac{λ}{3}$

$Δ x = d \sin θ$

$\frac{λ}{3} = d \sin θ$

$\sin θ = \frac{λ}{3 d}$

$θ = \sin^{- 1} (\frac{λ}{3 d})$

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