In Young's double slit experiment the fringe pattern is observed on a screen placed at a distance D. The slits are illuminated by light of wavelength $λ$ . The distance from the central point where the intensity falls to half the maximum is

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$\frac{λD}{3 d}$

$\frac{λD}{2 d}$

$\frac{λD}{4 d}$

$\frac{λD}{d}$

answer is D.

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Detailed Solution

From $I = 4 I_{0} \cos^{2} \frac{φ}{2}$ , intensity will fall to half of max when phase diff $ϕ = \frac{π}{2}$ or path diff
$∴ x = \frac{λ}{4}, dsin θ = \frac{λ}{4} \Rightarrow d \frac{Y}{D} = \frac{λ}{4} \Rightarrow Y = \frac{λD}{4 d}$

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