In YDSE intensity on the screen at a distance y from the centre of central bright fringe is found to be 25% of maximum intensity of bright fringe. If $β$ be the fringe width, then $\frac{y}{β}$ is

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$\frac{1}{4}$

$\frac{1}{3}$

$\frac{3}{4}$

$\frac{2}{3}$

answer is A.

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Detailed Solution

$I = I_{m} \cos^{2} \frac{ϕ}{2}$

$\Rightarrow \frac{I_{m}}{4} = I_{m} \cos^{2} \frac{ϕ}{2} \Rightarrow \frac{ϕ}{2} = \frac{π}{3}$

Now, $\frac{ϕ}{2} = \frac{π}{λ} x d sinθ = \frac{π}{λ} d tanθ = \frac{π}{λ} d . \frac{y}{D} = π \frac{y}{β}$

$∴ π \frac{y}{β} = \frac{π}{3} \Rightarrow \frac{y}{β} = \frac{1}{3}$

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