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In a Young’s double slit experiment, $I_{0}$ is the intensity at the central maximum and $β$ is the fringe width. The intensity at point P at distance x from central maxima will be

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a

$I_{0} \cos \frac{π x}{β}$

b

$4 I_{0} \cos^{2} \frac{π x}{β}$

c

$I_{0} \cos^{2} \frac{π x}{β}$

d

$\frac{I_{0}}{4} \cos^{2} \frac{π x}{β}$

answer is C.

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Detailed Solution

Path difference at point $P = \frac{x d}{D}$
Phase deff. At $P = \frac{2 π}{λ} \frac{x d}{D} = \frac{2 π x}{β}$
Intensity at $P = I + I + 2 I \cos \frac{2 π x}{β} = 2 I [1 + \cos \frac{2 π x}{β}]$
$= 4 I \cos^{2} \frac{π x}{β} = I_{0} \cos^{2} \frac{π x}{β} [∵ I_{0} = 4 I]$

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