JEE Main 2015 · 11 Apr · Shift 1 · Physics Q25

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In a Young's double slit experiment with light of wavelength λ the separation of slits is d and distance of screen is D such that D >> d >> λ. If the Fringe width is β, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is :

\dfrac\beta2

\dfrac\beta4

\dfrac\beta3

\dfrac\beta6

Answer and worked solution

Official answer: B

$I=I_{max}\cos^2\left(\dfrac{\pi x}\beta\right)$ . Half intensity needs $\dfrac{\pi x}\beta=\dfrac\pi4$ , i.e. $x=\dfrac\beta4$ .

Topics:Interference and Young's double slit

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JEE Main 2015 · 11 Apr · Shift 1, Physics question 25