JEE Advanced 2013 · Paper 1 · Physics Q3

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PhysicsSingle correct+2Question 3 of 60

In the Young's double slit experiment using a monochromatic light of wavelength $\lambda$ , the path difference (in terms of an integer $n$ ) corresponding to any point having half the peak intensity is

(2n+1)\dfrac\lambda2

(2n+1)\dfrac\lambda4

(2n+1)\dfrac\lambda8

(2n+1)\dfrac\lambda{16}

Answer and worked solution

Official answer: B

Intensity. $I=I_{\max}\cos^2\frac\phi2=\frac{I_{\max}}2$ gives $\phi=(2n+1)\frac\pi2$ .

Path difference. $\Delta=\frac\lambda{2\pi}\phi=(2n+1)\frac\lambda4$ .

(B)

Topics:Interference and Young's double slit

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JEE Advanced 2013 · Paper 1, Physics question 3