Wave Optics

JEE Main level, single correct, medium

115 of 191

Question 115 of 191

In Young's double slit experiment, one of the slit is wider than other, so that the amplitude of the light from one slit is double of that from other slit. If ImI_m be the maximum intensity, the resultant intensity I when they interfere at phase difference ϕ\phi is given by

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Answer: (D) Im9(1+8cos⁡2ϕ2)\frac{I_m}{9}\left(1+8 \cos ^2 \frac{\phi}{2}\right)

Let A1=A0,A2=2A0A_1=A_0, A_2=2 A_0
If amplitude of resultant wave is AA then
A2=A12+A22+2 A1A2cos⁡ϕ\mathrm{A}^2=\mathrm{A}_1^2+\mathrm{A}_2^2+2 \mathrm{~A}_1 \mathrm{A}_2 \cos \phi
For maximum intensity,
Amax⁡2=A12+A22+2 A1A2\mathrm{A}_{\max }^2=\mathrm{A}_1^2+\mathrm{A}_2^2+2 \mathrm{~A}_1 \mathrm{A}_2
∴A2Amax⁡2=A12+A22+2A1A2cos⁡ϕA12+A22+2A1A2\therefore\frac{A^2}{A_{\max }^2}=\frac{A_1^2+A_2^2+2 A_1 A_2 \cos \phi}{A_1^2+A_2^2+2 A_1 A_2}
=A02+4A02+2(A0)(2A0)cos⁡ϕA02+4A02+2(A0)(2A0)=\frac{A_0^2+4 A_0^2+2\left(A_0\right)\left(2 A_0\right) \cos \phi}{A_0^2+4 A_0^2+2\left(A_0\right)\left(2 A_0\right)}
IIm=5+4cos⁡ϕ9=1+8cos⁡2(ϕ/2)9\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{m}}}=\frac{5+4 \cos \phi}{9}=\frac{1+8 \cos ^2(\phi / 2)}{9}

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