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JEE Main 2012 (AIEEE) · Superposition, Coherence & Intensity

Superposition, Coherence & Intensity question from JEE Main 2012 (AIEEE)

Q99JEE Main 2012 · AIEEEMedium

Superposition, Coherence & Intensity

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

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Answer: (d)

Amplitudes $2a$ and $a$ : $I_{max} = I_m = (3a)^2 = 9a^2$ .
$I = 4a^2 + a^2 + 2(2a)(a)\cos\phi = 5a^2 + 4a^2\cos\phi$
Use $\cos\phi = 2\cos^2\frac{\phi}{2} - 1$ : $I = a^2\left(1 + 8\cos^2\frac{\phi}{2}\right)$
$I = \frac{I_m}{9}\left(1 + 8\cos^2\frac{\phi}{2}\right)$ . So the answer is (d).

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