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 \rm I_m be the maximum intensity, the resultant intensity \rm I when they interfere at phase difference \phi is given by :

I_m \over 9( 1 + 8\cos ^2\,\phi \over 2 )

I_m \over 9( 4 + 5\cos \,\phi )

I_m \over 3( 1 + 2\cos ^2\,\phi \over 2 )

I_m \over 3( 1 + 4\cos ^2\,\phi \over 2 )

Wave Optics PYQ — JEE Main 2012 · Shift 0 · Q51 with Solution

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Correct answer: D

Let the amplitudes from the two slits be
$\qquad a_2 = a$$ since one slit has double the amplitude of the other.$
Intensity in terms of amplitudes

Resultant intensity for two coherent sources with phase difference $\phi$ is $I = a_1^2 + a_2^2 + 2a_1a_2\cos\phi$

Substituting $a_1=2a$ and $a_2=a$ : $I = (2a)^2 + a^2 + 2(2a)(a)\cos\phi$ $I = 4a^2 + a^2 + 4a^2\cos\phi$ $I = 5a^2 + 4a^2\cos\phi$
Maximum intensity

Maximum intensity occurs when $\cos\phi = 1$ : $I_m = 5a^2 + 4a^2 = 9a^2$

Hence, $a^2 = \frac{I_m}{9}$
Express $I$ in terms of $I_m$

$I = 5a^2 + 4a^2\cos\phi$ $I = a^2(5+4\cos\phi)$ $I = \frac{I_m}{9}(5+4\cos\phi)$
Convert using $\cos\phi = 2\cos^2\frac{\phi}{2} - 1$

Then, $5 + 4\cos\phi = 5 + 4\left(2\cos^2\frac{\phi}{2}-1\right)$ $= 5 + 8\cos^2\frac{\phi}{2} - 4$ $= 1 + 8\cos^2\frac{\phi}{2}$

Therefore, $I = \frac{I_m}{9}\left(1 + 8\cos^2\frac{\phi}{2}\right)$
Compare with options

This matches Option D.

Final Answer: $\boxed{I=\frac{I_m}{9}\left(1+8\cos^2\frac{\phi}{2}\right)}$