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(C) Let the amplitudes be $A_1$ and $A_2$. Given $A_2 = 2A_1$.Since intensity $I \propto A^2$,we have $I_2 = 4I_1$.The maximum intensity $I_m$ is give

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$\frac{I_m}{9} \left( 1 + 8 \cos^2 \frac{\phi}{2} \right)$

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(C) Let the amplitudes be $A_1$ and $A_2$. Given $A_2 = 2A_1$.Since intensity $I \propto A^2$,we have $I_2 = 4I_1$.The maximum intensity $I_m$ is given by $I_m = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{I_1} + 2\sqrt{I_1})^2 = (3\sqrt{I_1})^2 = 9I_1$. Thus,$I_1 = \frac{I_m}{9}$.The resultant intensity $I$ at phase difference $\phi$ is $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.Substituting $I_2 = 4I_1$,we get $I = I_1 + 4I_1 + 2\sqrt{4I_1^2} \cos \phi = 5I_1 + 4I_1 \cos \phi$.Using the identity $1 + \cos \phi = 2 \cos^2 \frac{\phi}{2}$,we can write $I = I_1 + 4I_1(1 + \cos \phi) = I_1 + 4I_1(2 \cos^2 \frac{\phi}{2}) = I_1(1 + 8 \cos^2 \frac{\phi}{2})$.Substituting $I_1 = \frac{I_m}{9}$,we get $I = \frac{I_m}{9} (1 + 8 \cos^2 \frac{\phi}{2})$.

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

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