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(A) Let the amplitudes be $a_1 = a$ and $a_2 = 2a$. The intensities are $I_1 = a^2$ and $I_2 = (2a)^2 = 4a^2 = 4I_1$.The resultant intensity $I$ is gi

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

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(A) Let the amplitudes be $a_1 = a$ and $a_2 = 2a$. The intensities are $I_1 = a^2$ and $I_2 = (2a)^2 = 4a^2 = 4I_1$.The resultant intensity $I$ is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.Substituting the values: $I = I_1 + 4I_1 + 2\sqrt{I_1(4I_1)} \cos \phi = 5I_1 + 4I_1 \cos \phi$.The maximum intensity $I_m$ occurs when $\cos \phi = 1$,so $I_m = (a_1 + a_2)^2 = (a + 2a)^2 = 9a^2 = 9I_1$.Thus,$I_1 = \frac{I_m}{9}$.Substituting $I_1$ into the expression for $I$:$I = \frac{5I_m}{9} + \frac{4I_m}{9} \cos \phi = \frac{I_m}{9}(5 + 4 \cos \phi)$.Using the identity $\cos \phi = 2\cos^2 \frac{\phi}{2} - 1$:$I = \frac{I_m}{9}(5 + 4(2\cos^2 \frac{\phi}{2} - 1)) = \frac{I_m}{9}(5 + 8\cos^2 \frac{\phi}{2} - 4) = \frac{I_m}{9}(1 + 8\cos^2 \frac{\phi}{2})$.

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 of the other. If $I_m$ is the maximum intensity,the resultant intensity $I$ when they interfere at a phase difference $\phi$ is given by

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