In Young's double-slit experiment, one slit i | vedclass

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(A) Let the amplitudes be $a_1$ and $a_2$. Given $a_1 = 2a_2$.Since intensity $I \propto a^2$, let $I_2 = I'$ and $I_1 = 4I'$.The maximum intensity $I

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

Vedclass · Answer

(A) Let the amplitudes be $a_1$ and $a_2$. Given $a_1 = 2a_2$.Since intensity $I \propto a^2$, let $I_2 = I'$ and $I_1 = 4I'$.The maximum intensity $I_m$ is given by $I_m = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{4I'} + \sqrt{I'})^2 = (2\sqrt{I'} + \sqrt{I'})^2 = (3\sqrt{I'})^2 = 9I'$.Thus, $I' = \frac{I_m}{9}$.The resultant intensity $I$ for a phase difference $\phi$ is $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.Substituting the values: $I = 4I' + I' + 2\sqrt{4I' \cdot I'} \cos \phi = 5I' + 4I' \cos \phi$.Using the identity $\cos \phi = 2 \cos^2 \frac{\phi}{2} - 1$:$I = I'(5 + 4(2 \cos^2 \frac{\phi}{2} - 1)) = I'(5 + 8 \cos^2 \frac{\phi}{2} - 4) = I'(1 + 8 \cos^2 \frac{\phi}{2})$.Substituting $I' = \frac{I_m}{9}$, we get $I = \frac{I_m}{9}(1 + 8 \cos^2 \frac{\phi}{2})$.

In Young's double-slit experiment, one slit is wider than the other, such that the amplitude of the light wave from one slit is twice that of the other. If the maximum intensity is $I_m$, then the resultant intensity $I$ when they interfere with a phase difference of $\phi$ is:

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