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In Young's double slit experiment, one of the slit is wider than other, so that the 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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$\frac{I_m}{9}\left(1+8 \cos ^2 \frac{\phi}{2}\right)$
$\frac{I_m}{9}\left(1+8 \cos ^2 \frac{\phi}{2}\right)$
Let $A_1=A_0, A_2=2 A_0$
If amplitude of resultant wave is $A$ then
$\mathrm{A}^2=\mathrm{A}_1^2+\mathrm{A}_2^2+2 \mathrm{~A}_1 \mathrm{A}_2 \cos \phi$
For maximum intensity,
$\mathrm{A}_{\max }^2=\mathrm{A}_1^2+\mathrm{A}_2^2+2 \mathrm{~A}_1 \mathrm{A}_2$
$\therefore\frac{A^2}{A_{\max }^2}=\frac{A_1^2+A_2^2+2 A_1 A_2 \cos \phi}{A_1^2+A_2^2+2 A_1 A_2}$
$=\frac{A_0^2+4 A_0^2+2\left(A_0\right)\left(2 A_0\right) \cos \phi}{A_0^2+4 A_0^2+2\left(A_0\right)\left(2 A_0\right)}$
$\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{m}}}=\frac{5+4 \cos \phi}{9}=\frac{1+8 \cos ^2(\phi / 2)}{9}$
If amplitude of resultant wave is $A$ then
$\mathrm{A}^2=\mathrm{A}_1^2+\mathrm{A}_2^2+2 \mathrm{~A}_1 \mathrm{A}_2 \cos \phi$
For maximum intensity,
$\mathrm{A}_{\max }^2=\mathrm{A}_1^2+\mathrm{A}_2^2+2 \mathrm{~A}_1 \mathrm{A}_2$
$\therefore\frac{A^2}{A_{\max }^2}=\frac{A_1^2+A_2^2+2 A_1 A_2 \cos \phi}{A_1^2+A_2^2+2 A_1 A_2}$
$=\frac{A_0^2+4 A_0^2+2\left(A_0\right)\left(2 A_0\right) \cos \phi}{A_0^2+4 A_0^2+2\left(A_0\right)\left(2 A_0\right)}$
$\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{m}}}=\frac{5+4 \cos \phi}{9}=\frac{1+8 \cos ^2(\phi / 2)}{9}$
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