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Two coherent monochromatic light beams of amplitude 3 and 5 units are superposed. The maximum and minimum possible intensities in the resulting beams are in the ratio
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The correct answer is:
$16: 1$
It is given that the amplitudes $A_1$ and $A_2$ are in the ratio $\frac{A_1}{A_2}=\frac{3}{5}$
$\therefore$ After superposition the maximum and minimum intensities will be in the ratio
$$
\frac{I_{\max }}{I_{\min }}=\frac{\left(A_1+A_2\right)^2}{\left(A_1-A_2\right)^2}=\frac{(3+5)^2}{(3-5)^2}=\frac{8^2}{(-2)^2}=\frac{64}{4}=\frac{16}{1}
$$
$\therefore$ After superposition the maximum and minimum intensities will be in the ratio
$$
\frac{I_{\max }}{I_{\min }}=\frac{\left(A_1+A_2\right)^2}{\left(A_1-A_2\right)^2}=\frac{(3+5)^2}{(3-5)^2}=\frac{8^2}{(-2)^2}=\frac{64}{4}=\frac{16}{1}
$$
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