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(C) The intensity $I$ of a wave is proportional to the square of its amplitude $A$, i.e.,$I \propto A^2$.Let the amplitudes of the two coherent source

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$7 : 5$

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(C) The intensity $I$ of a wave is proportional to the square of its amplitude $A$, i.e.,$I \propto A^2$.Let the amplitudes of the two coherent sources be $A_1$ and $A_2$.The maximum intensity is $I_{max} \propto (A_1 + A_2)^2$ and the minimum intensity is $I_{min} \propto (A_1 - A_2)^2$.Given the ratio of maximum to minimum intensity is $\frac{I_{max}}{I_{min}} = \frac{36}{1}$.Therefore, $\frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} = \frac{36}{1}$.Taking the square root on both sides, we get $\frac{A_1 + A_2}{A_1 - A_2} = \frac{6}{1}$.By applying componendo and dividendo, $\frac{(A_1 + A_2) + (A_1 - A_2)}{(A_1 + A_2) - (A_1 - A_2)} = \frac{6 + 1}{6 - 1}$.This simplifies to $\frac{2A_1}{2A_2} = \frac{7}{5}$.Thus, the ratio of the amplitudes is $\frac{A_1}{A_2} = \frac{7}{5}$.

In Young's experiment, the intensity ratio of the maxima and minima in an interference pattern produced by two coherent sources is $36 : 1$. The ratio of the amplitudes of the two individual sources will be

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