Light waves producing interference have their amplitudes in the ratio 3:2. The intensity ratio of maximum and minimum of interference fringes is

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6:4

36:1

9:4

25:1

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{A_{1}}{A_{2}} = \frac{3}{2} \\ \frac{I_{max}}{I_{min}} = {(\frac{A_{1} + A_{2}}{A_{1} - A_{2}})}^{2} = {(\frac{3 A_{0} + 2 A_{0}}{3 A_{0} - 2 A_{0}})}^{2} = \frac{25}{1} \end{array}$

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