If an interference pattern has maximum and mi | vedclass

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Question

(D) The ratio of maximum intensity to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2$. Given

Vedclass · Accepted Answer

$7:5$

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(D) The ratio of maximum intensity to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2$. Given $\frac{I_{\max}}{I_{\min}} = \frac{36}{1}$. Taking the square root on both sides,we get $\frac{a_1 + a_2}{a_1 - a_2} = \sqrt{\frac{36}{1}} = 6$. 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 $\frac{2a_1}{2a_2} = \frac{7}{5}$. Therefore,the ratio of amplitudes is $\frac{a_1}{a_2} = 7:5$.

If an interference pattern has maximum and minimum intensities in a $36:1$ ratio,then what will be the ratio of the amplitudes?

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