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Q.

In youngs double slit experiment, the slit widths are in the ratio 1 : 9. Then the ratio of the intensity of maxima to that of the mimima is (Assume intensity $\propto$ width)

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a

81 : 1

b

9 : 1

c

4 : 1

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{I_{1}}{I_{2}} = \frac{W_{1}}{w_{2}} = \frac{1}{9}; \frac{I_{max}}{I_{min}} = {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2} \\ \frac{I_{max}}{I_{min}} = {(\frac{\sqrt{I} + \sqrt{9 I}}{\sqrt{I} - \sqrt{9 I}})}^{2} = {(\frac{4}{2})}^{2} = \frac{4}{1} \end{array}$

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In youngs double slit experiment, the slit widths are in the ratio 1 : 9. Then the ratio of the intensity of maxima to that of the mimima is (Assume intensity∝width)