The ratio of the intensities at minima to maxima in the interference pattern is 9 : 25. What will be the ratio of the widths of the two slits in the young’s double slit experiment ?

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8 : 1

9 : 1

16 : 1

4 : 1

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{I_{max}}{I_{min}} = \frac{25}{9} \Rightarrow {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2} = \frac{25}{9} \\ \Rightarrow \frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}} = \sqrt{\frac{25}{9}} = \frac{5}{3} \\ \Rightarrow \frac{\sqrt{I_{1}}}{\sqrt{I_{2}}} = \frac{5 + 3}{5 - 3} = \frac{8}{2} = \frac{4}{1} \Rightarrow \frac{I_{1}}{I_{2}} = {(\frac{4}{1})}^{2} = \frac{16}{1} \end{array}$
As intensity is proportional to the slit width, The ratio of slit widths also $\frac{16}{1}$