The width of one of the two slits in a Young’s doubles slit experiment is three times the other slit. If the amplitude of the light coming from a slit is proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is x: 4 where x is …………..

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Detailed Solution

amplitude (A) $\propto$ slit width

Also intensity $\propto$ (Amplitude)² $\propto$ (Slit width)²
$\begin{array}{l} \frac{I_{1}}{I_{2}} = {(\frac{w_{2}}{w_{1}})}^{2} = {(\frac{3 w_{1}}{w_{1}})}^{2} = {(\frac{3}{1})}^{2} = 9 (∵ w_{2} = 3 w) \\ \Rightarrow I_{1} = 9 I_{2} \\ \Rightarrow \frac{I_{min}}{I_{max}} = {(\frac{\sqrt{I_{1}} - \sqrt{I_{2}}}{\sqrt{I_{1}} + \sqrt{I_{2}}})}^{2} = {(\frac{3 - 1}{3 + 1})}^{2} = \frac{1}{4} = \frac{x}{4} \Rightarrow x = 1 .00 \end{array}$