Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is :

<p>In Young's double-slit interference experiment, the width of one slit is half that of the other. The intensity, $ \mathrm{I} $, is proportional to the slit width. This gives us: </p> <p>$ \begin{align*} \mathrm{I}_1 &= \mathrm{I}_0, \\ \mathrm{I}_2 &= 2 \mathrm{I}_0. \end{align*} $</p> <p>The maximum intensity, $ \mathrm{I}_{\max} $, occurs when the amplitudes add constructively:</p> <p>$ \mathrm{I}_{\max} = \left(\sqrt{\mathrm{I}_1} + \sqrt{\mathrm{I}_2}\right)^2. $</p> <p>The minimum intensity, $ \mathrm{I}_{\min} $, occurs when the amplitudes interfere destructively:</p> <p>$ \mathrm{I}_{\min} = \left(\sqrt{\mathrm{I}_1} - \sqrt{\mathrm{I}_2}\right)^2. $</p> <p>Now, using $ \mathrm{I}_1 = \mathrm{I}_0 $ and $ \mathrm{I}_2 = 2 \mathrm{I}_0 $, we find:</p> <p>$ \frac{\mathrm{I}_{\max}}{\mathrm{I}_{\min}} = \frac{(\sqrt{2} + 1)^2}{(\sqrt{2} - 1)^2} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}. $</p>