In Young's double-slit interference experimen | vedclass

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(C) The intensity of light $I$ is directly proportional to the width of the slit $w$,so $I_1/I_2 = w_1/w_2 = 1/25$.Let $I_1 = I_0$ and $I_2 = 25I_0$.

Vedclass · Accepted Answer

$9:4$

Vedclass · Answer

(C) The intensity of light $I$ is directly proportional to the width of the slit $w$,so $I_1/I_2 = w_1/w_2 = 1/25$.Let $I_1 = I_0$ and $I_2 = 25I_0$. Then the amplitudes are $a_1 = \sqrt{I_0}$ and $a_2 = \sqrt{25I_0} = 5\sqrt{I_0}$.The maximum intensity $I_{max}$ is given by $(\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{I_0} + 5\sqrt{I_0})^2 = (6\sqrt{I_0})^2 = 36I_0$.The minimum intensity $I_{min}$ is given by $(\sqrt{I_2} - \sqrt{I_1})^2 = (5\sqrt{I_0} - \sqrt{I_0})^2 = (4\sqrt{I_0})^2 = 16I_0$.The ratio of maximum to minimum intensity is $I_{max}/I_{min} = 36I_0 / 16I_0 = 36/16 = 9/4$ or $9:4$.

In Young's double-slit interference experiment,the ratio of the slit widths is $1:25$. Find the ratio of the maximum to minimum intensity in the interference pattern.

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