• # question_answer A long horizontal slit is placed 1 mm above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen 1m away from the slit. If the mirror reflects only $64%$ of the light falling on it, the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen is- A) $8\text{ }:\text{ }1$                 B) $3\text{ }:\text{ }1$C) $81\text{ }:\text{ }1$               D) $9\text{ }:\text{ }1$

$Intensity\text{ }of\text{ }direct\text{ }ray=\,\,{{I}_{0}}\,\,=\,\,kA_{0}^{2}$ Intensity of reflected ray $=\frac{64}{100}{{I}_{0}}\,=\,\,k{{\left( \frac{8{{A}_{0}}}{10} \right)}^{2}}$ $\therefore \,\,\frac{{{\operatorname{I}}_{max}}}{{{\operatorname{I}}_{min}}}\,\,=\,\frac{{{({{A}_{0}}+0.8{{A}_{0}})}^{2}}}{{{({{A}_{0}}-0.8{{A}_{0}})}^{2}}}={{\left( \frac{1.8}{0.2} \right)}^{2}}=\frac{81}{1}$?