question_answer

In a Young's double slit experiment, the ratio of the slit's width is 4:1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be :                                         [JEE Main 10-4-2019 Afternoon]

A) \[{{(\sqrt{3}+1)}^{4}}:16\]     

B)      \[9:1\]

C) \[4:1\]                D)      \[25:9\] Correct Answer: B Solution : [b] \[{{I}_{1}}=4{{I}_{0}}\]

\[{{I}_{2}}={{I}_{0}}\]

\[{{I}_{\max }}={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}\]

\[={{(2\sqrt{{{I}_{0}}}+\sqrt{{{I}_{0}}})}^{2}}=9{{I}_{0}}\]

\[{{I}_{\min }}={{(\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}})}^{2}}\]

\[={{(2\sqrt{{{I}_{0}}}-\sqrt{{{I}_{0}}})}^{2}}={{I}_{0}}\]

\[\therefore \]\[\frac{{{I}_{\max }}}{{{\operatorname{I}}_{\min }}}=\frac{9}{1}\]