NEET Sample Paper NEET Sample Test Paper-51

  • 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) \[9\text{ }:\text{ }1\]

    Correct Answer: C

    Solution :

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


You need to login to perform this action.
You will be redirected in 3 sec spinner