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JEE Main & Advanced Sample Paper JEE Main Sample Paper-21

  • question_answer
    Two beam of light having intensities \[I\] and \[4I\] interfere to produce of fringe pattern on a screen. The phase difference between the beams is \[\pi /2\] at point A and \[\pi \]  at point B. Then the difference between resultant intensities at A and B is :

    A)  \[2I\]                           

    B)  \[4I\]

    C)  \[5I\]                           

    D)  \[7I\]

    Correct Answer: B

    Solution :

    \[{{I}_{1}}=I\,\,and\,{{I}_{2}}=4I\] \[I={{I}_{1}}+{{I}_{2}}\,+2\sqrt{{{I}_{1}}{{I}_{2}}}\,\cos \phi \] At point \[A(\phi \,=\pi /2),\] \[{{I}_{A}}=I+4I=5I\] At point \[B(\phi =\pi )\] \[{{I}_{B}}=I+4I\,-2\,\sqrt{4I.I}=I\] \[{{I}_{A}}-{{I}_{B}}\,=4I\]


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