CET Karnataka Medical CET - Karnataka Medical Solved Paper-2000

  • question_answer
    Two coherent light beams of intensity \[I\]and \[4I\] are superimposed. The maximum and minimum possible intensities in the resulting beam are:

    A)  \[9I\] and \[3I\]

    B)  \[9I\] and \[I\]

    C)  \[5I\] and \[3I\]

    D)  \[5I\] and \[I\]

    Correct Answer: B

    Solution :

     Given : Intensity of first coherent beam \[=I\] Intensity of second coherent beam \[=4I\] The amplitude of first beam is \[{{A}_{1}}=\sqrt{I}\] Similarly, \[{{A}_{2}}=\sqrt{4I}=2\sqrt{I}\] The ratio of maximum and minimum intensities is \[\frac{{{\operatorname{I}}_{max}}}{{{\operatorname{I}}_{min}}}=\frac{{{({{A}_{1}}+{{A}_{2}})}^{2}}}{{{({{A}_{1}}-{{A}_{2}})}^{2}}}=\frac{{{(\sqrt{I}+2\sqrt{I})}^{2}}}{{{(\sqrt{I}-2\sqrt{I})}^{2}}}=\frac{9I}{I}\] Hence, the possible maximum and minimum intensities are \[9I\]and \[I\].


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