Manipal Engineering Manipal Engineering Solved Paper-2008

  • question_answer
    Intensity of wave A is 91, while of wave B is 7. What is maximum and minimum intensity in YDSE?

    A)  82\[I\], 80\[I\]                 

    B)         8\[I\], 10\[I\]

    C)  16\[I\], 4\[I\]                   

    D)        4\[I\],\[I\]

    Correct Answer: C

    Solution :

    For two coherent sources, the resultant intensity is given by                 \[I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \phi \] For maximum intensity,\[\cos \phi =+1\] \[\therefore \]  \[{{I}_{\max }}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\]                         \[={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}\] For minimum intensity,\[\cos \phi =-1\] \[\therefore \]  \[{{I}_{\min }}={{I}_{1}}+{{I}_{2}}-2\sqrt{{{I}_{1}}{{I}_{2}}}\]                        \[={{(\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}})}^{2}}\] Hence, \[{{I}_{\max }}={{(\sqrt{9I}+\sqrt{I})}^{2}}\]                        \[={{(3\sqrt{I}+\sqrt{I})}^{2}}=16I\] and        \[{{I}_{\min }}={{(\sqrt{9I}-\sqrt{I})}^{2}}\]                        \[={{(3\sqrt{I}-\sqrt{I})}^{2}}=4I\] Note: In an interference pattern, maximum intensity is obtained for constructive interference and minimum intensity is obtained for destructive interference.


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