JCECE Medical JCECE Medical Solved Paper-2006

  • question_answer
    If the intensity of fringe at wavelength\[\lambda \] is \[K,\]then its intensity at wavelength\[\lambda /2\]is:

    A) \[\frac{K}{2}\]

    B)  \[K\]

    C)  zero

    D) \[\sqrt{2}K\]

    Correct Answer: C

    Solution :

     Key Idea: Resultant intensity is given by \[{{I}_{R}}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \,o|\] Resultant intensity due to two waves of intensities \[{{I}_{1}},{{I}_{2}}\]is given by \[{{I}_{R}}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \text{o }\!\!|\!\!\text{ }\] when path difference is \[\lambda ,\text{o }\!\!|\!\!\text{ =}\,\text{2}\pi \] when path difference is\[\frac{\lambda }{2},\text{o }\!\!|\!\!\text{ =}\pi \] \[\therefore \] \[{{I}_{{{R}_{1}}}}=I+I+2I\cos 2\pi \] \[I{{R}_{1}}=4I=K\] \[{{I}_{{{R}_{2}}}}=I+I+2I\cos \pi \] \[{{I}_{{{R}_{0}}}}=0\]


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