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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2010

  • question_answer
    The wavelength of the light used in Youngs double slit experiment is\[\lambda \]. The intensity at a point on the screen is\[I\], where the path difference is \[\frac{\lambda }{6}.\]. If \[{{I}_{0}}\]denotes the maximum intensity, then the ratio of \[I\]and \[{{I}_{0}}\]is

    A)  0.866           

    B)  0.5

    C)  0.707           

    D)  0.75

    Correct Answer: D

    Solution :

    Phase difference, \[\phi =\frac{2\pi }{\lambda }\times \] path difference \[\phi =\frac{2\pi }{\lambda }\times \frac{\lambda }{6}=\frac{\pi }{3}={{60}^{o}}\] Intensity, \[I={{I}_{0}}{{\cos }^{2}}\left( \frac{\phi }{2} \right)\] \[\frac{I}{{{I}_{0}}}={{\cos }^{2}}({{30}^{o}})={{\left( \frac{\sqrt{3}}{2} \right)}^{2}}=0.75\]


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