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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2010

  • question_answer
    The wavelength of the light used in Young's 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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