JEE Main & Advanced Physics Wave Optics / तरंग प्रकाशिकी Question Bank Self Evaluation Test - Wave Optics

  • question_answer
    In an interference arrangement similar to Young's double-slit experiment, the slits \[{{S}_{1}}\] and \[{{S}_{2}}\] are illuminated with coherent microwave sources, each of frequency 106 Hz. The sources are synchronized to have zero phase difference. The slits are separated by a distance d = 150.0 m. The intensity I \[\left( \theta  \right)\] is measured as a function of \[\theta \], where \[\theta \] is defined as shown. If \[{{I}_{0}}\] is the maximum intensity, then I\[\left( \theta  \right)\] for \[\] is given by  

    A) \[I(\theta )={{I}_{0}}/2\] for \[\theta ={{30}^{o}}\]

    B) \[I(\theta )={{I}_{0}}/4\] for\[\theta ={{90}^{o}}\]

    C) \[I(\theta )={{I}_{0}}\] for \[\theta ={{0}^{o}}\]

    D) \[I=\theta \] is constant for all values of \[\theta \]

    Correct Answer: C

    Solution :

    [c] We know that \[I(\theta )={{I}_{0}}{{\cos }^{2}}\frac{\delta }{2}\] where \[\delta =\frac{2\pi d\tan \theta }{\lambda }\] \[I(\theta )={{I}_{0}}{{\cos }^{2}}\left( \frac{\pi d\tan \theta }{\pi } \right)={{I}_{0}}{{\cos }^{2}}\left( \frac{\pi \times 150\times \tan \theta }{3\times {{10}^{8}}/{{10}^{6}}} \right)\]\[={{I}_{0}}{{\cos }^{2}}\left( \frac{\pi }{2}\tan \theta  \right)\] For \[\theta ={{30}^{o}}\]; \[I(\theta )={{I}_{o}}{{\cos }^{2}}\left( \frac{\pi }{2\sqrt{3}} \right)\] For \[\theta ={{90}^{o}}\]; \[I(\theta )={{I}_{o}}{{\cos }^{2}}(\infty )\] For \[\theta ={{0}^{o}}\]  \[I(\theta )={{I}_{0}}\] \[I(\theta )\] is not constant. Alternatively, when \[\theta \] is zero the path difference between wave originating from\[{{S}_{1}}\]and that from \[{{S}_{2}}\]will be zero. This corresponds to a maxima.


You need to login to perform this action.
You will be redirected in 3 sec spinner