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NEET Sample Paper NEET Sample Test Paper-48

  • 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 \[{{\operatorname{I}}_{o}}\]denotes the maximum intensity, then the ratio of I and I is:

    A) 0.866                           

    B) 0.5    

    C) 0.707                           

    D) 0.75

    Correct Answer: D

    Solution :

    \[\operatorname{Phase} difference \phi =\frac{2\pi }{\lambda }\Delta x\] \[\phi =\frac{2\pi }{\lambda }\frac{\lambda }{6}\] \[\phi =\frac{\pi }{3}\] \[\operatorname{I}={{I}_{o}}co{{s}^{2}}\left[ \frac{\phi }{2} \right]\] \[\frac{\operatorname{I}}{{{I}_{o}}}=co{{s}^{2}}\left[ \frac{\pi }{3\times 2} \right]\] \[\frac{\operatorname{I}}{{{I}_{o}}}=co{{s}^{2}}\frac{\pi }{6}\] \[\frac{\operatorname{I}}{{{I}_{o}}}=\frac{3}{4}=0.75\] \[\therefore \cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}\]


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