JEE Main & Advanced Sample Paper JEE Main - Mock Test - 23

  • question_answer
    In Young's double slit experiment with light of wavelength \[K=600\text{ }nm,\]intensity of central fringe is \[{{I}_{0}}\]- One of the slits is covered by glass plate of refraction index \[1.4\] and thickness \[t=5\mu m,\] the new intensity at the same point on screen will be

    A)  \[\frac{{{I}_{0}}}{4}\]            

    B) \[\frac{3{{I}_{0}}}{4}\]                    

    C) \[{{I}_{0}}\]                

    D)        \[\frac{{{I}_{0}}}{2}\]

    Correct Answer: A

    Solution :

    [a] Optical path diff. \[\Delta x=(\mu -1)t\] Phase diff.  \[\Delta \phi =\frac{2\pi }{\lambda }\Delta x\] \[=\frac{2\pi }{600\times {{10}^{-9}}}\times 0.4\times 5\times {{10}^{-6}}\] \[\Delta \phi =\frac{20\pi }{3}\] \[{{I}_{res}}={{I}_{0}}{{\cos }^{2}}\left( \frac{\Delta \phi }{2} \right)={{I}_{0}}{{\cos }^{2}}\left( \frac{10\pi }{3} \right)\] \[\left[ {{I}_{res}}=\frac{{{I}_{0}}}{4} \right]\] 

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