JEE Main & Advanced Sample Paper JEE Main Sample Paper-45

  • question_answer
    In Young's experiment, the distance between two slits is halved and the distance between the screen and slit is made three times. Then width of the fringe

    A)  becomes half              

    B)  remains the same

    C)  becomes 6 times       

    D)  becomes 4 times

    Correct Answer: C

    Solution :

     Given, the distance between two slits \[d'=\frac{d}{2}\] and the distance between screen and slit \[D'=3D\] We know that fringe width \[\beta =\frac{D\lambda }{d}\] Now, according to question \[\beta '=\frac{D'\lambda }{d'}\] \[\Rightarrow \]            \[\beta '=\frac{3D\cdot \lambda }{d/2}\] \[\therefore \]    \[\beta '=6\,\frac{D\lambda }{d}=6\beta \] \[\Rightarrow \] Final fringe width \[=6\times \] initial fringe width.

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