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JEE Main & Advanced Sample Paper JEE Main Sample Paper-28

  • question_answer
    In the Young's slit experiment, when a glass plate \[(\mu =1.5)\] of thickness t is introduced in the path of one of the interfering beams (wavelength\[=\lambda \]), the intensity at the position where central maxima occurred previous remains unchanged. The minimum thickness of the glass plate is:

    A)  \[2\lambda \]

    B)  \[\lambda \]

    C)  \[\frac{2}{3}\lambda \]                                

    D)  \[\frac{\lambda }{3}\]

    Correct Answer: A

    Solution :

    For the minimum thicknesses \[{{t}_{\min }}\]of the glass plate, for a maxima at centre, the optical path difference must be\[\lambda \]. \[\Rightarrow t(\mu -1)=\lambda \] \[\therefore t=\frac{\lambda }{(\mu -1)}=\frac{\lambda }{(1.5-1)}=2\lambda \]


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