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JEE Main & Advanced Sample Paper JEE Main Sample Paper-34

  • question_answer
    Monochromatic light of wavelength \[5000\overset{o}{\mathop{\text{A}}}\,\] is incident on two slits separated by a distance of \[5\times {{10}^{-4}}\text{ }m\]. Interference pattern is seen on the screen placed at a distance of 1 m from the slits. A thin glass plate of thickness \[1.5\times {{10}^{-6}}m\] and refractive index 1.5 is placed between one of the slits and the screen. If intensity in the absence of plate was \[{{I}_{0}}\], then new intensity at the centre of the screen will be:

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

    B)  \[2{{I}_{0}}\]

    C)  \[{{I}_{0}}/2\]                  

    D)  zero

    Correct Answer: D

    Solution :

    At the centre, geometrical path difference is zero. The optical path difference due to glass plate is:             \[\Delta x\,=t\,(\mu -1)=1.5\,\times {{10}^{-6}}\,(1.5-1)\]             \[=0.75\,\times {{10}^{-6}}\,m\] \[\frac{\Delta x}{\lambda }\,=\frac{0.75\,\times {{10}^{-6}}}{5000\,\times {{10}^{-10}}}\,=1.5\Rightarrow \,\Delta x\,=3\lambda /2\] Hence, there will be a destructive interference and the new intensity at the centre of screen will be zero.


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