JEE Main & Advanced Physics Wave Optics / तरंग प्रकाशिकी Question Bank Young's Double Slit Experiment and Biprism

  • question_answer
    If prism angle \[\alpha =1{}^\circ ,\ \mu =1.54,\]distance between screen and prism \[(b)=0.7\,m,\] distance between prism and source \[a=0.3\,m,\ \lambda =180\pi \ nm\] then in Fresnal biprism find the value of \[\beta \] (fringe width)                                                                       [RPMT 2002]

    A)            \[{{10}^{-4}}m\]                 

    B)            \[{{10}^{-3}}mm\]

    C)            \[{{10}^{-4}}\times \pi m\]    

    D)             \[\pi \times {{10}^{-3}}m\]

    Correct Answer: A

    Solution :

                       By using \[\beta =\frac{(a+b)\lambda }{2a(\mu -1)\alpha }=\frac{(0.3+0.7)\times 180\pi \times {{10}^{-9}}}{2\times 0.3(1.54-1)\times \left( 1\times \frac{\pi }{180} \right)}\]= \[{{10}^{-4}}m\]


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