MGIMS WARDHA MGIMS WARDHA Solved Paper-2004

  • question_answer
    A prism of refractive index \[\sqrt{2}\] has a refracting angle of \[60{}^\circ C\]. At what angle a ray must be incident on it so that it suffers a minimum deviation?

    A) \[45{}^\circ C\]

    B) \[60{}^\circ C\]

    C) \[90{}^\circ C\]

    D) \[180{}^\circ C\]

    Correct Answer: A

    Solution :

    The relation for refractive index of prism is \[\mu =\frac{\sin i}{\sin r}\]                               ...(i) The condition for minimum deviation is \[r=\frac{A}{2}=\frac{{{60}^{o}}}{2}={{30}^{o}}\] Putting the given values of\[\mu =\sqrt{2}\]and\[r=30{}^\circ \]in eq (i), we get \[\sqrt{2}=\frac{\sin i}{\sin {{30}^{o}}}\] or        \[\sin i=\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}}\]                 \[\sin i=\sin {{45}^{o}}\] \[\therefore \]  \[i={{45}^{o}}\]


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