RAJASTHAN PMT Rajasthan - PMT Solved Paper-2006

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

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

    B)        \[60{}^\circ \]

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

    D)        \[180{}^\circ \]

    Correct Answer: A

    Solution :

    In 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 \[r={{30}^{o}}\]in Eq. (i), we get                 \[\sqrt{2}=\frac{\sin i}{\sin {{30}^{o}}}\] or            \[\sin i=\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}}\]                 \[\sin i=\sin {{45}^{o}}\]                       \[i={{45}^{o}}\]


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