Manipal Medical Manipal Medical Solved Paper-2013

  • question_answer
    What is the refractive index of a prism whose angle\[A=60{}^\circ \]and angle of minimum deviation \[{{d}_{m}}=30{}^\circ \]?

    A)  \[\sqrt{2}\]

    B)  \[{{\sin }^{-1}}(\sqrt{3})\]

    C)  \[{{\tan }^{-1}}(\sqrt{2})\]

    D)  \[{{\tan }^{-1}}(\sqrt{3})\]

    Correct Answer: A

    Solution :

     Refractive index of prism \[\mu =\frac{\sin \frac{A+\delta m}{2}}{\sin A/2}\] \[=\frac{\sin \frac{60{}^\circ +30{}^\circ }{2}}{\sin \frac{60}{2}}\] \[=\frac{\sin 45{}^\circ }{\sin 30{}^\circ }=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}=2\]


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