Manipal Medical Manipal Medical Solved Paper-2009

  • question_answer
    If refractive index of a material of equilateral prism is\[\sqrt{3},\]then angle of minimum deviation of the prism is

    A)  \[30{}^\circ \]             

    B)  \[45{}^\circ \]

    C)  \[60{}^\circ \]              

    D)  \[75{}^\circ \]

    Correct Answer: C

    Solution :

    \[\mu =\frac{\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \frac{A}{2}}\] \[\sqrt{3}=\frac{\sin \left( \frac{60{}^\circ +{{\delta }_{m}}}{2} \right)}{\sin \frac{60{}^\circ }{2}}\] \[\Rightarrow \] \[\frac{\sqrt{3}}{2}=\sin \left[ 30{}^\circ +\frac{{{\delta }_{m}}}{2} \right]\] \[\Rightarrow \] \[{{\delta }_{m}}=60{}^\circ \]


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