Manipal Medical Manipal Medical Solved Paper-2006

  • question_answer
    A prism of refractive index \[\sqrt{2}\]has a refracting angle of\[60{}^\circ \]. At 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 :

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


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