VMMC VMMC Medical Solved Paper-2002

  • question_answer
    A prism is filled with liquid of refractive index of \[\sqrt{2}\] If angle of prism is \[60{}^\circ \], find angle of minimum deviation

    A) \[75{}^\circ \]                                   

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

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

    D)        \[30{}^\circ \]

    Correct Answer: D

    Solution :

    Applying the formula for refractive index \[(\mu )\] of a prism having angle of prism A \[\mu =\frac{\sin \frac{A+\delta m}{2}}{\sin \frac{A}{2}}\]                                                           ?(1) Putting the given values of\[\mu =\sqrt{2}\] and angle of prism \[\text{A = 6}{{\text{0}}^{\text{o}}}\] in the Eq. (1), we get \[\sqrt{2}=\frac{\sin \frac{{{60}^{o}}+\delta m}{2}}{\sin \frac{{{60}^{o}}}{2}}=\frac{\sin \frac{{{60}^{o}}+\delta m}{2}}{\sin {{30}^{o}}}\] or            \[\sin \left( \frac{{{60}^{o}}+\delta m}{2} \right)=\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}}\] or            \[\sin \frac{{{60}^{o}}+\delta m}{2}=\sin {{45}^{o}}\] or            \[{{60}^{o}}+\delta m={{90}^{o}}\]or \[{{\delta }_{m}}={{90}^{o}}-{{60}^{o}}={{30}^{o}}\]


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