VMMC VMMC Medical Solved Paper-2012

  • question_answer
    The angle of minimum deviation \[{{\delta }_{m}}\] for an equilateral glass prism is \[\text{3}{{\text{0}}^{\text{o}}}\text{.}\] Refractive index of the prism is

    A) \[1/\sqrt{2}\]

    B)  \[\sqrt{2}\]

    C)  \[2\sqrt{2}\]

    D)  \[1/2\sqrt{2}\]

    Correct Answer: B

    Solution :

     Given, \[A={{60}^{o}}\]and \[{{\delta }_{m}}={{30}^{o}}\] As, \[\mu =\sin \frac{\left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \left( \frac{A}{2} \right)}\] \[\Rightarrow \] \[\mu =\frac{\sin ({{45}^{o}})}{\sin ({{30}^{o}})}\] \[=\frac{1/\sqrt{2}}{1/2}=\sqrt{2}\]


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