Manipal Engineering Manipal Engineering Solved Paper-2009

  • question_answer
    The angle of minimum deviation for an incident light ray on an equilateral prism is equal to its refracting angle. The refractive index of its material is

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

    B)  \[\sqrt{3}\]

    C) \[\frac{\sqrt{3}}{2}\]                     

    D)         \[\frac{3}{2}\]

    Correct Answer: B

    Solution :

    For an equilateral prism, angle of prism of refracting angle\[A={{60}^{o}}\] Here,     \[{{\delta }_{m}}=A={{60}^{o}}\] \[\therefore \]Refractive index, \[\mu =\frac{\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \frac{A}{2}}=\frac{\sin \left( \frac{{{60}^{o}}+{{60}^{o}}}{2} \right)}{\sin \left( \frac{{{60}^{o}}}{2} \right)}\] \[=\frac{\sin {{60}^{o}}}{\sin {{30}^{o}}}=\frac{\sin {{60}^{o}}}{\cos {{60}^{o}}}=\tan {{60}^{o}}=\sqrt{3}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner