BHU PMT BHU PMT (Mains) Solved Paper-2010

  • question_answer
    The angle of minimum deviation for a glass prism is equal to its refracting angle. The refractive index of glass is 1.5. Then the angle of prism is

    A)  \[2{{\cos }^{-1}}(3/4)\]

    B)  \[{{\sin }^{-1}}(3/4)\]

    C)  \[2{{\sin }^{-1}}(3/2)\]                 

    D)  \[{{\cos }^{-1}}(3/2)\]

    Correct Answer: A

    Solution :

                     According to question, the angle of minimum deviation \[{{\delta }_{m}}=\]Refracting angle Refractive index, \[\mu =\frac{\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \left( \frac{A}{2} \right)}\]                                 \[=\frac{\sin \left( \frac{A+A}{2} \right)}{\sin \left( \frac{A}{2} \right)}\]                                 \[=\frac{\sin A}{\sin \left( \frac{A}{2} \right)}\]                                 \[=\frac{2\sin \left( \frac{A}{2} \right).\cos \left( \frac{A}{2} \right)}{\sin \left( \frac{A}{2} \right)}\]                 \[\frac{3}{2}=2\cos \left( \frac{A}{2} \right)\]                 \[\cos \left( \frac{A}{2} \right)=\frac{3}{4}\]                 \[\frac{A}{2}={{\cos }^{-1}}\left( \frac{3}{4} \right)\]                 \[A=2{{\cos }^{-1}}\left( \frac{3}{4} \right)\]


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