J & K CET Medical J & K - CET Medical Solved Paper-2007

  • question_answer
    The least angle of 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) \[2{{\cos }^{-1}}\left( \frac{3}{4} \right)\]                            

    B) \[{{\sin }^{-1}}\left( \frac{3}{4} \right)\]

    C) \[2{{\sin }^{-1}}\left( \frac{3}{2} \right)\]                             

    D) \[{{\cos }^{-1}}\left( \frac{3}{2} \right)\]

    Correct Answer: A

    Solution :

                     \[\mu =\frac{\sin \frac{(A+{{\delta }_{m}})}{2}}{\sin \frac{A}{2}}=\frac{\sin (A+A)/2}{\sin (A/2)}\] \[=2\cos A/2\] Or           \[A=2{{\cos }^{-1}}\left( \frac{\mu }{2} \right)\]                 \[=2{{\cos }^{-1}}\left( \frac{1.5}{2} \right)=2{{\cos }^{-1}}\left( \frac{3}{4} \right)\]


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