J & K CET Engineering J and K - CET Engineering 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)  \[si{{n}^{-1}}\left( \frac{3}{4} \right)\]

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

    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 \[A=2{{\cos }^{-1}}\left( \frac{\mu }{2} \right)\] \[=2\,{{\cos }^{-1}}\,\left( \frac{1.5}{2} \right)=2\,{{\cos }^{-1}}\left( \frac{3}{4} \right)\]c


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