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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2005

  • question_answer
    A prism has a refracting angle \[60{}^\circ \]. A ray of given monochromatic light suffers minimum            deviation of \[38{}^\circ \] in passing through prism refractive index of the material of prism is:

    A)  1.5094                                 

    B)  1.3056

    C)  0.7849                 

    D)         2.425               

    Correct Answer: A

    Solution :

    Here: Refracting angle A = \[A=60{}^\circ \], minimum   deviation \[\delta m=\] \[38{}^\circ \]. The refractive index is given by the formula \[\mu =\frac{\frac{A+\delta m}{2}}{\sin \frac{A}{2}}=\frac{\frac{60{}^\circ +38{}^\circ }{2}}{\sin \frac{60{}^\circ }{2}}\] \[=\frac{\sin 49{}^\circ }{\sin 30{}^\circ }=\frac{07547}{0.5}=1.5094\]


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