JEE Main & Advanced Physics Question Bank Prism Theory and Dispersion of Light

  • question_answer
    Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism. The angle of prism is (cos 41° = 0.75)                                                            [MP PET/PMT 1988]

    A)            62°

    B)            41°

    C)            82°

    D)            31°

    Correct Answer: C

    Solution :

                       By prism formula \[n=\frac{\sin \frac{A+A}{2}}{\sin \frac{A}{2}}=\frac{2\sin \frac{A}{2}\cos \frac{A}{2}}{\sin \frac{A}{2}}\]            \[\therefore \] \[\cos \frac{A}{2}=\frac{n}{2}=\frac{1.5}{2}=0.75=\cos {{41}^{o}}\] Þ \[A={{82}^{o}}\]


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