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

  • question_answer
    For a prism of refractive index 1.732, the angle of minimum deviation is equal to the angle of the prism.  The angle of the prism is                                                       [CBSE PMT 2001]

    A)            80o                                            

    B)            70o

    C)            60o                                            

    D)            50o

    Correct Answer: C

    Solution :

                       \[\mu =\frac{\sin \frac{A+{{\delta }_{m}}}{2}}{\sin \frac{A}{2}}=\frac{\sin \frac{A+A}{2}}{\sin \frac{A}{2}}=\frac{\sin A}{\sin \frac{A}{2}}\]                    = \[\frac{2\sin \frac{A}{2}\cos \frac{A}{2}}{\sin \frac{A}{2}}=2\cos \frac{A}{2}\]            So, \[\sqrt{3}=2\cos \frac{A}{2}\Rightarrow \frac{\sqrt{3}}{2}=\cos \frac{A}{2}\Rightarrow A={{60}^{o}}\]


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