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

  • question_answer
    A prism of refractive index m and angle A is placed in the minimum deviation position.  If the angle of minimum deviation is A, then the value of A in terms of m is              [EAMCET 2003]

    A)            \[{{\sin }^{-1}}\left( \frac{\mu }{2} \right)\]               

    B)            \[{{\sin }^{-1}}\sqrt{\frac{\mu -1}{2}}\]

    C)            \[2{{\cos }^{-1}}\left( \frac{\mu }{2} \right)\]            

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

    Correct Answer: C

    Solution :

                       Given \[{{\delta }_{m}}=A,\] as \[\mu =\frac{\sin \,\left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \,\left( \frac{A}{2} \right)}\]            \[\Rightarrow \]\[\mu =\frac{\sin \left( \frac{A+A}{2} \right)}{\sin \left( \frac{A}{2} \right)}=2\cos \frac{A}{2}\Rightarrow A=2{{\cos }^{-1}}\left( \frac{\mu }{2} \right)\]


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