JEE Main & Advanced Sample Paper JEE Main Sample Paper-24

  • 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)  \[{{\sin }^{-1}}\left( \frac{3}{4} \right)\]

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

    D)  \[{{\cot }^{-1}}\left( \frac{3}{2} \right)\]

    Correct Answer: A

    Solution :

    Refractive index of prism \[\mu =\frac{\sin \frac{(A+{{\delta }_{m}})}{2}}{\sin \frac{A}{2}}\] \[=\frac{\sin (A+A/2)}{\sin (A/2)}=2\cos A/2\] or \[A=2\,{{\cos }^{-1}}\,\left( \frac{\mu }{2} \right)\] \[=2\,{{\cos }^{-1}}\,\left( \frac{1.5}{2} \right)=2\,{{\cos }^{-1}}\left( \frac{3}{4} \right)\]


You need to login to perform this action.
You will be redirected in 3 sec spinner