question_answer

Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of the prism. The angle of the prism is : (given \[cos\text{ }41{}^\circ -24-36\text{ }=\text{ }0.75\]}

A)  \[82{}^\circ -{{49}^{}}-12\]                         

B)  \[72{}^\circ -48-30\]

C)  \[41{}^\circ -24-36\]                      

D)  \[{{31}^{0}}-49-30\]

Correct Answer: A

Solution :

                Let A be the angle of prism, and 5 the angle of minimum deviation, then refractive index of the medium of prism is given by                 \[\mu =\frac{\sin \left( \frac{A+\delta }{2} \right)}{\sin \left( \frac{A}{2} \right)}\] Given,     \[\delta =A,\text{ }\mu =1.5\] \[\therefore \]  \[1.5=\frac{\sin \left( \frac{A+A}{2} \right)}{\sin \left( \frac{A}{2} \right)}\] Also,   \[\sin 2\theta =2\sin \theta \cos \theta \] \[\therefore \]  \[1.5=\frac{2\sin \frac{A}{2}\cos \frac{A}{2}}{\sin \frac{A}{2}}\] \[\Rightarrow \]               \[\cos \frac{A}{2}=\frac{1.5}{2}=0.75\] \[\Rightarrow \]               \[\frac{A}{2}={{41}^{o}}-24-36\] \[\Rightarrow \]               \[A={{82}^{o}}-49-12\]