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 \[\cos \,{{41}^{o}}-24'-36''=0.75\])

A)  \[{{82}^{o}}-49'-12''\]    

B)  \[{{72}^{o}}-48'-30''\]

C)  \[{{41}^{o}}-24'-36''\]

D)  \[{{31}^{o}}-49'-30''\]

Correct Answer: A

Solution :

Let A be the angle of prism, and \[\delta \] 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, \[\delta =A,\,\,\,\mu =1.5\] \[\therefore \] \[1.5=\frac{\sin \left( \frac{A+A}{2} \right)}{\sin \left( \frac{A}{2} \right)}\] Also,  \[\sin \,2\theta =2\sin \theta \,\,\cos \theta \] \[\therefore \] \[1.5=\frac{2\sin \frac{A}{2}\cos \frac{A}{2}}{\sin \frac{A}{2}}\] \[\Rightarrow \] \[\cos \frac{A}{2}=\frac{1.5}{2}=0.75\] \[\Rightarrow \] \[\frac{A}{2}={{41}^{o}}-24'-36''\] \[\Rightarrow \] \[A={{82}^{o}}-49'-12''\]