Answer to Question #71285 in Optics for Jasper

Question

A 60 degree prism of refractive index is 1.5. what is the angle minimum deviation? calculate the angle of refraction of light passing through the prism at minimum deviation and the angle of incidence on the prism at minimum deviation

Answer 1

The angle minimum deviation D can be found from the equation:
n=sin⁡〖(A+D)/2〗/sin⁡〖A/2〗
sin⁡〖(A+D)/2〗=n sin⁡〖A/2〗=1.5 sin⁡30
sin⁡〖(A+D)/2〗=n sin⁡〖A/2〗=1.5 1/2=3/4
(A+D)/2=sin^(-1)⁡〖3/4〗=48.59°
D=2(48.59)-60=37°.
The refracting angle is
r=A/2=30°.
The angle of incidence is
i=(A+D)/2=(60+37)/2=48.5°