Question

If a 60 degree glass prime has a refractive index of 1.5 what is the angle of minimum deviation

Answer 1

Answer:

Let δm be the angle of minimum deviation A=δm

A=δm

μ=sin(2A)sin2(A+δm)

1.5=sin(2A)sin(2A+A)

1.5=sin(2A)sin2(2A)=sin(2A)2sin2Acos2A

1.5=2cos2A

0.75=cos2A

2A=cos<

Answer 2

Answer:

your answer is in attachment