A 60° prism has a refractive index of 1.5 Calculate the angle of incidence for minimum deviation, the angle of emergence of light at maximum deviation and angle of maximum deviation.

_{m} is minimum angle of deviation by substituting A = 60º, μ=1.5 and solving for D_{m} we get, D_{m} = 37.2°.

Maximum angle of deviation occurs when incident angle i is 90°.

As shown in figure, angle of deviation δ is given by, δ = i + e - A …(1)

also we have, r + r’ = A …(2)

for maximum angle of deviation,

from eqn.(2), r’ = A-r = 60 - 41.8 = 18.2º

to get angle of emergence e, we have

from eqn.(1), we get maximum angle of deviation, δ = 90+27.94-60 = 57.94 º