Question

a ray of light suffers minimum deviation while passing through a prism of refractive index 1.5 and refracting angle 60 degree. calculate angle of deviation and angle of incidence.

Answer 1

here R.I= 1.5
A=60 degree
so r= A/2
60/2=30
R.I= sin i/sin r
1.5= sin i/sin 30
sin i= 1.5*0.5= 0.75
i= sin^-1(0.75)
= 48.6 degree
angle of deviations= 2i - A
2*48.6-60
=37.2 degree

Answer 2

Angle of deviation is 37.2° and angle of incidence is 48.6°

For a prism, refractive index μ is given as:

μ = [sin ([A + δ]/2) ] / sin (A/2)

A is the angle of prism = 60° and

δ is the angle of deviation.

and we know μ = 1.5. So,

⇒ [sin ([60 + δ]/2) ] / sin (60/2) = 1.5

⇒  [sin ([60 + δ]/2) ] = 1.5 × 0.5 = 0.75

⇒ ([60 + δ]/2) = sin⁻¹ (0.75) = 48.6°

⇒ 60° + δ = 2(48.6°) = 97.2°

⇒ δ = 97.2° - 60° = 37.2°

Angle of deviation is 37.2°.

So, angle of incidence i = (A + δ)/2 = (60° + 37.2°)/2 = 48.6°