Question

62. A ray of light incident on the face AB of
an isosceles triangular prism makes an
angle of incidence i and deviates by angle
B as shown in the figure. Show that in the
position of minimum deviation ZB = Za.
Also find out the condition, when the
refracted ray QR suffer total internal
reflection
C All India 2019​

Answer 1

Position of Minimum deviation ZB = Za

i.e ∠β = ∠α = 60°

Explanation: Here α = 60° (for isosceles triangle)

• Now r₁ = 90°- β from diagram
• and r₂ = β - 30° from diagram
• and for minimum deviation angle
• r₁ = r₂
• 90°-β = β -30°
• 2β = 120°
• β = 60° = α Hence proved i.e ZB =Za.
• For TIR
• r₂= Ic
• r₂ = 30°
• now according to condition of TIR
• 1/ SinIc ≤ u
• 1/Sin30° ≤ u
• u≥ 2