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
Answers
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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
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