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CBSE Class 12 Physics 2019 Outside Delhi Set 2 Paper

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Question : 8 of 9
Marks: +1, -0
A ray of light incident on the face ABA B of an isosceles triangular prism makes an angle of incidence (i) and deviates by angle β\beta as shown in the figure. Show that in the position of minimum deviation ∠β=∠α\angle \beta = \angle \alpha. Also find out the condition when the refracted ray QRQ R suffers total internal reflection.
Solution:  
Proving α=β\alpha = \beta
Finding ici_c
For minimum deviation,
  r1+r2=A;    r1=r2\; r_1 + r_2 = A ; \; \; r_1 = r_2
  (90∘−β)+(90∘−β)=A\; (90^{\circ} - \beta) + (90^{\circ} - \beta) = A
  180∘−2β=A\; 180^{\circ} - 2 \beta = A
  2β=180∘−A\; 2 \beta = 180^{\circ} - A
  2β=2α\; 2 \beta = 2 \alpha
  β=α\; \beta = \alpha
  r1+r2=A\; r_1 + r_2 = A
  r1+ic=A\; r_1 + i_c = A
  ic=A−r1\; i_c = A - r_1
  ic=A−(90∘−β)\; i_c = A - (90^{\circ} - \beta)
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