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CBSE Class 12 Physics 2022 Term 2 Delhi Set 1 Solved Paper

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Question : 10 of 12
Marks: +1, -0
A ray of light passes through a prism of refractive index 2\sqrt{2} as shown in the figure. Find: 3
(i) The angle of incidence (∠r2)(\angle r_2) at face AC.
(ii) The angle of minimum deviation for this prism.
Solution:  
(i) Since at point N, the angle of reflection is 90∘90^{\circ}, then ∠r2\angle r_2 is the critica angle for the glass - air pair of media.
sin⁡∠r2=1μ=12\sin \angle r_2 = \frac{1}{\mu} = \frac{1}{\sqrt{2}}
∠r2=sin⁡−1(12)=45∘\angle r_2 = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^{\circ}
(ii) μ=sin⁡(A+δm2)sin⁡A2\mu = \frac{\sin\left(A+\frac{\delta_m}{2}\right)}{\sin\frac{A}{2}}
Or, 2=sin⁡(60∘+δm2)sin⁡60∘2\sqrt{2} = \frac{\sin\left(60^{\circ}+\frac{\delta_m}{2}\right)}{\sin\frac{60^{\circ}}{2}}
Or, 2=sin⁡(30∘+δm2)12\sqrt{2} = \frac{\sin\left(30^{\circ}+\frac{\delta_m}{2}\right)}{\frac{1}{2}}
Or, 0.7=sin⁡(30∘+δm2)0.7 = \sin\left(30^{\circ}+\frac{\delta_m}{2}\right)
Or, sin⁡−10.7=30∘+δm2\sin^{-1} 0.7 = 30^{\circ} + \frac{\delta_m}{2}
Or, 44.4∘=30∘+δm244.4^{\circ} = 30^{\circ} + \frac{\delta_m}{2}
∴δm=28.8∘\therefore \delta_m = 28.8^{\circ}
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