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CBSE Class 12 Physics 2019 Outside Delhi Set 1 Paper

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Question : 22 of 27
Marks: +1, -0
A triangular prism of refracting angle 60∘60^{\circ} is made of a transparent material of refractive index   23\;\frac{2}{\sqrt{3}}.
A ray of light is incident normally on the face KLKL as shown in the figure. Trace the path of the ray as it passes through the prism and calculate the angle of emergence and angle of deviation.
Solution:  
If iCi_C is the critical angle for the prism/material,
μ  =  1sin⁡  iC\mu\;=\;\frac{1}{\sin\;i_C}
∴    sin⁡  iC  =  1μ  =  32\therefore\;\; \sin\;i_C\;=\;\frac{1}{\mu}\;=\;\frac{\sqrt{3}}{2}
⇒    iC  =60∘\Rightarrow\;\; i_C\;=60^{\circ}
Angle of incidence at face ACAC of the prism
=60∘=60^{\circ}
Hence, refracted ray grazes the surface AC.
⇒   Angle of emergence   =90∘\Rightarrow\; \text{ Angle of emergence }\;=90^{\circ}
⇒   Angle of deviation     =30∘\Rightarrow\; \text{ Angle of deviation }\;\;=30^{\circ}
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