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Q.At what angle should a ray of light be incident on the face of an equilateral prism, so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.5.

Karnataka PUCKarnataka II PUC Board 2020Subjective· 5mImportance★★★★★
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TIR at 2nd face needs r2=Cr_2=C where sin⁡C=1/μ⇒C=41.8∘\sin C=1/\mu\Rightarrow C=41.8^\circ. Prism relation r1+r2=A=60∘r_1+r_2=A=60^\circ gives r1=18.2∘r_1=18.2^\circ. Snell at 1st face: sin⁡i=μsin⁡r1=1.5×0.312=0.468⇒i≈27.9∘\sin i=\mu\sin r_1=1.5\times0.312=0.468\Rightarrow i\approx27.9^\circ.

Data. Equilateral prism ⇒A=60∘\Rightarrow A=60^\circ; refractive index μ=1.5\mu=1.5.

Step 1 — Critical angle. For total internal reflection at the second face, the ray must hit it at the critical angle CC:

sin⁡C=1μ=11.5=0.6667  ⇒  C=41.8∘.\sin C=\frac{1}{\mu}=\frac{1}{1.5}=0.6667\;\Rightarrow\; C=41.8^\circ.

Step 2 — Refraction angle at the first face. For a prism, the two internal refraction angles satisfy

r1+r2=A.r_1+r_2=A.

To just suffer TIR, r2=C=41.8∘r_2=C=41.8^\circ, so …

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