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Asked in GUJCET 2013 · Total internal reflection and optical fibres
Given: prism glass of index n_g=√3, a liquid of index n on face AC, and light entering AB along its normal.
Idea: a ray that enters AB normally is undeviated, so it travels straight to AC. The normals of AB and AC are as far apart as the faces themselves, so the angle of incidence at AC equals the angle of the prism at A, which is 60°.
Total internal reflection at AC needs that angle to exceed the critical angle C, where sin C=n/(n_g).
So sin 60°>n/(n_g), that is n<n_g sin 60°=√3×(√3)/2.
Hence n<3/2 — any ordinary liquid, water for instance, keeps the ray trapped inside the prism.
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