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Asked in CBSE AIPMT 1998 · Total internal reflection and optical fibres
Given: light enters the flat end of a rod of index μ at some angle i, and must not escape through the curved side whatever i is.
Idea: the refracted ray makes r with the rod's axis, so it strikes the lateral face at 90°-r; that angle has to exceed the critical angle for every possible r.
sin(90°-r)>sin C, that is cos r>1/μ.
The hardest case is grazing entry, i=90°, which gives the largest r: sin r=1/μ.
cos r=√1-1/(μ²)>1/μ
1-1/(μ²)>1/(μ²), so μ²>2
μ>√2
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