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Asked in GSEB Board July 2017 · Total internal reflection and optical fibres
Given: light inside the denser medium, critical angle C for the pair.
Idea: the ray behaves in two quite different ways, so find the largest deviation in each range and compare.
While i < C the ray refracts out, bending away from the normal to r > i, and its deviation is δ = r - i. As i climbs towards C, r climbs towards 90°, so the biggest deviation available here is 90° - C.
Once i ≥ C nothing emerges: the ray is totally internally reflected, and a reflected ray is turned through δ = π - 2i. That is largest at the smallest permitted i, namely i = C, giving π - 2C.
Compare the two: π - 2C = 2(90° - C), exactly twice the refraction maximum, so the reflected branch wins.
The maximum possible deviation is π - 2C, suffered by the ray that strikes the boundary at exactly the critical angle.
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