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Asked in GUJCET 2022 · Brewster's law and polarisation by reflection
Given: unpolarised light on a plane glass surface, with the reflected and the refracted ray at 90° to each other. Glass is taken as μ = 1.5.
Idea: that right angle is exactly the Brewster condition, so the angle asked for is the polarising angle i_B.
With the reflected ray at i_B on one side of the normal and the refracted ray at r on the other, i_B + 90° + r = 180°, so r = 90° - i_B.
Snell's law then gives μ = (sin i_B)/(sin r) = (sin i_B)/(cos i_B) = tan i_B, which is Brewster's law.
i_B = tan⁻¹(1.5) = 56.3°.
The nearest printed value is 56°.
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