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Asked in GSEB Board March 2020 · Brewster's law and polarisation by reflection
Given: unpolarised light on a plane glass surface, with the reflected and the refracted rays at right angles. Glass is taken as μ = 1.5.
Idea: the reflected and refracted rays are perpendicular only at the polarising angle, so the angle of incidence asked for is the Brewster angle i_B.
At that angle i_B + r = 90°, so r = 90° - i_B.
Snell's law μ = (sin i_B)/(sin r) becomes μ = (sin i_B)/(cos i_B) = tan i_B, which is Brewster's law.
i_B = tan⁻¹(1.5) = 56.3°.
Of the values offered that is 57°, and the refracted ray then leaves at 33°.
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