θ 1 = tan ⁡ ( 1.5 1.003 ) = 56.302 ∘ {\displaystyle {\begin{aligned}\theta _{1}&=\tan \left({\frac {1.5}{1.003}}\right)\\[5pt]&=56.302^{\circ }\end{aligned}}}

This is a refraction problem with the case where the refracted ray is perpendicular to the normal line. Thus, it is appropriate to use Snell's Law n 1 sin ⁡ ( θ 1 ) = n 2 sin ⁡ ( θ 2 ) , {\displaystyle n_{1}\sin(\theta _{1})=n_{2}\sin(\theta _{2}),} where θ 2 = 90 ∘ . {\displaystyle \theta _{2}=90^{\circ }.}

(1) The Brewster's angle is calculated using the formula θ B = tan ⁡ ( n 2 n 1 ) . {\displaystyle \theta _{B}=\tan \left({\frac {n_{2}}{n_{1}}}\right).} See derivation here.

Let n 1 {\displaystyle n_1} represent the refractive index above of air, and n 2 {\displaystyle n_2} the refractive index above of glass. The refractive index is n 1 = 1.0003 {\displaystyle n_{1}=1.0003} for air and n 2 = 1.5 {\displaystyle n_{2}=1.5} for glass. Substitute these two values into the Brewster's angle formula θ B = tan ⁡ ( n 2 n 1 ) {\displaystyle \theta _{B}=\tan \left({\frac {n_{2}}{n_{1}}}\right)} to determine θ 1 . {\displaystyle {\theta }_{1}.}

Part 2: Suppose the angle of incidence is θ 1 = 60 ∘ , {\displaystyle {\theta }_{1}=60^{\circ },} but the glass is immersed in a solution with refractive index n 1 > 1. {\displaystyle n_{1}>1.} Determine the upper bound of n 1 {\displaystyle n_1} that the refracted light can be transmitted through the right side of the glass. Give the answer to three significant figures.

Part 1: If the incident light is unpolarized, find the angle of incidence θ 1 {\displaystyle {\theta }_{1}} that the light reflected from the top is horizontally polarized. Give the answer to five significant figures.