Given below are two statements :
Statement I : If the Brewster's angle for the light propagating from air to glass is $θ_{B}$ , then Brewster's angle for the light propagating from glass to air is $\frac{π}{2} - θ_{B}$ .

Question

Statement II : The Brewster's angle for the light propagating from glass to air is $\tan^{- 1} (μ_{g})$ , where $μ_{g}$ is the refractive index of glass.
In the light of the above statements, choose the correct answer from the options given below :

TestHub · Accepted Answer

Applying Snell's law, we get μ a sin i = μ g sin 90 ° - i tan i = μ g μ a When going from glass to air tan i ' = μ a μ g = cot i Clearly, i ' = π 2 - i & hence Statement I is true but Statement II is false.

Physics - Wave Optics Question with Solution | TestHub

Options:

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