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 $\frac{π}{2} - θ_{B} .$
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:

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Both statements I and statement II are true.

Statement I is true but statement II is false

Both statement I and Statement II are false

Statement I is false but statement II is true

answer is B.

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Detailed Solution

From air to glass,

$μ_{a} \sin i_{1} = μ_{g} \sin (90 - i_{1})$

$\tan i_{1} = \frac{μ_{g}}{μ_{a}}$

From glass to air,

$\tan i_{2} = \frac{μ_{a}}{μ_{g}} = \cot i_{1}$

$\Rightarrow i_{2} = \frac{π}{2} - i_{1}$

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