Unpolarised light is incident on the boundary between two dielectric media , whose dielectric constants are 2.8 (medium-1) and 6.8 (medium -2), respectively .To satisfy the condition ,so that the reflected and refracted rays are perpendicular to each other , the angle of incidence should be $\tan^{- 1} {(1 + \frac{10}{θ})}^{\frac{1}{2}}$ the value of $θ$ is ___
(Given for dielectric media $μ_{r} = 1)$

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answer is 7.

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Detailed Solution

Angle of incidence should be equal to Brewster's angle $e (θ_{B}) .$

$as θ_{B} = \tan^{- 1} (\frac{μ_{2}}{μ_{1}})$
$\begin{array}{l} As C^{1} = \frac{1}{\sqrt{μ_{0} ϵ_{r}}} = \frac{C}{μ} \\ \Rightarrow \frac{C}{μ_{1}} = \frac{1}{\sqrt{μ_{0} K_{1} ϵ_{0}}} and \frac{C}{μ_{2}} = \frac{1}{\sqrt{μ_{0} K_{2} ϵ_{0}}} \\ \Rightarrow \frac{μ_{2}}{μ_{1}} = \sqrt{\frac{K_{2}}{K_{1}}} \Rightarrow θ_{B} = \tan^{- 1} {(\frac{6.8}{2.8})}^{1 / 2} \\ θ_{B} = \tan^{- 1} {(1 + \frac{4}{2.8})}^{1 / 2} = \tan^{- 1} {(1 + \frac{10}{7})}^{1 / 2} \end{array}$