Unpolarised light is incident on the boundary between two...

As given in the question, we can write

${\mu}_{1} = \sqrt{{ε}_{r}{\mu}_{r}} = \sqrt{2 . 8 \times 1} = \sqrt{2 . 8}$ and

${\mu}_{2} = \sqrt{{ε}_{r}{\mu}_{r}} = \sqrt{6 . 8 \times 1} = \sqrt{6 . 8}$

Now applying Snell's law and considering $r = 90 ° - i$, we get

${\mu}_{1}sini = {\mu}_{2}cosi$

$⇒ tani = \dfrac{{\mu}_{2}}{{\mu}_{1}} = \sqrt{\dfrac{6 . 8}{2 . 8}}$

$⇒ tani = \left(\dfrac{2 . 8 + 4}{2 . 8}\right)\dfrac{1}{2}$

$⇒ i = \tan^{- 1}\left(1 + \dfrac{10}{7}\right)\dfrac{1}{2}$

Therefore, $\theta = 7$.