Unpolarised light is incident on the boundary | vedclass

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(B) For dielectric media,the refractive index is given by $\mu = \sqrt{\epsilon_r \mu_r}$. Given $\mu_r = 1$,we have $\mu_1 = \sqrt{2.8}$ and $\mu_2 =

Vedclass · Accepted Answer

$7$

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(B) For dielectric media,the refractive index is given by $\mu = \sqrt{\epsilon_r \mu_r}$. Given $\mu_r = 1$,we have $\mu_1 = \sqrt{2.8}$ and $\mu_2 = \sqrt{6.8}$.When the reflected and refracted rays are perpendicular to each other,the angle of incidence $i$ is the Brewster's angle,satisfying $	an i = \frac{\mu_2}{\mu_1}$.Substituting the values,$	an i = \sqrt{\frac{6.8}{2.8}}$.We can rewrite the fraction as $\frac{6.8}{2.8} = \frac{2.8 + 4}{2.8} = 1 + \frac{4}{2.8} = 1 + \frac{40}{28} = 1 + \frac{10}{7}$.Thus,$	an i = \sqrt{1 + \frac{10}{7}}$.Comparing this with the given expression $	an i = \sqrt{1 + \frac{10}{	heta}}$,we get $	heta = 7$.

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