The Brewster's angle i_b for an interface sho | vedclass

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(D) According to Brewster's law,$	an i_b = \frac{\mu_2}{\mu_1}$.Assuming the first medium is air,$\mu_1 = 1$.Thus,$	an i_b = \mu_2$.Since the refrac

Vedclass · Accepted Answer

$45^{\circ} < i_b < 90^{\circ}$

Vedclass · Answer

(D) According to Brewster's law,$ an i_b = \frac{\mu_2}{\mu_1}$.Assuming the first medium is air,$\mu_1 = 1$.Thus,$ an i_b = \mu_2$.Since the refractive index of any denser medium $\mu_2$ is greater than $1$ (i.e.,$\mu_2 > 1$),we have $ an i_b > 1$.Since $ an 45^{\circ} = 1$,for $ an i_b > 1$,the angle $i_b$ must be greater than $45^{\circ}$.Also,the angle of incidence must be less than $90^{\circ}$ for the interface.Therefore,$45^{\circ} < i_b < 90^{\circ}$.

The Brewster's angle $i_b$ for an interface should be

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