Question

The Brewster’s single for air to water interface is

• A. $\tan^{-1} (1.33)$
• B. $\sin^{-1} (1.33)$
• C. $\cos^{-1} (1.33)$
• D. $\tan^{-1} (\frac{1}{1.33})$
• E. $\sin^{-1} (\frac{1}{1.33})$

Answer 1

Answer: (A)

As we know, when partially reflected and refracted rays makes $90^{\circ}$ angle then both get polarised, i.e.
$i_{p}=\tan ^{-1}(\mu)$ (Brewester's law)
$\Rightarrow i_{p}=\tan ^{-1}\left(\frac{\mu_{\text {water }}}{\mu_{\text {air }}}\right) $
where, $ i_{p}=$ Brewester's angle.
$\Rightarrow i_{p} =\tan ^{-1}\left(\frac{1.33}{1}\right)$
$ i_{p}=\tan ^{-1}(1.33)$