Explanation:
: As we know, when partially reflected and refracted rays makes \(90^{\circ}\) angle then both get polarized, i,e,
According to Brewster's law -
\(\mathrm{i}_{\mathrm{p}}=\tan ^{-1}(\mu)\)
We know that,
Refractive index \(\mu=\frac{\mu_{\text {water }}}{\mu_{\text {air }}}\)
Putting the value of from equation (ii) in equation (i), we get -
\(i_p=\tan ^{-1}\left[\frac{\mu_{\text {water }}}{\mu_{\text {air }}}\right]\)
Where, \(\mathrm{i}_{\mathrm{p}}=\) Brewster's angle
\(\mathrm{i}_{\mathrm{p}}=\tan ^{-1}\left[\frac{1.33}{1}\right]\)
\(\mathrm{i}_{\mathrm{p}}=\tan ^{-1}(1.33)\)