CUET Preparation Today
The Brewster's angle for air to water transition is (Refractive index of water $=\frac{4}{3}$) |
$\sin ^{-1}\left(\frac{4}{3}\right)$ $\sin ^{-1}\left(\frac{3}{4}\right)$ $\tan ^{-1}\left(\frac{4}{3}\right)$ $\tan ^{-1}\left(\frac{3}{5}\right)$ |
$\tan ^{-1}\left(\frac{4}{3}\right)$ |
The correct answer is Option (3) → $\tan ^{-1}\left(\frac{4}{3}\right)$ By Brewster’s law $i_c=\tan ^{-1}(\mu)$ $i_c=\tan ^{-1}\left(\frac{4}{3}\right)$ |
