Q143 / 160 · TS EAMCET 2024 (10 May Shift 2) · Physics
The Brewster angle for air to glass transition of light is (Refractive index of glass $=1.5$ ).
- A$\sin ^{-1}\left(\frac{3}{2}\right)$
- B$\cos ^{-1}\left(\frac{3}{2}\right)$
- Correct answerC$\tan ^{-1}\left(\frac{3}{2}\right)$
- D$\cos ^{-1}\left(\frac{2}{3}\right)$
Solution
Brewster's angle, $i_p=\tan ^{-1}(\mu)=\tan ^{-1}(1.5)$
$=\tan ^{-1}\left(\frac{3}{2}\right)$
Hence, the correct answer is option: 'C'.