CUET Preparation Today
A diver at a depth of 1.2 m in clear water $(n = \frac{4}{3})$ sees the sky in a cone of semi-vertical angle: |
$\sin^{-1}\frac{4}{3}$ $\tan^{-1}\frac{4}{3}$ $\sin^{-1}\frac{3}{4}$ $\tan^{-1}\frac{3}{4}$ |
$\sin^{-1}\frac{3}{4}$ |
The correct answer is Option (3) → $\sin^{-1}\frac{3}{4}$ |
