A diver at a depth of (12, m) in water (\bigg(\mu = \frac{4}... | Physics | SolveeIt | Solveeit

Today, August 24

Question

A diver at a depth of $12\, m$ in water $\bigg(\mu = \frac{4}{3}\bigg)$ sees the sky in a cone of semi vertical angle

A

$\sin^{-1} \bigg(\frac{4}{3}\bigg)$

B

$\tan^{-1} \bigg(\frac{4}{3}\bigg)$

C

$\sin^{-1} \bigg(\frac{3}{4}\bigg)$

D

$90^\circ$

Answer

$\sin^{-1} \bigg(\frac{3}{4}\bigg)$

Explanation

Solution

We can represent this diagrammatically as

Thus, we see that $c=\sin^{-1} \bigg(\frac{1}{\mu}\bigg)=\sin^{-1} \bigg(\frac{3}{4}\bigg)$