Question

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

• A. $si{n}^{-1}\left(\frac{4}{3}\right)$
• B. $ta{n}^{-1}\left(\frac{4}{3}\right)$
• C. $si{n}^{-1}\left(\frac{3}{4}\right)$
• D. ${90}^{\circ }$

Answer 1

The correct option is C $si{n}^{-1}\left(\frac{3}{4}\right)$

We can represent this diagrammatically as

Thus; we see that $\theta =si{n}^{-1}\left(\frac{1}{\mu }\right)=si{n}^{-1}\left(\frac{3}{4}\right)$.