Question

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

• A. ${\sin }^{-1}\left(\frac{4}{3}\right)$
• B. ${\tan }^{-1}\left(\frac{4}{3}\right)$
• C. ${\sin }^{-1}\left(\frac{3}{4}\right)$
• D. ${90}^o$

Answer 1

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

Given : $\mu =\frac{4}{3}$ and ${\mu }_{air}=1$

The ray emerges out of water making an angle ${90}^o$ with the water surface, i.e $r={90}^o$

Using Snell's law of refraction at point A:

$\Rightarrow \mu sin\theta ={\mu }_{air}sinr$

$\Rightarrow$$\frac{4}{3}\times sin\theta =1\times sin{90}^o$

$\Rightarrow \frac{4}{3}\times sin\theta =1\times 1$

$\Rightarrow$ $sin\theta =\frac{3}{4}$

$\Rightarrow$ $\theta =si{n}^{-1}(\frac{3}{4})$