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:

$\sin^{-1}\frac{3}{4}$

The correct answer is Option (3) → $\sin^{-1}\frac{3}{4}$

Using Snell's law,

$μ_{water}\sin(θ)=μ_{air}\sin(r)$

$⇒\frac{4}{3}\sin(θ)=1.\sin 90°$

$⇒\sin(θ)=\frac{3}{4}$

$⇒θ=\sin^{-1}\left(\frac{3}{4}\right)$