A diver at a depth of 12 m in water left(mu | vedclass

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(C) When a diver is inside the water, they see the outside world through a cone of light due to the phenomenon of total internal reflection. The semi-

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

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(C) When a diver is inside the water, they see the outside world through a cone of light due to the phenomenon of total internal reflection. The semi-vertical angle of this cone is equal to the critical angle $c$ for the water-air interface. The formula for the critical angle is given by $\sin c = \frac{1}{\mu}$. Given the refractive index of water $\mu = \frac{4}{3}$, we have: $\sin c = \frac{1}{4/3} = \frac{3}{4}$. Therefore, the semi-vertical angle is $c = \sin ^{-1}\left(\frac{3}{4}\right)$.

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

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