A small light fixture is on the bottom of a swimming pool 1.00 m below surface The emerging from water forms circle still What diameter this circle?

$$n_1\sin\theta_1=n_2\sin\theta_2$$

$$1.00\sin90\degree=1.33\sin\theta_2$$

$$\theta_2=48.7\degree$$

So angle between the extreme rays is

$$2\theta_2=97.5\degree$$

or

$$\tan48.75\degree=\frac h{R}$$

and we have

$$R=\frac{1.00\text{ m}}{\tan48.75\degree}=\boxed{1.14\text{ m}}$$