A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm.

Actual depth of the bulb in water, d1 = 80 cm = 0.8 m

Refractive index of water, 

The given situation is shown in the following figure:

Where,

i = Angle of incidence

r = Angle of refraction = 90°

Since the bulb is a point source, the emergent light can be considered as a circle of radius, $R=\frac{A C}{2}=A O=O B$

Using Snell' law, we can write the relation for the refractive index of water as:

$\mu=\frac{\sin r}{\sin i}$

$1.33=\frac{\sin 90^{\circ}}{\sin i}$

$\therefore i=\sin ^{-1}\left(\frac{1}{1.33}\right)=48.75^{\circ}$

Using the given figure, we have the relation:

$\tan i=\frac{O C}{O B}=\frac{R}{d_{1}}$

∴R = tan 48.75° × 0.8 = 0.91 m

∴Area of the surface of water = πR2 = π (0.91)2 = 2.61 m2

Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61 m2.