A light source is located in the cell with a liquid at a depth. At the bottom of the cell there is a flat mirror

A light source is located in the cell with a liquid at a depth. At the bottom of the cell there is a flat mirror. The liquid layer in the cell is 6 cm. A black disk with an area of 314 cm ^ 2 floats on the surface of the liquid above the light source. At what depth should the light source be located so that it was visible to an outside observer if the refractive index of the liquid is 1.14? should come out 0.5cm

Area of a circle S = π * r2;

Disc area S = 314cm2;
disk radius r = (S / π) = 314 / 3.14 = 10 cm.
In order for the light to escape from under the disk, the angle of incidence from the liquid must be less than the angle of reflection.

n = 1.14;
α = arcsin (1 / n) = arcsin (1 / 1.14) = 61.305590;
So the depth of the light source = R / tg 61.305590 = 5.473573 ≈ 5.5 cm.
If the light source is at a depth of 5.5 cm and deeper, an outside observer will see 2 luminous light sources. If the source is at a depth of less than 5.5 cm, then the observer will see 1 light source – a reflection of the source in a flat mirror.