The conditions of the problem are represented in given figure below: Here it is a point source. When the angle of incidence is equal to the critical angle θ for water-air boundary, no light will be refracted into air. Instead, it will graze the water-air interface. Hence, the light will appear to emerge out from a circle whose edges make angle θ with the vertical. If r is the radius of this circle, then area of surface of water through which light can come out is πr2.Now, sin Therefore, Referring to figure,r/AC = tan  = tan 48.76oTherefore, r = AC tan  = tan 48.76o = 7 tan 48.76o = 8 cmTherefore, Required area =