State the laws of refraction of light - Leaving Cert Physics - Question (b) - 2011

In the ray diagram, illustrate the lamp at the bottom of the swimming pool emitting light rays upwards toward the surface. Show the incident rays hitting the water surface, the refracted rays bending as they pass into the air, and the apparent position of the lamp as seen by the observer at the edge of the pool. Indicate the correct emergent ray and the image position clearly.

The area of water surrounding the illuminated disc appears dark because light from the lamp refracts when it exits the water into the air, creating an angle of incidence that exceeds the critical angle for water. As a result, total internal reflection occurs for the rays that would normally illuminate areas outside the disc. Thus, no light emerges from these areas, leading to a dark appearance.

To calculate the area of the illuminated disc:

1. First, find the critical angle using the refractive index of water ($n = 1.33$): $n = \frac{1}{\sin i_c} \Rightarrow \sin i_c = \frac{1}{n} = \frac{1}{1.33} \Rightarrow i_c \approx 48.76^\circ$

2. Using trigonometry to find the radius of the illuminated disc ($r$): $r = 1.8 m \tan 48.76^\circ \approx 2.053 m$

3. Calculate the area ($A$): $A = \pi r^2 \approx \pi (2.053 m)^2 \approx 13.24 m^2$