As light passes from water into air the critical angle may be exceeded and total internal reflection may occur - Leaving Cert Physics - Question c - 2019

Total internal reflection occurs when the angle of incidence is greater than the critical angle, resulting in all the incident light being reflected back into the denser medium instead of transmitting into the less dense medium.

To find the area of the circular disc of light visible to the diver, we first need to determine the radius using Snell's Law:

1. Determine the angle of refraction using the critical angle:

   • Given the refractive index of water (n = 1.33), the sine relationship is:

   $n = \frac{1}{\sin C} \Rightarrow \sin C = \frac{1}{1.33} \Rightarrow C \approx 48.8^{\circ}$

2. Using the tangent relation, we find the radius:

   • From geometry, we have:

   $\tan \theta = \frac{r}{d} \Rightarrow r = d \cdot \tan \theta$

   where $d = 12$ m.

   • Therefore,

   $r = 12 \cdot \tan 48.8^{\circ} \approx 12 \cdot 1.14 \approx 13.7 \text{ m}$

3. Calculate the area:

   • The area $A$ of the disc is given by:

   $A = \pi r^2 \approx \pi (13.7)^2 \approx 590 \text{ m}^2$

In the diagram, draw the diver below the water surface, indicating the path of light rays as they move from the water to the air. Mark the point of total internal reflection and indicate the apparent position of the diver when viewed from above the water:

• The light rays coming from the diver bend away as they exit the water, which results in the diver appearing closer to the surface than his actual depth of 12 m.
• Label the correct refracted ray, the correct position of the image above the water, and the apparent position from the observer's perspective. This illustrates that due to refraction, the diver seems to be at a shallower depth.