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

To find the area of the circular disc of light seen by the diver, we first need to determine the radius of the circle formed at the water's surface.

Using Snell's law: $n = \frac{1}{\sin \theta_c}$ where $n$ is the refractive index of water (1.33). We can find the critical angle ( \theta_c ): $\theta_c = \sin^{-1}\left(\frac{1}{1.33}\right) \approx 48.8^{\circ}$

Next, we calculate the radius ( r ) of the circle at the water's surface using the tangent function: $r = h \tan \theta$ where ( h = 12 ) m (depth of the diver) and ( \theta = 48.8^{\circ} ).

Calculating gives: $\tan(48.8^{\circ}) = \frac{r}{12}$ Thus, $r \approx 12 \tan(48.8^{\circ}) \approx 13.7$ m.

Now, the area ( A ) of the disc can be calculated using the formula: $A = \pi r^2 \approx \pi (13.7)^2 \approx 590$ m^2.

When viewed from outside the pool, the diver appears at a different depth due to the effects of refraction. Light rays emanating from the diver bend as they travel from the water (denser medium) to air (less dense medium). This bending of light makes the diver appear closer to the surface than they actually are.

• Correct refracted ray: Draw the incident ray coming from the diver hitting the water surface, showing the angle of incidence and then bending away towards the observer.
• Correct position of image: Indicate the position of the diver below the actual depth of 12 m, showing how the observer perceives them closer to the surface.

In constructing the diagram, label all relevant angles, the critical angle, the position of the diver, and the direction of the refracted ray.