The problem involves refraction of light as it passes from water to air. The diver sees the sun at an angle of \(54^\circ\) above the horizontal. We need to find the angle at which the sun appears to a fisherman above the water.
Snell's Law relates the angles of incidence and refraction for light passing between two media with different refractive indices. It is given by:
\[
n_1 \sin(\theta_1) = n_2 \sin(\theta_2)
\]
where:
- \(n_1\) is the refractive index of water (\(n_1 \approx 1.33\)),
- \(n_2\) is the refractive index of air (\(n_2 \approx 1.00\)),
- \(\theta_1\) is the angle of incidence (angle in water, \(54^\circ\)),
- \(\theta_2\) is the angle of refraction (angle in air, which we need to find).
Rearrange Snell's Law to solve for \(\theta_2\):
\[
\sin(\theta_2) = \frac{n_1}{n_2} \sin(\theta_1)
\]
Substitute the known values:
\[
\sin(\theta_2) = \frac{1.33}{1.00} \sin(54^\circ)
\]
Calculate \(\sin(54^\circ)\):
\[
\sin(54^\circ) \approx 0.8090
\]
Substitute and solve for \(\sin(\theta_2)\):
\[
\sin(\theta_2) = 1.33 \times 0.8090 \approx 1.0759
\]
Since \(\sin(\theta_2)\) cannot be greater than 1, this indicates a calculation error. Re-evaluate the setup: the angle in water should be less than the angle in air due to refraction. Correct the approach by considering the angle of incidence as the complement of the angle above the horizontal.
The angle of incidence is actually the complement of the angle above the horizontal:
\[
\theta_1 = 90^\circ - 54^\circ = 36^\circ
\]
Recalculate using Snell's Law:
\[
\sin(\theta_2) = \frac{1.33}{1.00} \sin(36^\circ)
\]
Calculate \(\sin(36^\circ)\):
\[
\sin(36^\circ) \approx 0.5878
\]
Substitute and solve for \(\sin(\theta_2)\):
\[
\sin(\theta_2) = 1.33 \times 0.5878 \approx 0.7828
\]
Now, find \(\theta_2\):
\[
\theta_2 = \arcsin(0.7828) \approx 51.6^\circ
\]
The angle of the sun above the horizon to a fisherman in a boat is approximately:
\[
\boxed{51.6^\circ}
\]
Answered
