Solution

a) Consider two rays: one goes vertically upward from a point on the bottom, A, so it does not refract and intersects the water surface at point O; the other goes at a small angle β to the vertical, intersects the water surface at point E, and emerges at an angle α.

The apparent bottom will be at the intersection of the vertical line and the emerging ray.

From Snell's law, using sin(x) ≈ tan(x) ≈ x:

α = n β

Express the segment EO in two triangles:

EO = h' α = h β

h = n h' = 4 m

b) The solution is analogous to part (a), except here the extension of the refracted ray will go beyond the true bottom, so the apparent depth will be greater:

h' = h n = 4 km