Geometrical optics. Photometry. Quantum nature of light > Refraction of light. The lens formula > Problem 13.2.2

Statement

$13.2.2.$ a) The apparent depth of a reservoir, when looking vertically downward, is $3$ m. What is its true depth?

b) An airplane flies over a submarine submerged at a small depth, at an altitude of $3$ km. What will the altitude of the airplane appear to be when observed from the submarine?

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 $\beta$ to the vertical, intersects the water surface at point $E$, and emerges at an angle $\alpha$.

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

From Snell's law, using $\sin(x) \approx \tan(x) \approx x$:

$$\alpha = n \beta$$

Express the segment $EO$ in two triangles:

$$EO = h' \alpha = h \beta$$

$$h = n h' = 4 \text{ m}$$

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

$$h' = h n = 4 \text{ km}$$

Answer

a) $$h = n h' = 4 \text{ m}$$

b) $$h' = h n = 4 \text{ km}$$