Ray Optics: Real-World Scenarios (4)

Question

A swimming pool appears 1.5 m deep when viewed from directly above. The refractive index of water is $n = 4/3$. Find the actual depth of the pool. Why does the pool look shallower than it really is?

Solution — Step by Step

Why This Works

The factor $1/n$ comes from a small-angle expansion of Snell’s law. For exact normal incidence ($\theta = 0$), there is no refraction, so the formula must be interpreted as a near-normal approximation — valid when you look almost straight down.

This is why a fish in a pond looks shifted toward the surface, why a coin in a glass of water appears raised, and why divers must compensate when grabbing objects underwater.

Alternative Method

Apply Snell’s law for a small angle of incidence $\theta_w$ from below:

$n\sin\theta_w = \sin\theta_a \approx \theta_a$

For a point on the bottom at horizontal offset $x$, $\tan\theta_a = x/d_{\text{apparent}}$ and $\tan\theta_w = x/d_{\text{real}}$. Setting up the small-angle ratio gives $d_{\text{apparent}} = d_{\text{real}}/n$.

Common Mistake