A scuba diver is $1.2 \mathrm {~m}$ beneath the surface of a still pond of water. At what angle relative to the normal at the surface must the diver shine a beam of light toward the surface in order for a person on a distant bank to be able to see it? The index of refraction for water is $1.33$.
A) $41 ^ { \circ }$
B) $90 ^ { \circ }$
C) $59 ^ { \circ }$
D) $0 ^ { \circ }$
E) $48 ^ { \circ }$

Question

Quizplus · Accepted Answer

To solve this problem, we use Snell's Law, which is $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$, where $n_1$ and $n_2$ are the indices of refraction of the first and second medium respectively, and $\theta_1$ and $\theta_2$ are the angles of incidence and refraction respectively. For a person on the bank to see the light, it must refract at $90^\circ$ relative to the normal upon exiting the water, meaning it travels along the surface. Given water's index of refraction is 1.33 and air's is approximately 1, setting $\theta_2 = 90^\circ$ and $n_1 = 1.33$, we find $\theta_1$ using the critical angle concept, where $\sin(\theta_1) = \frac{n_2}{n_1}$. Substituting $n_2 = 1$ (for air) and $n_1 = 1.33$ (for water), we get $\sin(\theta_1) = \frac{1}{1.33}$. Solving for $\theta_1$ gives an angle of approximately $48^\circ$, which matches choice E.

A Scuba Diver Is 1.2 m1.2 \mathrm {~m}1.2 m Beneath the Surface of a Still Pond of Water

A Scuba Diver Is $1.2 \mathrm {~m}$ Beneath the Surface of a Still Pond of Water