When light crosses into a material with a different index of refraction at an angle, it bends. Snell's law gives the exact angle. You will undo a sine with inverse trig to find it.
Key idea
$$n_1 \sin\theta_1 = n_2 \sin\theta_2,$$ with both angles measured from the normal. Going into a higher index, light bends toward the normal; going into a lower index, it bends away. Light that hits head-on ($\theta_1 = 0$) goes straight through.
Why it works
Picture a row of marchers crossing from pavement into mud at a slant. The end that reaches the mud first slows down first, so the row swings around. Light's wavefronts do the same. In one period, the wavefront moves $\lambda_1$ in material 1 and $\lambda_2$ in material 2, and the two pieces share the same stretch $s$ of the boundary: $\lambda_1 = s\sin\theta_1$ and $\lambda_2 = s\sin\theta_2$. Divide: $\dfrac{\sin\theta_1}{\sin\theta_2} = \dfrac{\lambda_1}{\lambda_2} = \dfrac{v_1}{v_2} = \dfrac{n_2}{n_1}$, which is Snell's law.
Common mistake
Measuring angles from the surface instead of from the normal. If a problem says the ray makes $25^\circ$ with the surface, the angle of incidence is $65^\circ$. Quick check: going into a denser material, your refracted angle (from the normal) must come out smaller than the incident angle.
Worked example: Air into water
Light in air hits water ($n = 1.33$) at $40^\circ$ from the normal. Find the angle of refraction.
- $(1.00)\sin 40^\circ = 1.33\sin\theta_2$.
- $\sin\theta_2 = \dfrac{0.643}{1.33} = 0.483$.
- $\theta_2 = \sin^{-1}(0.483) = 28.9^\circ$. Smaller than $40^\circ$: it bent toward the normal.
Answer: $28.9^\circ$
Worked example: Glass into water
Light in glass ($n = 1.50$) hits a boundary with water ($n = 1.33$) at $30^\circ$. Find the angle in the water.
- $1.50\sin 30^\circ = 1.33\sin\theta_2$, so $\sin\theta_2 = \dfrac{0.750}{1.33} = 0.564$.
- $\theta_2 = 34.3^\circ$. Larger: going into a lower index, it bent away from the normal.
Answer: $34.3^\circ$
Worked example: Measuring an index
A ray enters a clear plastic from air at $50^\circ$ and travels at $31^\circ$ inside. Find the plastic's index.
- $n_2 = \dfrac{n_1\sin\theta_1}{\sin\theta_2} = \dfrac{\sin 50^\circ}{\sin 31^\circ}$.
- $n_2 = \dfrac{0.766}{0.515} = 1.49$.
Answer: $n \approx 1.49$