A ray in air (n = 1.00) hits the surface of water (n = 1.33) at an angle of 41.0° to the normal. What is the angle to the normal after the light enters the water?

What you are solving

When light crosses from one medium to another, its direction changes (refraction). The angle is always measured from the normal line (a line perpendicular to the surface).

Apply Snell’s law

Snell’s law is $$n_1\sin\theta_1=n_2\sin\theta_2.$$ Here:

• $n_1=1.00$ (air)
• $n_2=1.33$ (water)
• $\theta_1=41.0^\circ$

Solve for $\theta_2$: $$\sin\theta_2=\frac{n_1}{n_2}\sin\theta_1=\frac{1.00}{1.33}\sin(41.0^\circ).$$ Compute: $$\sin(41.0^\circ)\approx0.6561$$ $$\sin\theta_2\approx\frac{1.00}{1.33}(0.6561)\approx0.4933$$

Convert back to an angle

$$\theta_2=\sin^{-1}(0.4933)\approx29.6^\circ.$$

Since water has a higher refractive index than air, the ray bends toward the normal, so the refracted angle is smaller than $41.0^\circ$, which matches $29.6^\circ$.