Which way does light bend at a boundary?

Measure both angles from the normal

Reset to air above and water below. The dashed line is perpendicular to the boundary. Measure both angles from that line. With the model indices 1.000 and 1.333, Snell’s law gives sin θ₂ = (1.000/1.333) sin 45°. The result is about 32.04°, so the transmitted ray moves closer to the normal.

Keep the slider at 45° and swap the media. Now the ratio in the equation is 1.333/1.000, and θ₂ grows to about 70.49°. Try 0° in either direction. The ray remains on the normal because both sines are zero.

Find the threshold for total internal reflection

With water above and air below, select Set critical angle. The incidence is about 48.61° and the transmitted direction is 90° from the normal, along the boundary. Substituting sin 90° = 1 into Snell’s law gives θc = arcsin(n₂/n₁), defined here when n₁ > n₂.

Raise the incidence to 50°. Solving Snell’s law would require sin θ₂ to exceed 1, so the model shows total internal reflection. Reduce the angle below 48.61° and a transmitted ray returns. Try the reverse air-to-water setup again. There is no critical angle for this direction.

Check your understanding

Set the upper medium to model glass and the lower medium to air. Predict what the diagram will show at 45°. Check your prediction by finding the critical angle.