The reflected ray is perpendicular to the refracted ray when tan θ1 = n, that is θ1 = arctan n. For your n = 1.33 this is θ1 = 53.061°. For water (n = 1.33) it is 53.1°.
Problem statement
At what angle of incidence θ1 is a shaft of light reflected from the surface of water perpendicular to the refracted shaft?
We assume the light goes from air into water through a flat surface. This means one refracted ray and one reflected ray leave the point of incidence.
We assume the index of air is 1 and the index of the medium is n. This means Snell's law reads sin θ1 / sin θ2 = n.
We assume the angle of reflection equals the angle of incidence. This means the reflected ray also makes θ1 with the normal.
Symbols: θ1 is the angle of incidence, measured from the normal N. θ2 is the angle of refraction, also from the normal. n is the refractive index of the medium. β is the angle between the reflected ray OB and the refracted ray OC.
The big idea in plain words
The reflected ray leaves at θ1 above the surface line, measured from the normal. The refracted ray goes down at θ2 from the normal. Together with the surface, the three angles θ1, β and θ2 fill the straight angle of 180°. So β = 180° − θ1 − θ2. If β must be 90°, then θ1 + θ2 = 90°. That fixes θ2, and Snell's law then gives θ1.
The solution in each frame
Lab frame
The surface is at rest. The full solution is in the next section.
Other frames
A moving surface would need a second frame, because the angles change for a moving mirror or medium. This problem has a fixed surface, so one frame is enough.
The solution step by step
Look at the angles at O on the air side and the medium side. The reflected ray OB makes θ1 with the normal. The refracted ray OC makes θ2 with the normal, on the other side of the surface. The angle between OB and OC is
The rays are perpendicular when β = 90°:
Snell's law for light going from air (index 1) into the medium (index n) is
Put θ2 = 90° − θ1. Since sin(90° − θ1) = cos θ1,
For water, n = 1.33, so θ1 = arctan 1.33 = 53.1°. For glass with n = 1.5 it is 56.3°. For your slider n = 1.33, θ1 = 53.061°. At your θ1 = 40° the refracted angle is 28.901° and the angle between the rays is 111.1°.
A separate check solves β(θ1) = 90° by bisection, using Snell's law at every step. It finds θ1 = 53.0612°, the same as arctan n = 53.061°.
Five methods
1. Force method
Not needed. This is a ray-optics problem about directions, not forces.
2. Energy method
Not needed. The split of energy between the two rays is not used.
3. Graph method
The first graph plots β against θ1. It falls from 180° at θ1 = 0 and crosses 90° at the answer. The third graph plots θ2 against θ1 for Snell's law and the line θ2 = 90° − θ1. They cross at the same θ1.
4. Calculus method
Not needed. The slope of the answer graph, dθ1/dn = 1 / (1 + n²), shows that the angle rises slowly with n: from 45° at n = 1 to 68° at n = 2.5.
5. Check method
Limits: with n = 1 nothing bends, θ2 = θ1, and the rays are perpendicular at 45°, which is arctan 1. As n grows, the angle approaches 90°, because the refracted ray bends toward the normal. Both are correct.
Fastest: tan θ1 = n. Most reliable: the numerical solution of β = 90°.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Refractive index n | Yes | θ2, β, the answer arctan n, the label of the medium and all graphs |
| Angle of incidence θ1 | Not the answer, but the picture | θ2, β, the arcs, the rays, the right-angle mark and the dots on the graphs |
| Wanted β (given-answer mode) | Sets θ1 | The model solves θ1 for the wanted angle between the rays |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Refractive index | n | 1.33 |
| Angle of incidence | θ1 | 40° |
| Angle of refraction | θ2 | 28.901° |
| Angle between reflected and refracted rays | β | 111.1° |
| Answer | arctan n | 53.061° |
Exam traps
- Measuring the angles from the surface instead of from the normal.
- Writing Snell's law upside down: for air to water it is sin θ1 / sin θ2 = n, with n greater than 1.
- Forgetting that the reflected angle equals θ1, so the angle between the rays is 180° − θ1 − θ2, not 90° − θ2.
- Giving tan θ1 = 1/n. That is the answer for light going from the medium into air.
- Calling this the critical angle. Total internal reflection needs light going from the denser medium, which is not the case here.