The law of reflection in vector form is e' = e − 2 n (e · n). For your sliders e' = (0.766 i + 0.643 j) and the beam turns by 80.01°.
Problem statement
Write the law of reflection of a light beam from a mirror in vector form, using the directing unit vectors e and e' of the incident and reflected beams and the unit vector n of the outside normal to the mirror surface.
We assume the mirror is flat and smooth. This means all the light of a narrow beam turns the same way, so one direction describes the beam.
We assume the beam is a straight ray. This means we follow only its direction, as a unit vector.
We assume n points out of the mirror on the side of the incoming beam. This means e · n is negative for a beam that hits the front of the mirror.
Symbols: e is the unit vector along the incident beam. e' is the unit vector along the reflected beam. n is the unit vector of the outside normal, which is perpendicular to the mirror. e · n is the dot product, the part of e along n. θi is the angle of incidence and θr the angle of reflection, both measured from the normal.
The big idea in plain words
A vector can be split into two parts at right angles: one along the normal and one along the mirror. The mirror cannot push the beam sideways, because the surface is smooth. So the part along the mirror stays as it is. The mirror turns the beam back from the surface, so the part along the normal reverses. Put the two parts together and you get the reflected beam. This short rule is the whole law, and it also contains the equal angles.
The solution in each frame
Lab frame
The mirror is at rest. The full solution is in the next section.
Other frames
A moving mirror would need a second frame, because the reflected frequency and direction change for a moving mirror. This problem has a fixed mirror, so one frame is enough.
The solution step by step
Split the unit vector e into a part along n and a part along the mirror:
The part e⊥ is the projection of e on n. The dot product e · n is the length of that part, with a sign, and the unit vector n gives it a direction.
The law of reflection says two things: the angle of reflection equals the angle of incidence, and the incident beam, the reflected beam and the normal lie in one plane. In terms of the parts, the part along the mirror does not change, and the part along the normal reverses:
So
Check the length. Since e∥ and e⊥ are at right angles, |e'|² = |e∥|² + |e⊥|² = |e|² = 1. The vector e' is a unit vector, as it must be.
Check the angles. Take the dot product with n: e' · n = e · n − 2 (e · n)(n · n) = − (e · n). The cosine of the angle with the normal has only changed sign, so θr = θi.
For your sliders e = (0.766 i − 0.643 j), n = (0.000 i + 1.000 j), e · n = -0.6428, so e' = (0.766 i + 0.643 j). Here θi = 50° and θr = 50°.
For any beam and any mirror tilt, the page finds |e'| = 1 and θr = θi to five digits. The reflected direction from the components also matches γ' = 2ν − γ, the rule for a mirror tilted by ν.
Five methods
1. Force method
This one fits. A smooth mirror pushes on the light only along the normal, so only the part of the momentum along n can change. The part along the mirror stays the same. The change along n is a reversal, so the momentum changes by −2 n (n · p).
2. Energy method
The mirror does not change the energy of the light, so the length of the vector stays 1. A change that reverses the part along n and keeps the part along the mirror does keep the length.
3. Graph method
The third graph plots the part of e along n, which is −cos θi, and the part of e' along n, which is +cos θi, against θi. The two curves are mirror images. The grey dashed curve is the part along the mirror, sin θi, the same for both beams.
4. Calculus method
Not needed. This is a result in vector algebra. The slope of the first graph, 2, shows a useful fact: rotating the mirror by ν turns the reflected beam by 2ν.
5. Check method
Limits: for normal incidence e = −n, so e' = −n − 2 n (−1) = n. The beam returns the way it came. For grazing incidence e · n is nearly 0, so e' is nearly e. Both are correct.
Fastest: the formula e' = e − 2 n (e · n). Most reliable: the length and angle checks in the step-by-step section.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| x in e = x i + y j | Yes | x and y on the incident ray, e · n, the angle of incidence, e' on the reflected ray and the picture |
| Mirror tilt ν | Yes | The normal n, e · n, e', and the reflected direction by 2ν |
| Wanted reflected direction (given-answer mode) | Sets the answer | Solves the mirror tilt |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Beam direction | γ | -40° |
| Mirror tilt | ν | 0° |
| Dot product | e · n | -0.6428 |
| Angle of incidence | θi | 50° |
| Angle of reflection | θr | 50° |
| Reflected direction | γ' | 40° |
| Turn of the beam | e to e' | 80.01° |
Exam traps
- Writing e' = e − n (e · n). That removes the normal part but does not reverse it. The factor is 2.
- Using a normal n that is not a unit vector. The formula needs n · n = 1.
- Mixing up the sign convention. If n points the other way, e · n changes sign, but e' = e − 2 n (e · n) gives the same e'.
- Measuring the angles from the mirror instead of from the normal.
- Forgetting that e, e' and n are all in one plane.
- Thinking a tilt ν turns the reflected beam by ν. It turns by 2ν.