Reflection and plane-mirror images

First year measured θᵢ and θᵣ from the normal and stopped. The working form is the vector one. Let k̂ be the unit direction of the incoming ray and n̂ the unit normal at the point the ray actually strikes. Split k̂ into a part along n̂, (k̂⋅n̂)n̂, and a part in the tangent plane. Reflection reverses the first and leaves the second alone: k̂′ = k̂ − 2(k̂⋅n̂)n̂. Try it on a mirror tilted 20° from horizontal, n̂ = (−sin 20°, cos 20°) = (−0.342, 0.940), with a ray k̂ = (0.60, −0.80). Then k̂⋅n̂ = −0.205 − 0.752 = −0.957, so k̂′ = (0.60, −0.80) + 1.914 × (−0.342, 0.940) = (−0.055, 0.999). Since k̂⋅n̂ = −0.957 and k̂′⋅n̂ = +0.957, the incident and reflected rays both make cos⁻¹ 0.957 = 16.9° with the normal line: equal angles were never an extra fact. The word local matters. On a curved mirror n̂ changes from point to point, and this one line, evaluated with the normal at each strike, is the entire content of the ray tracer the rest of the section relies on. No small-angle step has been taken anywhere.

Put a point source S a distance s in front of a plane mirror and pick any point P on the surface. Drop the perpendicular from S to the mirror, meeting it at O, and mark S′ on the far side with OS′ = OS. Triangles SOP and S′OP share OP, have equal legs SO and S′O and right angles at O, so they are congruent and angle S′PO equals angle SPO. So the line from S′ through P is the reflected ray, leaving at the angle SP arrived. Nothing about P was special, so every reflected ray extrapolates back to the same S′. The plane mirror is stigmatic for all rays, which no spherical surface manages. Sign convention: distances are measured from the surface and are positive on the side the light occupies, so s > 0 and s′ = −s. The image is virtual: reflected light diverges from S′ but never visits it, and a screen placed there records nothing. The eye cannot tell a virtual source from a real one and focuses at its own distance from S′, so a person 0.80 m from a mirror focuses at 1.60 m. Limiting case: the spherical-mirror equation 1/s + 1/s′ = 2/R with R → ∞ gives exactly s′ = −s.

Point by point, the image of an extended object is r′ = r − 2[(r − r₀)⋅n̂]n̂. This map is an isometry: it preserves every length and every angle, so m = +1 and nothing is enlarged, shrunk or tilted. Its determinant is −1, the one thing a rotation cannot do: components in the plane of the mirror are copied and only the component along n̂ is reversed. A right-handed set of axes becomes left-handed, a right hand becomes a left hand, and print reads backwards. The construction also tells you how much mirror you need. To see your feet you need a ray from feet to eye, and by the congruent-triangle argument it reflects at the height halfway between them. With eyes 1.63 m up and a crown at 1.75 m, the working strip of glass runs from 0.815 m to 1.69 m above the floor: 0.875 m, exactly half your height, wherever you stand. The general move is to trace from the image, not the object: draw S′, join it to the eye with a straight line, and where that line crosses the mirror is where the reflection happens.

Apply the vector law twice, at mirrors whose planes meet at the dihedral angle α. Each reflection has determinant −1, so the pair has +1: two reflections are a rotation, by 2α about the line where the mirrors meet, for any ray in the plane perpendicular to that line. Mirrors at α = 60°, the first horizontal with n̂₁ = (0, 1), the second with inward normal n̂₂ = (sin 60°, −cos 60°) = (0.866, −0.500). A ray at 30° incidence on the first, k̂ = (−0.500, −0.866), leaves it as (−0.500, 0.866); at the second, k̂⋅n̂₂ = −0.866, so k̂″ = (−0.500, 0.866) + 1.732 × (0.866, −0.500) = (1.000, 0.000). It arrived at 240° from the x axis and leaves at 360°: a turn of 120° = 2α. Repeat at 50° incidence and the turn is again 120°, though the second incidence has changed from 30° to 10°. Set α = 90° and the turn is 180°: two perpendicular mirrors return any in-plane ray antiparallel to itself, and three mutually perpendicular ones return any ray at all, the corner-cube reflector. One mirror with a rotating normal doubles the same way: turn it by β and the ray swings through 2β, the optical lever.

Position maps compose the same way. An object between two mirrors has an image in each; each image is an object for the other mirror; and the chain continues until an image lands behind both mirrors, where neither can see it. In the plane perpendicular to the line of intersection, the object and all its images lie on one circle centred on that line, because reflection in a plane containing the line preserves distance from it. Reflecting alternately in the two mirrors advances a point around the circle in steps of 2α, so when 360°/α is an even integer the points close up after 360°/α steps with the object as one of them: N = 360°/α − 1 images. Two mirrors at 90° give 3, and the middle one, made by two reflections, is the one that is not handedness-reversed; 60° gives 5, a hexagonal ring; 45° gives 7. As α → 0 the count diverges into the corridor between parallel mirrors a distance L apart, successive images in either mirror 2L apart. When 360°/α is odd or non-integral the count depends on where the object sits, and a drawing settles it faster than a formula.

The law assumes a surface with one normal at each point on the scale of a wavelength. Real surfaces have bumps. For wavelets reflected from the top and the bottom of a bump of height h, the extra path is 2h cos θᵢ, so their phase difference is Δφ = 4πh cos θᵢ/λ. Rayleigh's working criterion counts Δφ < π/2 as still specular, giving h < λ/(8 cos θᵢ). For λ = 550 nm at normal incidence that is h < 69 nm: a polished metal mirror passes, paper at roughly 1 μm fails and scatters diffusely, each microscopic facet obeying the law in its own direction. The cos θᵢ is the interesting part. At 85° incidence the allowance grows to 790 nm, so a wet road or a page held nearly edge-on to a lamp turns specular and glares. Two things the model never predicts: how much light returns, and its phase. Both come from the Fresnel treatment of the wave at the boundary, which gives about 4% reflectance for glass at normal incidence and a phase reversal of π for light reflecting off a denser medium. This topic is about direction only.