Law of Reflection

Stand at a still pond at dusk and the far bank hangs upside-down beneath your feet, sharp enough to count the trees. Clap in a canyon and the sound returns a beat later from the rock wall. Both are the same phenomenon — a wave meeting a boundary and turning back into the medium it came from — and both obey one simple rule a Greek geometer wrote down two thousand years ago: the ray leaves at the angle it arrived, on the far side of the normal. That is the law of reflection, the first place the abstract ray of geometrical optics earns its keep. The law has two clauses, and it pays to keep them separate. The angular clause says the angle of reflection equals the angle of incidence, both measured from the surface normal — never from the surface itself. The coplanarity clause says the incident ray, the reflected ray, and the normal all lie in a single plane, the plane of incidence; reflection never swings a ray sideways out of it, because nothing in the geometry prefers left to right. Together they fix the outgoing ray completely once you know the incoming ray and the local normal. What makes the law worth a whole article is how much falls out of it. Reflect the ray about the normal and you have the vector formula a graphics engine runs billions of times a second. Meet two mirrors at a right angle and any ray comes straight back — the trick behind the reflectors on the Moon and on your bicycle. Apply it at every point of a flat mirror and a virtual twin of the world appears an equal distance behind the glass, which is why the smallest mirror that shows your whole body is exactly half your height, whatever your distance. One equal-angle rule, an entire wing of optics.

Key takeaways

The law of reflection has two parts: the reflected ray lies in the plane of incidence, and the angle of reflection equals the angle of incidence, $\theta_r=\theta_i$ , both measured from the surface normal.
It is one of the two founding rules of geometrical optics; Snell's law of refraction is the other, and both share the "measure from the normal" convention so they combine cleanly at an interface.
It is a theorem, not a postulate: Hero's shortest path, Fermat's least time, Huygens' wavelets, and Maxwell's boundary conditions all deliver $\theta_r=\theta_i$ .
In vector form it is one line, $\hat{d}_r=\hat{d}_i-2(\hat{d}_i\cdot\hat{n})\hat{n}$ — keep the along-surface part, flip the along-normal part — powering ray tracing, corner-cube retroreflectors, and flat-mirror imaging.
Specular (mirror-like) reflection needs surface roughness far below the wavelength; a rough page reflects diffusely, yet the same law holds locally on every microfacet.
Key idea: reflection reverses only the component of the ray along the normal and leaves the tangential component untouched — one operation that encodes equal angles, coplanarity, the antiparallel corner cube, and the equal-distance virtual image all at once.

The core insight

Reflection reverses the normal component of a ray and preserves its tangential component: $\hat{d}_r=\hat{d}_i-2(\hat{d}_i\cdot\hat{n})\hat{n}$ . That single geometric move contains both clauses of the law — equal angles because the normal component merely flips sign, and coplanarity because no new direction is ever introduced. Everything else (the antiparallel corner reflector, the virtual image an equal distance behind a flat mirror, the half-height mirror) is this one operation applied again or applied to position instead of direction.

Where this lives today

The vector law is one of the busiest lines of code anywhere. Every rendered film frame and video-game scene traces rays that bounce by $\hat{d}_r=\hat{d}_i-2(\hat{d}_i\cdot\hat{n})\hat{n}$ , and the specular/diffuse split is what makes a rendered chrome sphere read as metal and a wall as plaster. The same formula on real photons aligns laser cavities and steers a galvanometer mirror: because a mirror turned by $\alpha$ swings the beam by $2\alpha$ , a tiny actuated mirror sweeps a laser across a barcode or a LIDAR scene. The corner-cube result — orthogonal mirrors returning any ray antiparallel — is everywhere. Corner cubes stud road signs, number plates, and bicycle pedals so headlights bounce back to the driver; surveyors' total stations range to a corner-cube prism to the millimetre; and the arrays that Apollo astronauts and the Soviet Lunokhod rovers left on the Moon still return laser pulses fired from Earth, letting Lunar Laser Ranging track the Earth–Moon distance to a few centimetres. At the opposite wavelength extreme, grazing-incidence reflection ( $\theta_i\to90^\circ$ , where even X-rays reflect efficiently) is how Wolter telescopes focus X-rays and how synchrotron beamlines bend hard radiation — the same equal-angle law with the normal nearly in the surface.

The physical picture

Picture a billiard ball rolling into a cushion with no spin. The velocity along the cushion sails through untouched; the part aimed into it reverses. Light does exactly this at a mirror — tangential direction survives, normal direction flips — and that single asymmetry forces the outgoing angle to match the incoming one. The ray's speed and wavelength do not change; it stays in the same medium, and only its direction folds across the normal. Two features of the fold are worth naming. First, all angles live off the normal, the line perpendicular to the surface — a ray skimming almost flat has $\theta_i$ near $90^\circ$ , not near $0^\circ$ . Second, the reflected ray cannot leave the plane of the incident ray and the normal; nothing in the geometry tips it to one side, so coplanarity is as much a symmetry statement as the equal-angle rule is a metric one.

The picture also says what reflection is not. A good mirror sends a parallel bundle to a single parallel bundle and forms an image; a wall does not, because its microscopic facets tilt every which way and fan the rays out — diffuse reflection, though each facet still obeys $\theta_r=\theta_i$ . The dividing line is roughness versus wavelength: the same aluminium, ground rough or polished below about $\lambda/8$ , is wall or mirror depending only on that comparison. Now the punchline the ancients found first. Why that angle? Hero of Alexandria answered with a ruler and no calculus: of all the broken paths that leave $A$ , touch the mirror, and reach $B$ , light takes the shortest. The trick is the mirror image. Reflect $B$ through the mirror to a phantom point $B'$ the same distance behind. Any $A\!\to\!P\!\to\!B$ is exactly as long as $A\!\to\!P\!\to\!B'$ , since $PB=PB'$ for every foot point $P$ . So the shortest is the shortest $A\!\to\!P\!\to\!B'$ — the straight line — which crosses the mirror precisely where the two angles are equal.

Figure

Hero's shortest path — reflect B to its image B′; the straight line A→B′ crosses the mirror at the reflection point, where the path A-P-B is shortest and θi = θr = 45°

The right-hand curve is Hero's argument as arithmetic for the concrete geometry $A=(0,2)$ , $B=(6,4)$ , mirror on the $x$ -axis. Slide the foot point $P$ and the path $A\!\to\!P\!\to\!B$ traces a shallow valley whose floor sits at $x=2$ , where both segments make $45^\circ$ with the normal and the length bottoms out at $6\sqrt{2}\approx 8.49$ . The equal-angle law is not a rule bolted onto optics — it is where a length is stationary, the same stationarity that reappears as Fermat's principle and, deeper still, as the phase-matching of Maxwell's waves along the interface.

The core equations

The law has two clauses, and both concern a single plane. Erect the surface normal $\hat{n}$ at the point where the ray strikes, and agree to measure every angle from that normal, never from the surface. The angular and coplanarity clauses then read together as:

\theta_r=\theta_i, \qquad \hat{d}_i,\ \hat{n},\ \hat{d}_r\ \text{coplanar} \qquad (1)

where: $\theta_i$ — angle of incidence, between incident ray and normal; $\theta_r$ — angle of reflection, between reflected ray and normal; "coplanar" — the incident ray, reflected ray, and normal all lie in one plane, the plane of incidence. Physically, $(1)$ folds a ray across the normal like a hinge: the fold is symmetric ( $\theta_r=\theta_i$ ) and never tips the ray out of the hinge plane. Angles are awkward to compute with; a single vector formula carries both clauses at once and runs inside every ray tracer. Split the incident direction into the piece along the surface and the piece along the normal; reflection keeps the first and reverses the second, and collecting terms gives:

\hat{d}_r=\hat{d}_i-2(\hat{d}_i\cdot\hat{n})\,\hat{n} \qquad (2)

where: $\hat{d}_i$ — unit vector along the incident ray; $\hat{d}_r$ — unit vector along the reflected ray; $\hat{n}$ — unit surface normal; $(\hat{d}_i\cdot\hat{n})$ — the signed normal component that gets doubled and subtracted. This is the billiard-ball bounce in symbols. Because only a sign flips, $|\hat{d}_r|=1$ automatically — reflection is an isometry, and equal angles are baked in. Written as an operator on any direction vector, that same reflection is the Householder matrix, and this form explains the corner reflector. Composing two reflections multiplies their matrices, and when the mirror normals are perpendicular the product is astonishingly simple:

\mathbf{R}=\mathbf{I}-2\,\hat{n}\hat{n}^{\mathsf{T}}, \qquad \mathbf{R}_2\mathbf{R}_1=-\mathbf{I}\ \ (\hat{n}_1\perp\hat{n}_2) \qquad (3)

where: $\mathbf{R}$ — the $3\times3$ reflection operator for a mirror with normal $\hat{n}$ ; $\mathbf{I}$ — identity; $\hat{n}\hat{n}^{\mathsf{T}}$ — outer product projecting onto the normal; $\mathbf{R}_2\mathbf{R}_1$ — mirror 1 then mirror 2. When $\hat{n}_1\perp\hat{n}_2$ the product is $-\mathbf{I}$ : every ray exits exactly antiparallel to how it entered, whatever the arrival angle. Add a third perpendicular mirror and $-\mathbf{I}$ holds in full 3D — the corner cube. Finally, turn the law loose on an extended flat mirror. Diverging rays from an object point $O$ , each reflected by $(2)$ , emerge as if from a single point directly behind the mirror — a virtual image — whose location follows from applying the same reflection to position instead of direction:

s_i=s_o \qquad (\text{image as far behind as the object is in front}) \qquad (4)

where: $s_o$ — object distance in front; $s_i$ — image distance behind. The image is virtual: no light converges there, but the backward extensions of the reflected rays cross there, so the eye reads an undistorted twin an equal distance beyond the glass.

Derivation AThe law of reflection from Hero's shortest path

We want to prove $\theta_i=\theta_r$ rather than assume it, starting from the oldest extremal principle in physics: of all the bent paths from a fixed point $A$ down to a mirror and back up to a fixed point $B$ , light takes the shortest. Because the medium is uniform, geometric length is proportional to optical path length, so "shortest path" and "stationary optical path" (Fermat's principle) coincide here. Write the length as a function of one variable — the foot point where the ray touches the mirror — set its derivative to zero, and recognise the two terms that appear as the sines of the two angles.

Step 1. Put the mirror on the $x$ -axis and place $A=(0,a)$ and $B=(d,b)$ above it, both on the same side. Let the ray touch the mirror at the free point $P=(x,0)$ . The total length is $L(x)=\sqrt{x^2+a^2}+\sqrt{(d-x)^2+b^2}$ . Why: the only freedom left, once $A$ and $B$ are fixed, is the foot point $x$ ; the whole physical content is in how $L$ depends on it.

Step 2. Impose stationarity, $\mathrm{d}L/\mathrm{d}x=0$ , giving $\dfrac{x}{\sqrt{x^2+a^2}}=\dfrac{d-x}{\sqrt{(d-x)^2+b^2}}$ . Why: Hero's shortest-path (equivalently Fermat's least-time) principle selects exactly the foot point at which the length stops changing to first order.

Step 3. Read the two sides geometrically. The left fraction is the horizontal run of the incoming segment over its length — that is $\sin\theta_i$ , the angle from the vertical normal; the right fraction is $\sin\theta_r$ . Stationarity therefore says $\sin\theta_i=\sin\theta_r$ . Why: each fraction is (opposite over hypotenuse) for the triangle whose vertical leg lies along the normal, i.e. the sine of the angle from the normal.

Step 4. With both angles acute, equal sines force $\theta_i=\theta_r$ . The image method makes the same point without calculus: reflect $B$ to $B'$ behind the mirror, note $PB=PB'$ so $A\!\to\!P\!\to\!B$ and $A\!\to\!P\!\to\!B'$ are equally long, and the shortest $A\!\to\!P\!\to\!B'$ is the straight line $AB'$ — which crosses the mirror at the equal-angle point. Coplanarity comes free: any sideways detour only lengthens $L$ . Why: the mirror-image trick converts a bent-path minimisation into a straight-line one, exposing the reflection point as a unique minimum. *Verified in verify_law-of-reflection.py: $\mathrm{d}L/\mathrm{d}x$ equals $\sin\theta_i-\sin\theta_r$ symbolically, the concrete geometry $A=(0,2),B=(6,4)$ gives the stationary foot point $x^\ast=2$ with $\theta_i=\theta_r=45^\circ$ , the image-method line crosses at the same $x^\ast$ , and $L''>0$ confirms a true minimum of $6\sqrt2\approx8.49$ .*

Go deeper · Derivation B · 2 minThe vector law, the virtual image, and why a mirror flips depth not sides

Derivation BThe vector law, the virtual image, and why a mirror flips depth not sides

Here we build all of flat-mirror reflection from one algebraic move — reverse the normal component of a vector — used twice: on the ray direction to get the vector law, on position to locate the image. The pay-off is the oldest puzzle about mirrors: why your reflection seems to swap left and right but not up and down. The algebra makes the answer unavoidable — a mirror swaps neither; it inverts only the axis pointing out of the glass, and our habit of imagining ourselves turned around does the rest.

Step 1. Resolve the incident direction into tangential and normal pieces, $\hat{d}_i=\hat{d}_{i\parallel}+(\hat{d}_i\cdot\hat{n})\hat{n}$ . Reflection keeps the tangential and reverses the normal, $\hat{d}_r=\hat{d}_{i\parallel}-(\hat{d}_i\cdot\hat{n})\hat{n}$ ; eliminating $\hat{d}_{i\parallel}$ gives $\hat{d}_r=\hat{d}_i-2(\hat{d}_i\cdot\hat{n})\hat{n}$ . Why: "keep along-surface, flip into-surface" is the entire physics of a mirror, and $2(\hat{d}_i\cdot\hat{n})\hat{n}$ is exactly twice the normal component removed and reversed.

Step 2. Check it is a proper reflection. Length is preserved, $|\hat{d}_r|^2=|\hat{d}_i|^2-4(\hat{d}_i\cdot\hat{n})^2+4(\hat{d}_i\cdot\hat{n})^2=1$ , and the normal component merely changes sign, $\hat{d}_r\cdot\hat{n}=-\hat{d}_i\cdot\hat{n}$ , so the angle from the normal is unchanged: $\theta_r=\theta_i$ . Why: a genuine reflection must be an isometry and must equalise the angles — both are now consequences, not assumptions.

Step 3. Apply the identical operation to position. With the mirror as the plane through the origin with normal $\hat{n}$ , an object point $\mathbf{r}_O$ maps to $\mathbf{r}_I=\mathbf{r}_O-2(\mathbf{r}_O\cdot\hat{n})\hat{n}$ , the same perpendicular distance on the far side. So $s_i=s_o$ : the virtual image is as far behind as the object is in front. Why: the backward extensions of every reflected ray from $O$ intersect at this mirrored point, so the eye sees an undistorted twin at equal depth.

Step 4. Now the famous puzzle. With the mirror in the $xy$ -plane and depth along $z$ , the image map is $(x,y,z)\mapsto(x,y,-z)$ : left–right and up–down are untouched, only depth is inverted. Your right hand maps directly across from it, on the same side — it only reads as a left hand because comparing the two figures means mentally rotating one by $180^\circ$ about the vertical, and that rotation, not the mirror, trades left for right. Why: the mirror anticommutes only with the outward normal, so "left–right reversal" is an artefact of the observer's imagined turn, not of the reflection. *Verified in verify_law-of-reflection.py: $|\hat{d}_r|=1$ and $\hat{d}_r\cdot\hat{n}=-\hat{d}_i\cdot\hat{n}$ symbolically; a $30^\circ$ ray reflects to $30^\circ$ numerically; the position map sends an object at $(D,h)$ to $(-D,h)$ so $s_i=s_o$ .*

Limiting cases

The equal-angle law is exact for a smooth interface, but its visible consequences change sharply with geometry and surface finish.

Case	Result	Meaning
Normal incidence	$\theta_i=0^\circ\Rightarrow\theta_r=0^\circ$	The ray retraces its path straight back; used in autocollimators and cavity end mirrors.
Grazing incidence	$\theta_i\to 90^\circ$	Reflectance climbs toward unity even for weak reflectors — the basis of grazing-incidence X-ray mirrors.
Specular vs diffuse	$\sigma_{\mathrm{rms}}\ll\lambda$	Mirror-like reflection needs roughness far below $\lambda$ ; a rougher surface scatters diffusely, though the law still holds facet by facet.
Two mirrors at $90^\circ$	$\hat{d}_r=-\hat{d}_i$	The composed reflection is $-\mathbf{I}$ in the plane; any ray returns antiparallel, independent of orientation (corner reflector).
Flat mirror, extended	$s_i=s_o$	Every object point maps to a virtual twin an equal distance behind — upright, same size, depth-reversed.

Common trap“A mirror reverses left and right”

It does not — and the belief hides the real geometry. A flat mirror in the $xy$ -plane performs the map $(x,y,z)\mapsto(x,y,-z)$ : it inverts only depth, the axis pointing out of the glass, and leaves left–right and up–down alone. What tempts everyone into "left–right reversal" is the unconscious step of imagining yourself turned around to face where the image faces; that $180^\circ$ turn about the vertical swaps left and right, and the mirror gets the blame. The tell: a mirror treats horizontal and vertical identically, so if it truly flipped left–right it would flip up–down too and stand your reflection on its head. It doesn't. Text and clocks look reversed only because you rotate them to hold them up. The law reverses the normal component, nothing else.

Worked examples

1 · The corner reflector — any ray sent straight back (confidence builder)

Two mirrors meet at a right angle: a horizontal floor (normal $\hat{n}_{\text{floor}}=(0,1)$ ) and a vertical wall (normal $\hat{n}_{\text{wall}}=(1,0)$ ). A ray enters with direction $\hat{d}=(-0.6,-0.8)$ — a 3-4-5 direction heading down and to the left. Find the two angles of incidence and the exit direction.

Step 1. Going downward, the ray meets the floor first. Its angle from the floor normal satisfies $\cos\theta_1=|\hat{d}\cdot\hat{n}_{\text{floor}}|=0.8$ , so $\theta_1=\arccos(0.8)=\mathbf{36.87^\circ}$ . The floor flips the $y$ -component: $\hat{d}\to(-0.6,+0.8)$ .

Step 2. Now travelling up and to the left, the ray meets the wall. Its angle from the wall normal satisfies $\cos\theta_2=0.6$ , so $\theta_2=\arccos(0.6)=\mathbf{53.13^\circ}$ . The wall flips the $x$ -component: $(-0.6,0.8)\to(0.6,0.8)$ .

Step 3. The exit direction is $(0.6,0.8)=-\hat{d}$ : the ray leaves exactly antiparallel to how it arrived. The net turn is $(180^\circ-2\theta_1)+(180^\circ-2\theta_2)=106.26^\circ+73.74^\circ=\mathbf{180^\circ}$ .

Step 4. Nothing in that total used the specific incidence angle: for any orthogonal pair the two angles obey $\theta_1+\theta_2=90^\circ$ , so the deviation is always $360^\circ-2(\theta_1+\theta_2)=180^\circ$ . This is why retroreflectors on the Moon and on every road sign throw light back to its source however they are tilted — the vector statement $\mathbf{R}_2\mathbf{R}_1=-\mathbf{I}$ , extended to full 3D by a third perpendicular mirror.

2 · The smallest mirror that shows your whole body (exam level)

You are $H=1.80\ \mathrm{m}$ tall with your eyes $0.10\ \mathrm{m}$ below the top of your head, so eye height $E=1.70\ \mathrm{m}$ . What is the shortest vertical mirror that lets you see yourself head to toe, where must it hang, and does the answer change if you step back?

Step 1. To see the top of your head, a ray must leave the crown, reflect, and reach your eye. By the equal-angle law the reflection point sits midway between crown and eye: $y_{\text{top}}=(H+E)/2=(1.80+1.70)/2=1.75\ \mathrm{m}$ . (Equivalently, the crown's image lies an equal distance behind the mirror, so the eye-to-image line crosses the glass at half height.)

Step 2. To see your feet, the reflection point sits midway between the floor and the eye: $y_{\text{feet}}=E/2=1.70/2=0.85\ \mathrm{m}$ .

Step 3. The mirror need only span $0.85\ \mathrm{m}$ to $1.75\ \mathrm{m}$ , a height of $1.75-0.85=\mathbf{0.90\ \mathrm{m}}$ — exactly $H/2$ . Top edge at 1.75 m (half the crown-to-eye gap above eye level), bottom edge at 0.85 m (half your eye height).

Step 4. Both midpoint heights carry a factor $\tfrac12$ that has nothing to do with how far you stand: the distance $D$ cancels because the eye-to-image crossing is always at the halfway point. Step forward, step back, press your nose to the glass — the minimum mirror is half your height, $0.90\ \mathrm{m}$ , independent of distance — one of the cleanest surprises in elementary optics, straight out of $\theta_i=\theta_r$ .

Physics of the portrait

Long before anyone could measure a refractive index, reflection was the one optical fact simple enough to be turned into pure geometry — and the person who did it set the template for treating light with theorems.

Origins · 1 minEuclid and how this was discovered

Euclid

fl. c. 300 BCE · Alexandria

Euclid, working at Alexandria around 300 BCE and best known for the Elements, also wrote the Optica, the earliest surviving treatise to treat vision with straight geometric rays. The companion Catoptrics (traditionally attributed to Euclid, on mirrors) states the law of reflection as a geometric proposition — the visual ray rebounds making equal angles — and applies it to plane and curved mirrors. Whether every proposition is Euclid's own hand or partly a later compilation in his school, the achievement is the same: reflection became a deductive subject, the equal-angle rule an axiom-like starting point from which image positions could be proved. That move — from "mirrors do this" to "here is the theorem" — is the seed of geometrical optics, and Hero, Ptolemy, Ibn al-Haytham, and Fermat all built on the ground Euclid cleared.

History

300 BCE

— Euclid's Catoptrics states the equal-angle law of reflection as a geometric proposition.

CE — Hero of Alexandria proves the reflected ray follows the shortest path, the first extremal principle in physics.

1021

— Ibn al-Haytham studies reflection from spherical and parabolic mirrors experimentally in the Book of Optics.

1662

— Fermat recasts reflection (and refraction) as paths of least time, giving the law a variational foundation.

Going further

References. Hecht, Optics, 5th ed., §4.2–4.4 (reflection, the plane mirror, and Fermat's route) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., Ch. 1 (ray optics and the vector/Householder form) · Born & Wolf, Principles of Optics, 7th ed., §1.5 (foundations, Huygens construction) · Feynman, Lectures on Physics, Vol. I, Ch. 26 (Hero's shortest path and least time).

Provenance. The angular and coplanarity statements follow Hecht §4.2; the vector and Householder forms are standard in Saleh & Teich Ch. 1. The Fermat/Hero derivation is verified by differentiating $L(x)$ and identifying the two sines as $\sin\theta_i,\sin\theta_r$ (Hecht §4.2), with the mirror-image construction giving the same reflection point. The half-height plane-mirror result and the corner-cube $-\mathbf{I}$ composition are elementary consequences of $\theta_i=\theta_r$ . A verifier should recompute $\mathrm{d}L/\mathrm{d}x=0\Rightarrow\sin\theta_i=\sin\theta_r$ , confirm $|\hat{d}_r|=1$ and $\hat{d}_r\cdot\hat{n}=-\hat{d}_i\cdot\hat{n}$ , check $\mathbf{R}_2\mathbf{R}_1=-\mathbf{I}$ for orthogonal normals, and obtain the $0.90\ \mathrm{m}$ minimum mirror independent of distance.

Builds on. Ray Model of Light. Leads to. Snell's Law of Refraction · Plane and Spherical Mirrors · Fermat's Principle of Least Time · Total Internal Reflection and Critical Angle.

Does a mirror really reverse left and right, or not?

Not left and right — front and back. A vertical mirror inverts only the depth axis pointing out of the glass, $(x,y,z)\mapsto(x,y,-z)$ , leaving horizontal and vertical exactly as they are. Your right hand images straight across from itself, on the same side of the room; it looks like a left hand only because comparing "you" with "the image" means imagining yourself turned around to face the way the image faces, and that half-turn about the vertical is what swaps left for right — the mirror did no such thing.

Why does a sheet of paper obey the law of reflection when it clearly doesn't act like a mirror?

It obeys the law at every point — it just isn't smooth. Under a microscope the paper is a jumble of fibres, and each tiny facet reflects specularly about its own local normal. Because those normals point every which way, the reflected rays leave in a broad fan (diffuse reflection) and no image forms; the surface looks equally bright from all sides. A mirror is the same physics with one difference: its roughness is far below the wavelength ( $\sigma_{\mathrm{rms}}\ll\lambda$ ), so all facets share one normal and the bundle stays together. Diffuse reflection does not break the law; it is the law, summed over a rough surface.

Is there really an image behind the mirror? What's back there?

Nothing optical is back there — the image is virtual. No light travels behind the glass or converges at the image point. The rays reflecting off the front surface diverge exactly as if they had come from a point an equal distance behind, so traced backwards they appear to meet there, and your eye — assuming straight-line propagation — places the source at that intersection. That is why you cannot catch a flat-mirror image on a screen behind the mirror, and why it sits precisely $s_i=s_o$ behind: a geometric extrapolation, not a real gathering of light.

Why measure the angles from the normal and not from the surface?

Because the normal is the one direction defined at every surface, flat or curved, and it is the axis reflection folds the ray across. Measuring from the surface is ambiguous on a curved mirror and breaks the clean $\theta_r=\theta_i$ . It also matches Snell's law, so reflection and refraction share one reference at any interface. The practical trap: a ray "grazing" a mirror is at $\theta_i\approx90^\circ$ , not $0^\circ$ ; a head-on ray is at $\theta_i=0$ . Read the angle from the wrong line and every reflection calculation inverts.

Why does turning a mirror by a small angle swing the beam by twice as much?

Because rotating the mirror rotates its normal by the same amount. Turn the mirror by $\alpha$ and the normal turns by $\alpha$ ; the incident ray is fixed, so its angle to the normal grows by $\alpha$ , and the equal-angle law puts the reflected ray a further $\alpha$ over on the other side — a total swing of $2\alpha$ . This "optical lever" is why a mirror on a galvanometer or a torsion fibre amplifies a tiny rotation into a large, easily read beam deflection, underlying laser scanners, oscillograph mirrors, and Cavendish-style experiments.

Does reflection change the light's colour, speed, or wavelength?

No. The reflected wave stays in the same medium, so its speed, frequency, and wavelength are unchanged — only the direction is folded across the normal. This is why the angle argument is so clean: with $|\vec{k}_r|=|\vec{k}_i|$ , matching the phase along the interface forces equal tangential wavevectors, $k\sin\theta_i=k\sin\theta_r$ , hence $\theta_i=\theta_r$ . (Reflection can shift the wave's phase — famously by $180^\circ$ off a denser medium — and alter its polarisation, but those are amplitude effects handled by the Fresnel equations, not by the geometric law.)