Visual walkthrough — Polarization — Malus's law, Brewster's angle derivation

---

Step 1 — What "reflecting off glass" even looks like

WHAT. A ray of light comes down through the air and strikes a flat sheet of glass. Two things happen at once: part of it bounces back into the air (the reflected ray), and part of it bends and continues into the glass (the refracted ray). See Reflection and Refraction at Interfaces.

WHY draw this first. Every symbol we will use is an angle measured on this picture. If we don't fix the picture, the symbols mean nothing.

PICTURE. The dashed vertical line is the normal — the line drawn perpendicular to the surface at the point where the light hits. Every angle in optics is measured from this normal, never from the surface.

---

Step 2 — Why the bounced light is already partly polarized

WHAT. The incoming sunlight is unpolarized — its electric field $\vec{E}$ wiggles in all directions across the ray (see Unpolarized vs Polarized Light). We split each wiggle into two independent pieces:

• s-piece: $\vec{E}$ wiggling perpendicular to the plane of the page (out of the paper, dots ⊙).
• p-piece: $\vec{E}$ wiggling within the plane of the page (arrows ↕ in the plane containing the rays and the normal — the plane of incidence).

WHY split it this way. The two pieces reflect by different rules. Nature treats "wiggle in the plane" and "wiggle out of the plane" differently at a surface — so we must track them separately. The plane of incidence is our natural reference.

PICTURE. On the reflected ray the ⊙ (s) survives strongly; the ↕ (p) is weak. That is why glare off water is already partly polarized. Our whole goal: find the special angle where the p-piece drops to exactly zero.

---

Step 3 — The re-radiation idea: electrons are tiny antennas

WHAT. Light doesn't literally "bounce". The refracted ray, travelling inside the glass, grabs the electrons and shakes them back and forth along the direction of its own $\vec{E}$. A shaking electric charge re-radiates light in all directions — and that re-radiated light is the reflected ray (see Electromagnetic Waves).

WHY this matters. There is one iron rule of a shaking charge:

PICTURE. The plum arrows show the doughnut of radiation: bright to the sides, dark along the axis. That dark direction is the loophole we will exploit.

---

Step 4 — The magic condition: reflected ⟂ refracted

WHAT. The p-piece electrons inside the glass wiggle along the p-direction of the refracted ray. The reflected p-light would have to be radiated by exactly those electrons. By the antenna rule, that reflected p-light vanishes if and only if the reflected ray points along the electrons' oscillation axis.

WHY that means a right angle. The electrons oscillate perpendicular to the refracted ray (an EM wave's $\vec{E}$ is perpendicular to its travel direction). So "reflected ray lies along the oscillation" means the reflected ray is perpendicular to the refracted ray:

$\underset{\begin{array}{c}\text{special incidence}\\ \text{angle}\end{array}}{\underbrace{{\theta }_B}}+\underset{\begin{array}{c}\text{refraction}\\ \text{angle}\end{array}}{\underbrace{{\theta }_r}}=90^{\circ }$

We renamed the incidence angle ${\theta }_i\to {\theta }_B$ (Brewster's angle) because it is now the one special value that makes this happen.

PICTURE. The teal reflected ray and the burnt-orange refracted ray meet at a clean $90^{\circ }$. Only at this incidence angle do they form that right angle.

---

Step 5 — Turn the geometry into an equation about angles

WHAT. From the right-angle condition, solve for ${\theta }_r$:

${\theta }_r=90^{\circ }-{\theta }_B$

WHY. We want to eliminate ${\theta }_r$ later. Snell's law will connect ${\theta }_B$ and ${\theta }_r$; if we can rewrite ${\theta }_r$ purely in terms of ${\theta }_B$, we'll get one clean equation in one unknown.

PICTURE. The two angles ${\theta }_B$ and ${\theta }_r$ are the two non-right angles of the right-angle "wedge" between the rays, so they must add to $90^{\circ }$. The figure shades that wedge.

---

Step 6 — Feed it into Snell's law

WHAT. Snell's law (from Snell's Law and Refraction) relates the two sides of the surface:

$\underset{\begin{array}{c}\text{index of}\\ \text{air}\end{array}}{\underbrace{n_1}}\sin \underset{\text{incidence}}{\underbrace{{\theta }_B}}=\underset{\begin{array}{c}\text{index of}\\ \text{glass}\end{array}}{\underbrace{n_2}}\sin \underset{\text{refraction}}{\underbrace{{\theta }_r}}$

WHY. Snell is the only law that ties an angle in medium 1 to an angle in medium 2. It's the bridge across the surface. The larger $n$, the more the ray bends toward the normal.

Now substitute ${\theta }_r=90^{\circ }-{\theta }_B$ from Step 5:

$n_1\sin {\theta }_B=n_2\sin (90^{\circ }-{\theta }_B)$

Here is the key trick — WHY a cosine appears. The angle $90^{\circ }-{\theta }_B$ is the complement of ${\theta }_B$. On a right triangle, the sine of one acute angle equals the cosine of the other (the "opposite" of one is the "adjacent" of the other). So:

$\sin (90^{\circ }-{\theta }_B)=\cos {\theta }_B$

Giving us:

$n_1\sin {\theta }_B=n_2\cos {\theta }_B$

PICTURE. The figure shows one right triangle where the same pair of sides is "opposite" to ${\theta }_B$ but "adjacent" to $90^{\circ }-{\theta }_B$ — that swap is exactly why $\sin$ turns into $\cos$.

---

Step 7 — Why divide by cosine → why tangent is the right tool

WHAT. We have $n_1\sin {\theta }_B=n_2\cos {\theta }_B$. Divide both sides by $\cos {\theta }_B$:

$n_1\frac{\sin {\theta }_B}{\cos {\theta }_B}=n_2⟹n_1\tan {\theta }_B=n_2$

WHY tangent, and not sine or cosine? We are stuck with both a $\sin$ and a $\cos$ of the same angle. The single tool that combines "sine over cosine" into one clean quantity is the tangent:

$\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{\text{opposite}}{\text{adjacent}}$

It measures the steepness of the angle. So dividing by $\cos {\theta }_B$ isn't a random algebra move — it's the deliberate step that collapses two trig functions into the one function that answers "what angle has this steepness?"

$\boxed{\tan {\theta }_B=\frac{n_2}{n_1}}$

For air → medium ($n_1=1$, $n_2=n$): $\tan {\theta }_B=n$.

PICTURE. A right triangle whose vertical side is $n_2$ and horizontal side is $n_1$: the angle at the base has $\tan =n_2/n_1$, and that base angle is ${\theta }_B$.

---

Step 8 — Plug in numbers, and check the edge cases

Degenerate / limiting cases — the reader must never be surprised:

---

The one-picture summary

Everything on one canvas: the incident ray strikes at ${\theta }_B$; reflected (teal, s-only ⊙) and refracted (orange) leave at a perfect $90^{\circ }$; Snell + the complement identity + the tangent collapse it all into $\tan {\theta }_B=n_2/n_1$.

Recall Feynman retelling — the whole walkthrough in plain words

Light hits a glass floor and splits: some bounces, some sinks in. The part that sinks in shakes the glass's electrons like tiny hands on a jump-rope. Shaking hands throw off new light in every direction — except straight along the shake, where they throw off nothing. That "nothing" direction is our loophole. Tilt the incoming light until the bounce-direction lines up exactly with the shake — now the in-plane wiggle simply can't be bounced, and the reflection comes out cleanly one-directional (pure glare). Geometrically, "bounce lines up with shake" means the bounced ray and the sunk-in ray make a perfect right angle: ${\theta }_B+{\theta }_r=90^{\circ }$. Feed that into Snell's law (the bridge between the two sides), notice that $\sin (90^{\circ }-{\theta }_B)$ is just $\cos {\theta }_B$, divide through, and the two trig pieces fuse into one tangent: $\tan {\theta }_B=n_2/n_1$. Plug in glass ($n=1.5$) → $56.3^{\circ }$. That's the fisherman's-sunglasses angle.

---

Connections

• Snell's Law and Refraction — Step 6 is Snell with the complement substituted.
• Reflection and Refraction at Interfaces — Steps 1–2 set the ray geometry.
• Electromagnetic Waves — the antenna rule of Step 3 is EM radiation.
• Unpolarized vs Polarized Light — why the raw ray has both s and p pieces.
• Wave Nature of Light — the transverse wiggle underpinning it all.