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

1. Setting up: what a polarizer does

WHY does only the parallel component survive? Think of the field vector as having two perpendicular pieces. The polarizer physically lets through one piece and kills the other — like a picket fence passing only vertical rope-wiggles.

If incoming polarized light has amplitude $E_0$ at angle $\theta$ to the transmission axis, the projected (surviving) amplitude is:

$E_{\text{trans}} = E_0\cos\theta$

WHY $\cos\theta$ ? Vector projection: the component of a vector along an axis is its magnitude times the cosine of the angle between them.

2. Malus's Law — derived from scratch

Derivation (HOW):

Intensity is proportional to the square of field amplitude: $I \propto E^2$ . Why? Energy of an EM wave $\propto E^2$ .
Surviving amplitude: $E = E_0\cos\theta$ (projection, step above).
Therefore $I \propto E_0^2\cos^2\theta$ . Writing $I_0 \propto E_0^2$ as the incoming intensity: $I = I_0\cos^2\theta.$

Special case: unpolarized light hits a polarizer

WHY $\tfrac12$ ? Unpolarized light is a random mix of all angles $\theta$ . We average $\cos^2\theta$ over a full circle: $\langle\cos^2\theta\rangle = \frac{1}{2\pi}\int_0^{2\pi}\cos^2\theta\,d\theta = \frac12.$ So exactly half the intensity passes — and the output is now polarized along the axis.

3. Worked examples (Malus)

Worked example Crossed polarizers + a middle one

Two crossed polarizers ( $90^\circ$ apart) pass nothing. Insert a third at $45^\circ$ between them. Input intensity (already polarized along first axis) is $I_0$ .

Step 1 (first→middle, 45°): $I_1 = I_0\cos^2 45^\circ = I_0\cdot\tfrac12 = \tfrac{I_0}{2}$ .
Step 2 (middle→last): last is $90^\circ$ from first, so $45^\circ$ from middle: $I_2 = I_1\cos^2 45^\circ = \tfrac{I_0}{2}\cdot\tfrac12 = \tfrac{I_0}{4}$ .
Answer: $\tfrac{I_0}{4}$ — light reappears! Why does this feel weird? The middle polarizer rotates the polarization in steps, so the final axis isn't fully perpendicular to the surviving field. That's the steel-manned surprise.

4. Brewster's Angle — derived from scratch

Figure — Polarization — Malus's law, Brewster's angle derivation

Derivation (HOW):

Key geometric condition: reflected ⟂ refracted, i.e. $\theta_B + \theta_r = 90^\circ \;\Rightarrow\; \theta_r = 90^\circ - \theta_B.$ Why this condition? The refracted ray drives the electrons' oscillation. Reflected light is radiated by those electrons. A charge radiates nothing along its oscillation axis. The reflected ray vanishes (for the in-plane polarization) precisely when it would point along that oscillation — which happens when reflected ⟂ refracted.
Apply Snell's law $n_1\sin\theta_B = n_2\sin\theta_r$ : $n_1\sin\theta_B = n_2\sin(90^\circ-\theta_B) = n_2\cos\theta_B.$
Divide by $\cos\theta_B$ : $n_1\tan\theta_B = n_2.$

5. Common mistakes (steel-manned)

6. Active recall

Recall Quick self-test (cover the answers)

State Malus's law and explain the square. → $I=I_0\cos^2\theta$ ; intensity $\propto E^2$ .
Why $\tfrac12$ for unpolarized light? → average of $\cos^2\theta$ over all angles.
Brewster condition between reflected & refracted rays? → they're perpendicular.
Derive $\tan\theta_B = n$ . → Snell + $\theta_r = 90-\theta_B$ .

Recall Feynman: explain to a 12-year-old

Light is like a wiggling jump-rope. A polarizer is a fence with vertical slots: only up-down wiggles get through, side-to-side wiggles get stopped. If your wiggle is tilted, only part of it sneaks through — tilt it more, less gets through. Malus's law measures exactly how much. Now, when light bounces off a shiny floor or lake, the bounced light becomes mostly side-to-side wiggling (glare). At one special tilt of looking (Brewster's angle), the bounce is perfectly one-direction — and that's exactly why fishermen's sunglasses kill the shine on water.

7. Connections

Snell's Law and Refraction — Brewster's derivation plugs directly into Snell.
Electromagnetic Waves — $I \propto E^2$ comes from EM energy density.
Reflection and Refraction at Interfaces — Fresnel equations give the full reflectance behind Brewster.
Wave Nature of Light — polarization proves light is transverse.
Unpolarized vs Polarized Light — the averaging that gives the $\tfrac12$ factor.

Malus's law statement

I = I_0\cos^2\theta

for polarized light at angle

\theta

to the transmission axis.

Why is intensity proportional to

\cos^2\theta

not

\cos\theta

Amplitude projects as

E_0\cos\theta

, and intensity

\propto E^2

, so squaring gives

\cos^2\theta

Intensity of unpolarized light after one polarizer

\tfrac12 I_0

(average of

\cos^2\theta

over all angles = 1/2).

Brewster's law

\tan\theta_B = n_2/n_1

; the reflected light is then completely polarized.

Geometric condition at Brewster's angle

Reflected and refracted rays are perpendicular:

\theta_B + \theta_r = 90^\circ

Derive Brewster from Snell

n_1\sin\theta_B = n_2\sin(90-\theta_B)=n_2\cos\theta_B \Rightarrow \tan\theta_B = n_2/n_1

Polarization direction of reflected ray at

\theta_B

Perpendicular to plane of incidence (parallel to surface).

Brewster's angle for glass (n=1.5)

\arctan(1.5) \approx 56.3^\circ

Physical reason reflected light can be 100% polarized

An oscillating charge radiates nothing along its oscillation axis; at

\theta_B

the reflected direction lies along that axis for the in-plane polarization.

Three polarizers (0°,45°,90°), polarized input

I_0

, output?

I_0/4

— middle polarizer rotates polarization so light reappears.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, light ek transverse wave hai — uska electric field travel direction ke perpendicular wiggle karta hai. Iss wiggle ki direction ko hi polarization kehte hain. Normal light unpolarized hoti hai (sab directions mein random). Polarizer ek filter hai jo sirf apne axis ke component ko pass karta hai. Agar light pehle se polarized hai aur axis se angle $\theta$ par aati hai, to sirf $E_0\cos\theta$ amplitude bachti hai. Lekin detector intensity padhta hai, aur intensity $\propto E^2$ , isliye square karke Malus's law banta hai: $I = I_0\cos^2\theta$ . Yaad rakho — square zaroori hai!

Ek important point: agar light unpolarized hai aur pehle polarizer pe aati hai, to half intensity nikalti hai ( $\tfrac12 I_0$ ). Ye half kahaan se aaya? Sab angles ka $\cos^2\theta$ ka average $1/2$ hota hai. Ye rule sirf unpolarized light ke liye pehle polarizer par lagता hai — uske baad har polarizer pe $\cos^2\theta$ use karo.

Ab Brewster's angle: jab light glass ya paani se reflect hoti hai, ek special incidence angle par reflected light 100% polarized ho jaati hai. Iska reason — reflected light glass ke electrons se re-radiate hoti hai, aur oscillating charge apni oscillation axis ke along kuch radiate nahi karta. Ye condition tab milti hai jab reflected aur refracted rays perpendicular ho ( $\theta_B + \theta_r = 90°$ ). Snell's law mein daalo to seedha aata hai $\tan\theta_B = n_2/n_1$ . Isiliye polarized sunglasses water aur road ki glare kaat dete hain — woh glare horizontally polarized hoti hai. Formula yaad rakhne ka trick: "Malus Squares, Brewster Tangles" (tan).

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

1. Setting up: what a polarizer does

2. Malus's Law — derived from scratch

Special case: unpolarized light hits a polarizer

3. Worked examples (Malus)

4. Brewster's Angle — derived from scratch

5. Common mistakes (steel-manned)

6. Active recall

7. Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Optics

Connections