Polarization

A transverse wave travelling along $z$ has its displacement confined to the $xy$-plane, perpendicular to the propagation direction. Write the two perpendicular components as SHMs of the same frequency and amplitude parameters $E_{0x}, E_{0y}$, differing only by a phase $\delta$:

where: $E_{0x}, E_{0y}$ are the component amplitudes, $k$ is the wavenumber, $\omega$ is the angular frequency, and $\delta$ is the phase difference between the two components. As $t$ advances, the resultant vector $\vec E = E_x\hat x + E_y\hat y$ sweeps out a curve in the $xy$-plane, and the shape of that curve names the polarization state.

• Linear polarization ($\delta = 0$ or $\pi$): the two components rise and fall in lockstep (or exactly opposed), so $\vec E$ never leaves a single fixed line, at angle $\theta = \arctan(\pm E_{0y}/E_{0x})$ to the $x$-axis. Nothing rotates; only the magnitude oscillates.
• Circular polarization ($E_{0x}=E_{0y}=E_0$, $\delta = \pm\pi/2$): then $E_x = E_0\cos(kz-\omega t)$ and $E_y = \mp E_0\sin(kz-\omega t)$, so $E_x^2+E_y^2=E_0^2$ at every instant — a circle of constant radius. The tip of $\vec E$ rotates at the steady rate $\omega$, and the sign of $\delta$ fixes whether that rotation is right- or left-handed.
• Elliptical polarization is the general case: any $E_{0x}, E_{0y}, \delta$ outside the two special conditions above traces an ellipse. Linear and circular polarization are its degenerate limits — collapse the ellipse to a line at $\delta=0,\pi$, or inflate it to a circle when $E_{0x}=E_{0y}$ and $\delta=\pi/2$.

Unpolarized light (an ordinary bulb, the Sun) has no fixed $\delta$ at all — it is an incoherent jumble of many independent polarization states, so $\vec E$'s direction wanders randomly and rapidly with no traceable pattern.

Malus's law. A linear polarizer transmits only the component of $\vec E$ along its transmission axis. If linearly polarized light of amplitude $E_0$ (intensity $I_0$) meets a polarizer with its axis at angle $\theta$ to $\vec E$, the transmitted amplitude is $E_0\cos\theta$, so intensity — proportional to amplitude squared — obeys

where: $I_0$ is the intensity of the already linearly polarized light striking the polarizer, and $\theta$ is the angle between the incoming polarization direction and the polarizer's transmission axis. When unpolarized light hits the first polarizer, there is no single $\theta$ to plug in — averaging $\cos^2\theta$ over all directions gives $\langle\cos^2\theta\rangle = 1/2$, so exactly half the intensity survives, and the transmitted beam emerges fully linearly polarized along that polarizer's axis. Any polarizer placed after that one then obeys Malus's law directly, since its input is now linearly polarized.

Unpolarized light of intensity $I_0 = 100\ \text{W/m}^2$ passes through two ideal polarizers in sequence: the first has a vertical axis, the second (the analyzer) is set at $30^\circ$ to the first. A third question asks what angle would instead reduce the final intensity to exactly $I_0/8$.

Step 1. Unpolarized light striking the first polarizer transmits half its intensity, emerging linearly polarized along the vertical: $I_1 = I_0/2 = 100/2 = 50\ \text{W/m}^2$.

Step 2. Apply Malus's law at the second polarizer with $\theta=30^\circ$ between the now-polarized beam and the analyzer's axis: $I_2 = I_1\cos^2(30^\circ) = 50\times(\sqrt3/2)^2 = 50\times(3/4) = 37.5\ \text{W/m}^2$.

Step 3. For the third part, require $I_2 = I_0/8 = 12.5\ \text{W/m}^2$ using $I_2 = (I_0/2)\cos^2\theta$: $I_0/8 = (I_0/2)\cos^2\theta \implies \cos^2\theta = 1/4 \implies \cos\theta = 1/2 \implies \theta = 60^\circ$.

The intensity after the first polarizer is 50 W/m², after the second (at 30°) is 37.5 W/m², and an analyzer angle of 60° would instead give exactly $I_0/8 = 12.5\ \text{W/m}^2$.

Competitive stretch

A wave has $E_x=3\cos(kz-\omega t)$ and $E_y=3\cos(kz-\omega t+\pi/2)$ (SI units, V/m). Equal amplitudes with $\delta=\pi/2$ is exactly the circular-polarization condition, so this beam is circularly polarized, and its tip sweeps through every orientation over one cycle — a linear polarizer placed in this beam transmits exactly half the intensity, independent of the polarizer's orientation, unlike the strongly angle-dependent linear case above.

• Treating "polarized" and "coherent" as the same idea. Polarization is about the direction $\vec E$ oscillates in; coherence is about phase stability over time between different waves — logically separate properties.
• Applying Malus's law to unpolarized input directly. $I=I_0\cos^2\theta$ needs $I_0$ to already be linearly polarized. For unpolarized light on the first polarizer, halve the intensity first, then apply Malus's law only from the second polarizer onward.
• Confusing amplitude falloff with intensity falloff. The field component drops as $\cos\theta$, but intensity — the measurable quantity — drops as $\cos^2\theta$, the square, as the derivation above makes explicit.
• Assuming sound can be polarized. Polarization needs a transverse wave with two independent perpendicular oscillation directions; sound is longitudinal and has none, so the concept simply does not apply to it.

Ask the tutor: want to work through three polarizers in sequence, or convert an amplitude-and-phase description into the resulting ellipse shape — ask the AI tutor to walk it with you.