= Solution
For propagation along $z$, a <transverse wave> has two perpendicular <displacement> components. With $\phi=kz-\omega t$, write $u_x=A\cos\phi$, $u_y=B\cos(\phi+\delta)$. Their trajectory at a fixed point obeys
$$
\boxed{\left(\frac{u_x}{A}\right)^2+\left(\frac{u_y}{B}\right)^2
-2\frac{u_xu_y}{AB}\cos\delta=\sin^2\delta.}
$$
The general nondegenerate trajectory is an ellipse: <elliptical polarization>. For $\delta=0$ or $\pi$, the components remain in phase or antiphase and the trajectory collapses to a line. A single nonzero component is also linearly polarized. Equal amplitudes $A=B$ with $\delta=\pm\pi/2$ give <circular polarization>. Unequal quadrature amplitudes give an ellipse aligned with the chosen axes, while general <phase differences> rotate its axes.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2015/paper-2-polarization.png]
{title=Linear, circular and elliptical transverse-wave trajectories at a fixed spatial point}
For the displayed convention and $\delta=+\pi/2$, at $z=0$ the vector starts at $+x$ and moves towards $+y$ as time increases. Reversing $\delta$ reverses its rotation. Right- and left-handed labels require a stated viewing and phase convention, so rotation is defined explicitly here. A fixed ellipse is a coherent relationship between transverse components; unpolarized light instead has fluctuating orientation and phase relationships on the measurement timescale.
A linear polarizer transmits a linearly polarized beam with <intensity> proportional to $\cos^2\theta$, allowing its direction to be measured by an analyser. <circular polarization> transmits equal <intensity> through any linear analyser, but that test alone does not distinguish it from unpolarized light. A <quarter-wave plate> converts a suitable coherent circular state to <linear polarization>, allowing the combination to discriminate them. For light the <electric field> defines the usual polarization; the <magnetic field> remains perpendicular to it and to propagation. Longitudinal sound in a homogeneous gas lacks the two transverse degrees of freedom needed for these polarization states.
Back to article page
