Past exam of the physics A course of the University of Cambridge / 2015 / Paper 2 / Section D / D13 / c / Solution

For propagation along $z$, a transverse wave has two perpendicular displacement components. With $\varphi =kz-\omega t$, write $u_x=A\cos \varphi$, $u_y=B\cos (\varphi +\delta )$. Their trajectory at a fixed point obeysThe 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.

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.