Past exam of the optics and electrodynamics course of the University of Cambridge / 2014 / Paper 1 / 1

For two-element polarization identification, assume ideal elements and use an intensity detector. For each source, test both element orders while rotating their relative axes. The circular source followed by the quarter-wave plate and then the polarizer is the combination that gives full intensity modulation. The plate converts circular light to linear light, and the downstream polarizer then obeys Malus's law.

If the polarizer comes first, it transmits half the circular source's intensity and the downstream lossless plate cannot change that intensity. Unpolarized light also gives constant half-intensity in either order: a wave plate leaves it unpolarized, while a polarizer followed by a wave plate has no final analyzer. Thus the modulating source is circular, its first element is the quarter-wave plate, and its second is the polarizer; the remaining source is unpolarized. Testing both orders avoids needing to know either source's original brightness.

For the oblique coupling of parallel Hertzian dipoles, a lossless Hertzian dipole has directional gain $G(\theta )=\frac{3}{2}{\sin }^2\theta$. The separation is $r=\sqrt{2}d$, and both transmitting and receiving patterns have ${\sin }^2\theta =1/2$, so $G_t=G_r=3/4$. Reciprocity gives matched receiving effective area $A_e={\lambda }^2{G}_r/(4\pi )$. The emitted polarization matches the transverse projection of the receiving dipole axis, so there is no additional polarization mismatch. ConsequentlyBoth the off-axis transmitting pattern and the off-axis receiving aperture matter. This assumes the stated far field, conjugate matching and negligible dissipative antenna losses.

Interpret the motion as uniform rectilinear motion: the charge is stationary in an inertial rest frame and has no radiation field. Constant speed alone is insufficient; a curved trajectory has acceleration and can radiate even in vacuum.

In a transparent medium, a charge can move faster than the light phase velocity $c/n$ without exceeding $c$. The induced polarization disturbance then adds coherently on a Cherenkov cone, with $\cos {\theta }_C=1/(n\beta )$. ThusThis is the kinematic threshold limit for the specified refractive index; appreciable emission requires exceeding it, and dispersion selects which frequencies satisfy the condition.