Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2015 / S12 / 2

Let the interface be $y=0$, with glass at $y<0$ and water at $y>0$, and choose $x$ parallel to the incident beam's tangential component. Define $k=\omega /c$ as the vacuum wavenumber, with glass and water refractive indices $n_g$ and $n_w$. Continuity of phase along the interface fixes $k_x=k{n}_g\sin \theta$. In water, the dispersion relation is $k_x^2+k_y^2=k^2{n}_w^2$, so above the critical angle,Choose the positive imaginary part because a field growing exponentially into the water would be unphysical. The evanescent field in total internal reflection is thereforeIt propagates in phase parallel to the interface but decays normally. Here $E_0$ is the transmitted field amplitude at the interface; relating it to the incident amplitude requires the appropriate polarization-dependent Fresnel equations. If the symbol $k$ instead denotes the incident glass wavenumber, replace every vacuum $k$ in the formula by $k/n_g$; equivalently $k_x=k\sin \theta$ and $\kappa =k\sqrt{{\sin }^2\theta -(n_w/n_g{)}^2}$. This states the wavenumber convention left implicit in the source.

The excitation rate in the weak, unsaturated fluorescence regime is proportional to the time-averaged squared electric field amplitude. Thus the illumination profile relevant to excitation isThe amplitude decay length $1/\kappa$ is twice the intensity decay length. This profile is often called the evanescent excitation intensity; it does not imply a propagating normal energy flux through an ideal lossless totally reflecting interface.

Total internal reflection fluorescence microscopy restricts excitation to a shallow layer adjacent to the glass, typically on the scale of hundreds of nanometres for visible wavelengths and suitable angles. Fluorophores near the basal cell membrane are excited while much of the cytoplasm and bulk solution remain dark. The resulting reduction in out-of-focus fluorescence and autofluorescence improves the signal-to-background ratio for individual molecules, allows camera-based parallel observation, and reduces unnecessary excitation of deeper regions. Its limitation is the same selectivity: it does not image arbitrary molecules deep inside a cell. This near-interface selectivity was directly demonstrated in cell–substrate contact imaging.^[ref]

Under the simplest interpretation, each spot contains two equally bright, independently bleaching green fluorescent protein emitters. Losing one halves the fluorescence and losing the other removes it. A stable protein dimer carrying one active label per subunit is consistent with this result. The conclusion is two active emitters in the observed spot; it does not by itself rule out unresolved coincident proteins, incomplete fluorescent maturation or blinking. Permanent membrane binding makes loss by dissociation less plausible, but lateral motion out of the observation region would still need experimental control.^[ref]

Let their independent photobleaching times be $T_A,T_B\sim \mathrm{Exp}(k_b)$. Then $t_1=\min (T_A,T_B)$ has rate $2k_b$, while the memoryless remaining lifetime $t_2-t_1$ has rate $k_b$ and mean $1/k_b$, matching the stated single-emitter calibration. The two-emitter photobleaching completion time is the maximum, not one single exponential and not a sum of two equal-rate exponential variables. Its cumulative distribution function isDifferentiating gives its probability density function:with zero density for $t<0$. It integrates to one, and its mean is $1/(2k_b)+1/k_b=3/(2k_b)$. This uses two active fluorophores at the start of observation, constant illumination and independent irreversible bleaching with the calibrated rate.