Past exam of the physics A course of the University of Cambridge / 2017 / Paper 2 / Section B / B9

Let the first ray reflect at the upper surface and the second traverse the film, reflect below, and emerge at a laterally displaced point. Its internal round-trip geometric length is $2d/\cos \theta$, but the external path to a common outgoing wavefront must also be included. The lateral separation is $2d\tan \theta$, whose projection along the external ray is $2d\tan \theta \sin {\theta }_i$. The net optical path difference is thereforeusing Snell's law $\sin {\theta }_i=n\sin \theta$. Multiplying by the vacuum wavenumber givesFor the usual index ordering $1<n<n_2$, including the stated later value $n=1.2$, both reflections are towards higher refractive index and both reverse the electric-field sign. Their relative reflection phase is therefore zero; $\delta$ is also their total relative phase modulo $2\pi$. A different index ordering or polarization-dependent reflection phase would require that additional term.

At the same output in air, use incident intensity $I_{\mathrm{inc}}\propto |A{|}^2$. The two specified output amplitudes, including their common reflection sign, give the thin-film interference resultThe output is a cosine with mean $0.0234I_{\mathrm{inc}}$, period $2\pi$, and extremaThe unequal amplitudes prevent a dark minimum; their interference visibility is $0.009/0.0234=5/13\simeq 0.385$. This is the stipulated two-beam model: the supplied amplitudes already include propagation and transmission factors, and internal re-reflections are excluded by the stated assumption. They need not be recomputed from the later index value.

Eliminate the internal angle using Snell's law:Neighbouring bright fringes differ in phase by $2\pi$. The local angular fringe spacing of a transparent film therefore givesWith $\Delta {\theta }_i=0.4\pi /180=6.981\times 10^{-3}\mathrm{rad}$, ${\theta }_i\simeq 10^{\circ }$, $n=1.2$ and $\lambda =500\mathrm{nm}$,The data specify a local spacing near $10^{\circ }$, so differentiating is appropriate; choosing particular neighbouring fringe angles slightly shifts this approximate result. The phase decreases as incidence angle increases.

At normal incidence, the propagation phase is $4\pi n_{\mathrm{oil}}d(x,y)/\lambda$. Different film thicknesses therefore produce bright and dark contours of equal thickness, with adjacent same-type fringes differing by $\boxed{\Delta d=\lambda /(2n_{\mathrm{oil}})}$. Under monochromatic illumination these are intensity variations of the incident colour, rather than a rainbow.

For the usual case $n_{\mathrm{oil}}>n_{\mathrm{water}}>1$, reflection at air-oil reverses phase while reflection at oil-water does not. The additional relative $\pi$ makes reflected minima occur at $2n_{\mathrm{oil}}d=m\lambda$ and maxima at $2n_{\mathrm{oil}}d=(m+1/2)\lambda$. If the oil index were below the water index, the phase-reversal pattern would change and these conditions would interchange. The thinnest regions need not be completely dark because the two reflected amplitudes need not match; at vanishing thickness the residual reflection approaches that of the air-water boundary. Surface motion changes the thickness contours and moves the fringes.