27 Jun 2025
Interference, Thin Films, and Newton's Rings
Coherent superposition, fringe visibility, division of wavefront and amplitude, equal-inclination and equal-thickness fringes, and Newton's rings.
Two-beam interference and visibility
At a point where two coherent fields have the same polarization,
\[E_1=E_{01}\cos\omega t, \qquad E_2=E_{02}\cos(\omega t+\delta).\]Since intensity is proportional to the time average of the squared field,
\[\left\langle(E_1+E_2)^2\right\rangle =\frac{E_{01}^2}{2}+\frac{E_{02}^2}{2} +E_{01}E_{02}\cos\delta.\]Writing the separate intensities as $I_1$ and $I_2$ gives
\[\boxed{I=I_1+I_2+2\sqrt{I_1I_2}\cos\delta}.\]Therefore
\[I_{\max}=(\sqrt{I_1}+\sqrt{I_2})^2, \qquad I_{\min}=(\sqrt{I_1}-\sqrt{I_2})^2,\]and the fringe visibility is
\[\boxed{\mathcal V=\frac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}} =\frac{2\sqrt{I_1I_2}}{I_1+I_2}}.\]Visibility reaches one only for equal intensities and complete mutual coherence. Division of wavefront makes coherent secondary sources from different portions of one wavefront, as in a two-slit arrangement. Division of amplitude splits one beam and later recombines its parts, as in thin films and interferometers.
Thin-film phase and reflection convention
Consider a film of refractive index $\mu$, thickness $t$, and internal ray angle $r$. The normal component of the wave vector in the film is
\[k_z=\frac{2\pi\mu}{\lambda}\cos r.\]A down-and-back traversal therefore adds propagation phase $2k_zt$, equivalent to the optical-path difference
\[\boxed{\Delta_g=2\mu t\cos r}.\]A reflection from a lower to a higher refractive index changes the field phase by $\pi$; reflection from higher to lower does not. Thus, when exactly one of the two reflected rays undergoes phase reversal,
\[\delta=\frac{2\pi}{\lambda}(2\mu t\cos r)+\pi.\]Reflected maxima require $\delta=2q\pi$, and reflected minima require $\delta=(2q+1)\pi$. Re-indexing with $m=0,1,2,\ldots$ gives
\[\boxed{2\mu t\cos r=m\lambda \quad\text{(reflected dark)}},\] \[\boxed{2\mu t\cos r=\left(m+\frac12\right)\lambda \quad\text{(reflected bright)}}.\]If there are zero or two reflection phase reversals, these bright and dark conditions interchange. Transmitted maxima and minima are complementary for a non-absorbing film.
Haidinger and Fizeau fringes
For a parallel film, $t$ is fixed. A given interference order therefore selects a fixed $r$. The resulting circular Haidinger fringes are fringes of equal inclination and are localized at infinity; a lens brings rays of equal inclination together in its focal plane.
For a wedge of small angle $\alpha$, the thickness at distance $x$ from the contact is
\[t(x)=x\tan\alpha\simeq x\alpha.\]At nearly normal incidence and with one phase reversal, the $m$th reflected dark fringe obeys
\[2\mu\alpha x_m=m\lambda.\]Hence neighboring fringes have constant spacing
\[\boxed{\beta=x_{m+1}-x_m=\frac{\lambda}{2\mu\alpha}}.\]These straight, parallel Fizeau fringes are fringes of equal thickness and are localized near the film.
Newton’s rings and measurements
A plano-convex lens of radius of curvature $R$ rests on a plane plate. At radial distance $r$ from the contact, geometry gives
\[R^2=(R-t)^2+r^2 =R^2-2Rt+t^2+r^2,\]so
\[r^2=2Rt-t^2.\]Because $t\ll R$, the $t^2$ term is negligible and
\[\boxed{t\simeq\frac{r^2}{2R}}.\]The film between lens and plate has one reflected phase reversal. Thus the contact is dark in reflected light. Combining $2\mu t=m\lambda$ with the geometrical thickness gives the dark-ring radii and diameters:
\[\boxed{r_m^2=\frac{m\lambda R}{\mu}}, \qquad \boxed{D_m^2=\frac{4m\lambda R}{\mu}}.\]For bright reflected rings,
\[\boxed{r_{m,{\rm bright}}^2 =\frac{(m+1/2)\lambda R}{\mu}}.\]For an air film, $\mu=1$. Subtracting the squared diameters of orders $m$ and $m+p$ removes the uncertain position of contact:
\[D_{m+p}^2-D_m^2=4p\lambda R.\]Therefore the wavelength is
\[\boxed{\lambda= \frac{D_{m+p}^2-D_m^2}{4pR}}.\]If a liquid fills the gap, then $D_{m,\ell}^2=4m\lambda R/\mu$. Comparing the same order in air and liquid gives
\[\boxed{\mu=\frac{D_{m,{\rm air}}^2}{D_{m,\ell}^2}}.\]More robustly, using two orders in each medium,
\[\boxed{\mu= \frac{D_{m+p,{\rm air}}^2-D_{m,{\rm air}}^2} {D_{m+p,\ell}^2-D_{m,\ell}^2}}.\]Every diameter must be measured in the same length unit; then the ratios are dimensionless and the wavelength formula returns the unit used for $D^2/R$.
The intensity, visibility, film phase, wedge spacing, and ring-diameter identities are checked in the Unit II Maxima worksheet.
Discussion