Optics 044 · Interference, coherence, cavities, and metrology

Interference Surface Metrology

How can interference fringes map the shape of a surface, down to tiny defects? Each fringe is a contour line of the air gap, half a wavelength higher than its neighbor. Evenly spaced straight fringes mean a uniform tilt, rings mean a curved surface, and a pit kinks the fringes locally. One image cannot tell a bump from a pit: that takes several frames with stepped phase plus a calibrated reference height.

A cutaway of the air gap shows how the exact sag of a sphere, a tilted flat and a smooth pit in the test surface produce monochromatic fringes. Four phase-stepped frames then recover the gap profile.

Shared centre-line probe / gap / normalized intensity

x_p,t_p,I_p

$0.50$

Recovered curvature, wedge or finite-window departure

\widehat R,\widehat\alpha,\mathrm{PV}_{\mathrm{res}}

$50\%$

Sampling grid / maximum adjacent phase jump

N,\max|\Delta\phi|/\pi

$0.00\pi$

Model regime $\text{valid model regime}$

Model checked against analytic results

Interference Surface Metrology

Monochromatic measurement wavelength

\lambda

600\,\mathrm{nm}

Spherical radius of curvature (not specimen radius)

R

4\,\mathrm m

Reference-flat tilt (zero means parallel)

\alpha

0.25\,\mathrm{mrad}

Base gap (not an automatic phase scan)

t_0

0\,\mathrm{nm}

Test-surface pit depth

d_{\mathrm{pit}}

200\,\mathrm{nm}

Shared centre-line probe position

x_p

1.50\,\mathrm{mm}

The contact point is dark. Increase the radius of curvature: the cyan reference surface flattens and the rings spread out, while the specimen width stays fixed. Try the wedge, then a pit: the pit depth moves the gold test surface and bends the fringes. Four captured phase steps recover the profile, using a known corner and a calibrated plane. Drag to orbit; the slider moves the shared probe. Play does not move the surface.

The model is scalar at normal incidence, with two balanced coherent reflections. Intensity is normalized to a fixed scale and is not the absolute reflectance. The vertical scale is magnified separately, the support is schematic, the phase steps are ideal, and the anchor point is calibrated. Noise is not modeled, and the Lab does not recover arbitrary defects or claim instrument accuracy.

Physics tutorial

Audit the physics of Interference Surface Metrology

BackgroundLight reflects once from each face of a thin air gap, and the two coherent reflections interfere. The round-trip phase across the gap is $\phi=4\pi t/\lambda$ . One reflection goes from glass to air and the other from air to glass, so the two have opposite signs. That is why the contact point is dark, not bright. The 3D surfaces, the camera images and the recovered profile all come from the same gap function.

Why it mattersHow can interference fringes map the shape of a surface, down to tiny defects?

Start with the essentials

Focus question: How can interference fringes map the shape of a surface, down to tiny defects?
One-sentence intuition: Neighbouring fringes of equal brightness differ in gap height by $\Delta t_{\mathrm{fringe}}=\lambda/2$ . One image cannot tell apart gaps that differ by a whole number of these steps, and it cannot tell a bump from a pit. Four ideal phase steps recover the wrapped phase. Dense sampling of a smooth surface, plus one corner calibrated independently, then fixes a continuous gap. These steps do not produce an absolute height out of nothing.

Core mathematical model

Why optical contact is dark

\phi=\frac{4\pi t}{\lambda},\qquad I_k=\frac{1-\cos(\phi+k\pi/2)}{2}

Intensity is divided by the fixed maximum that two equal coherent reflections reach when they add in phase. The minus sign comes from the reversed reflection. The four frames are taken with four separately applied reference phases; the surface does not change between them. The model ignores multiple reflections and the exact Fresnel imbalance, so it does not compute absolute reflectance.

Curvature radius is not specimen radius

\begin{aligned}t(r)&=t_0+\frac{r^2}{R+\sqrt{R^2-r^2}}\\r_m^2&=2Rs_m-s_m^2,\quad s_m=m\lambda/2-t_0\end{aligned}

The specimen stays an 8 mm square. Its radius of curvature is on the scale of metres, and changing it reshapes the spherical face itself. A larger radius means flatter glass and more widely spaced rings. The dark-ring radii use the exact sag, not only the small-sag approximation. The first dark ring outside the centre, which the readout lists, can be cut off by the square window.

A tilted flat becomes a thickness ruler

t(x)=t_0+(x+a)\tan\alpha,\qquad\Delta x=\frac{\lambda}{2\tan\alpha}

The window extends 4 mm on each side of centre. Tilting moves the reference surface itself; changing the wavelength changes the pattern without moving either face. At zero tilt the fringe spacing is infinite and the image is uniform. The wedge angle comes from the gaps recovered at the two edges, not from the slider setting.

Four images, one wrapped phase

\phi_w=\operatorname{atan2}(I_1-I_3,I_2-I_0),\qquad\widehat t=\frac{\lambda}{4\pi}\phi_u

The unwrapping step compares only neighbouring pixels and uses only the four intensity arrays, the wavelength and the calibrated corner height. The readout shows the largest phase jump between neighbours, which must stay below half a cycle. On the 257-square grid, every allowed control setting, including the extremes, meets this condition. Without the corner anchor, one overall integer order stays unknown. Noise, discontinuous surfaces and phase-step errors are not simulated.

A pit bends the fringes; subtracting a calibrated plane isolates it

\begin{aligned}t(x,z)&=t_0+(x+a)\tan\alpha+h(x,z)\\h(x,z)&=d\exp\!\left[-\frac{(x-x_c)^2+z^2}{2\sigma^2}\right]\end{aligned}

The test surface has a smooth pit 0.85 mm wide, centred 0.8 mm to the right of centre. A positive depth widens the air gap and lowers the test surface. Subtracting the independently calibrated piston and tilt leaves the recovered departure from the plane. The colour scale is fixed and does not rescale itself. Peak-to-valley is measured over the finite sampled window, so it is not simply the input depth.

Common difficulties

More rings do not mean more height information

Typical misconceptionA single photograph of the fringes gives the absolute height uniquely, accurate to the nanometre.

Better mental modelIntensity repeats as the phase grows, and a phase and its negative give the same intensity. Known phase steps remove the sign ambiguity, but without calibration they cannot fix the integer order. Here the forward and inverse calculations agree without noise; that agreement checks the code, not the accuracy of a real instrument. Errors in the reference face, vibration, finite coherence, Fresnel imbalance and imaging resolution each need their own model.

Run the experiment

01
Scene 1: Newton rings
Start with the surfaces in contact. Increase the curvature radius: the cyan face flattens, the square specimen keeps its size, and the dark rings move outward.
What to observe: Changing the wavelength never changes the shape of the measured surfaces.
02
Scene 2: Air wedge
Choose the air wedge, then lower the tilt to zero. Change the base gap by half a wavelength: the surfaces move, but the intensity pattern looks exactly the same.
What to observe: Two gaps that differ by half a wavelength can give identical images.
03
Scene 3: Fizeau optical-flat test
Choose the optical-flat pit. Increase its depth and watch the fringes bend. Then set the tilt to zero to see why one image loses the sign. The four-step inversion still recovers the calibrated profile.
What to observe: The curvature remains after the piston is removed, while a flat wedge disappears once the calibrated plane is subtracted.