Interferometry

Interference of light

When two light waves of the same wavelength (coherent) meet, they add. If the path difference is a whole number of wavelengths ($n\lambda$) the waves reinforce (bright band); if it is an odd number of half wavelengths ($(2n+1)\lambda/2$) they cancel (dark band). A thin air wedge between two surfaces gives alternate dark and bright bands called fringes.

Optical flat

It is a disc of quartz or fused silica (or glass), optically polished flat to $0.05\ \mu$m or better. It is placed on the surface to be tested at a slight angle, forming a thin air wedge, and illuminated by monochromatic light (sodium or helium, $\lambda = 0.5893\ \mu$m). Light reflected from the lower face of the flat and from the work surface interferes. A dark fringe appears wherever the air gap is

$$t = \frac{n\lambda}{2}$$

so adjacent fringes correspond to a height difference of $\lambda/2$ (about $0.3\ \mu$m).

If the fringes bend away from the line joining the edges by $a$, with spacing $b$, flatness error is $(a/b)(\lambda/2)$.

NPL flatness interferometer

It is used for testing the flatness of precision surfaces such as gauge blocks and measuring faces. A mercury-vapour lamp gives light that is passed through a filter and a slit, collimated by a lens, and reflected by a prism towards the optical flat resting on the test surface (on a table with three tilting screws). The reflected rays pass back through the prism to an eyepiece with a pinhole. The tilt screws adjust the wedge so that straight fringes appear; the pattern is observed and the error is calculated. It gives much sharper and brighter fringes than a simple optical flat in daylight.

Each fringe represents a height change of $\lambda/2$:

$$\frac{\lambda}{2} = \frac{0.5893}{2} = 0.2947\ \mu\text{m}$$

(a) Wedge

$$t = 8 \times 0.2947 = 2.357\ \mu\text{m}$$ $$\theta = \frac{t}{\text{length}} = \frac{2.357\times10^{-3}}{40} = 5.89\times10^{-5}\ \text{rad} \approx 12.2''$$

The fringe spacing is $40/8 = 5$ mm. In general, spacing $= \lambda/(2\theta)$.

(b) Curved fringes

$$\text{flatness error} = 0.4 \times 0.2947 = 0.118\ \mu\text{m}$$

The surface is flat to within about $0.12\ \mu$m.

(c) Gauge block

Each fringe of displacement equals $\lambda/2$ of height:

$$\Delta L = 2.5 \times 0.2947 = 0.737\ \mu\text{m}$$

Answer: (a) $t = 2.357\ \mu$m, $\theta = 5.9\times10^{-5}$ rad; (b) 0.118 $\mu$m; (c) 0.737 $\mu$m.