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Fringe measurement of flatness error

Short answer

Each fringe represents λ/2 = 0.2947 µm. (a) Air gap height = 8×0.2947 = 2.357 µm; wedge angle = 2.357×10⁻³/40 = 5.9×10⁻⁵ rad. (b) Flatness error = 0.4×0.2947 = 0.118 µm. (c) Block length difference = 2.5×0.2947 = 0.737 µm.

A surface 40 mm long is tested with an optical flat in sodium light (λ=0.5893 μ\lambda = 0.5893\ \mum). Eight equally spaced straight fringes are seen along the 40 mm length. (a) Find the height of the air gap at the far end above the touching end, and the wedge angle. (b) On a second surface the fringes are curved, with maximum deviation equal to 0.4 of the fringe spacing. Find the flatness error. (c) A gauge block is wrung on a flat platen and the same flat is placed over both. The fringes over the block are displaced by 2.5 fringes relative to those over the platen. Find the difference between the block length and the height of the reference.

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

λ2=0.58932=0.2947 μm\frac{\lambda}{2} = \frac{0.5893}{2} = 0.2947\ \mu\text{m}

(a) Wedge

t=8×0.2947=2.357 μmt = 8 \times 0.2947 = 2.357\ \mu\text{m} θ=tlength=2.357×10−340=5.89×10−5 rad≈12.2′′\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=540/8 = 5 mm. In general, spacing =λ/(2θ)= \lambda/(2\theta).

(b) Curved fringes

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

The surface is flat to within about 0.12 μ0.12\ \mum.

(c) Gauge block

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

ΔL=2.5×0.2947=0.737 μm\Delta L = 2.5 \times 0.2947 = 0.737\ \mu\text{m}

Answer: (a) t=2.357 μt = 2.357\ \mum, θ=5.9×10−5\theta = 5.9\times10^{-5} rad; (b) 0.118 μ\mum; (c) 0.737 μ\mum.

✓ Answer checked by Claude Opus 5.5 (high effort)

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