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Optical lever comparator magnification

Short answer

Magnification M = 2L/x = (2×250)/5 = 100. Least count = 0.5/100 = 0.005 mm. Deviation = 3.2×0.005 = 0.016 mm, so shaft size = 25.016 mm; it exceeds the upper limit 25.015 mm and is rejected (oversize by 0.001 mm).

A mechanical-optical comparator uses a mirror pivoted at one end and pushed by the plunger at a distance of 5 mm from the pivot. The reflected beam falls on a scale 250 mm from the mirror. Derive the magnification of this single optical lever and find it. If the scale divisions are 0.5 mm apart, what is the least count? The instrument is set to zero with a 25.000 mm slip gauge. A shaft is then put in and the reading is +3.2 divisions. Find the size of the shaft, and say whether it is within the limits 25±0.01525 \pm 0.015 mm.

Derivation

Let the plunger move by δ\delta and the mirror rotate by θ\theta about the pivot, with the plunger acting at distance xx from the pivot:

θ≈δx\theta \approx \frac{\delta}{x}

A reflected ray turns through 2θ2\theta when the mirror turns by θ\theta. On a scale at distance LL the spot moves by

y=2θL=2Lx δy = 2\theta L = \frac{2L}{x}\,\delta Magnification M=yδ=2Lx=2(250)5=100\text{Magnification } M = \frac{y}{\delta} = \frac{2L}{x} = \frac{2(250)}{5} = 100

Least count

LC=scale divisionM=0.5100=0.005 mm\text{LC} = \frac{\text{scale division}}{M} = \frac{0.5}{100} = 0.005\ \text{mm}

Shaft size

The deviation from the master is +3.2×0.005=+0.016+3.2 \times 0.005 = +0.016 mm, so

size=25.000+0.016=25.016 mm\text{size} = 25.000 + 0.016 = 25.016\ \text{mm}

Limits are 24.98524.985 to 25.01525.015 mm. The shaft is larger than the upper limit by 0.0010.001 mm, therefore it is not acceptable (oversize, can be reworked).

Answer: M=100M = 100, LC = 0.005 mm, size = 25.016 mm, rejected (oversize by 0.001 mm).

✓ Answer checked by Claude Opus 5.5 (high effort)

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