Length — Question 10

Question 10

More pixels on the display, or more information about length?

A microscope image has the calibration and object span shown. Assume uniform magnification, no distortion, and an exact calibration factor. Each object edge can lie anywhere within $\pm2$ pixels of its indicated position, with every pair of endpoint errors allowed. These are hard bounds, not standard deviations.

1. Find the central object length and its full allowed interval in millimetres. Explain why the edge-separation half-width is not merely 2 pixels.

2. The same stored image is enlarged digitally by a factor of four in each linear direction. Find its new calibration span, object span, and edge-localization bounds in displayed pixels. Does its physical-length interval improve?

3. Instead, a new optical image is acquired with four times the original linear magnification. The calibration is now 800 pixels per millimetre, and the indicated object span is 1680 pixels. Suppose testing establishes that each newly measured edge again has a hard bound of $\pm2$ pixels at this new acquisition setting. Find the new length interval.

4. A task requires a valid length half-width no greater than $0.010\ \mathrm{mm}$. Which image meets it under the stated assumptions? Explain why optical magnification alone, without the new edge-bound evidence, would not guarantee that result.

Question 10 - Solution

More pixels on the display, or more information about length?

(a) Opposite edge errors can add in the separation.

The central length is $(420\ \mathrm{pixels})/(200\ \mathrm{pixels/mm})=2.100\ \mathrm{mm}$. If the right edge moves 2 pixels right and the left edge 2 pixels left, the separation increases by 4 pixels; opposite errors decrease it by 4. Therefore $$L\in\left[\frac{416}{200},\frac{424}{200}\right]\ \mathrm{mm} =\boxed{[2.080,\ 2.120]\ \mathrm{mm}}.$$ The half-width is $0.020\ \mathrm{mm}$, not $0.010\ \mathrm{mm}$.

(b) Digital enlargement rescales existing uncertainty too.

$$\delta L=\frac{\epsilon_L+\epsilon_R}{k},\qquad\delta L_{\mathrm{digital}}=\frac{4\epsilon_L+4\epsilon_R}{4k}=\delta L.$$

The calibration span becomes 800 displayed pixels per millimetre and the object span 1680 pixels. Each edge bound becomes $\pm8$ displayed pixels; the separation bound is $\pm16$. Thus $$L\in[1664/800,1696/800]\ \mathrm{mm} =\boxed{[2.080,2.120]\ \mathrm{mm}}.$$ The image looks larger, but it contains the same localization information.

(c) Use the new acquisition’s independently supported bound.

With a new $\pm2$-pixel bound at each edge, the new separation half-width is 4 pixels, giving $$L\in[1676/800,1684/800]\ \mathrm{mm} =\boxed{[2.095,2.105]\ \mathrm{mm}}.$$ Its half-width is $0.005\ \mathrm{mm}$, four times smaller than before.

(d) The tighter inference needs more than display scale.

Only the new optical image meets the $0.010\ \mathrm{mm}$ requirement under the supplied conditions. Increased optical magnification can also enlarge blur or introduce other errors. The improvement here relies on the tested edge bound, exact calibration, and absence of distortion; magnification alone does not prove those assumptions.