How does changing image distance affect resolution when adapting a scanner lens to a Raspberry Pi camera?
Asked 7/28/2023
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I’m considering adapting the lens from a Minolta Dimage Scan Elite 5400 to a Raspberry Pi HQ/HD camera. The lens is described as a 39mm compound lens designed for a much larger image circle than the Pi sensor.
My question is about image distance and resolution:
- If I mount the lens at a longer image distance, the sensor captures only the center of the lens’s image circle.
- If I mount it much closer so the projected image better matches the small sensor, the field of view changes.
Assuming the subject is placed at the corresponding focus distance for each setup, does changing the image distance change the optical resolving power at the image plane? In other words, if the larger projected image can resolve a certain object detail size, should the smaller projected image be expected to resolve the same detail, or does magnification/image scale change the effective resolution?
Originally by Photography Stack Exchange contributor. Source · Licensed CC BY-SA 4.0
Photography Stack Exchange contributor
2y ago
2 Answers
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There's a general answer to your question. But based on something specific you said, there's another answer based on a misunderstanding.
Specific issue: Mounting the lens too close to the image plane
The lens has a focal length of 39 mm. Let's rearrange the thin lens formula, and see what happens when the image plane distance \$d_\text{i}\$ (i.e., how close the lens is mounted to the sensor) is less than the focal length. The object plane distance \$d_\text{o}\$ is the distance to the object being photographed:
$$ \begin{align} {1\over f} &= {1\over d_\text{o}} + {1\over d_\text{i}} \\ {1\over d_\text{o}} &= {1\over f} - {1\over d_\text{i}} \\ d_\text{o} &= \frac{f d_\text{i}}{d_\text{i} - f} \end{align} $$
Notice that if \$d_\text{i} < f\$, the denominator is negative, meaning that the lens cannot focus on anything in front of it; the plane of object focus is behind the lens, within the lens's focal length.
Said another way: for converging lenses, the region between the lens's rear and front 'focal planes' (roughly, within 1 focal length in front of or behind the lens) cannot be focused on or imaged.
In fact, mathematically speaking, using the thin lens formula, the focal length is defined as the image distance \$d_\text{i}\$ when the object is at infinity:
$$ \begin{align} {1\over f} &= {1\over d_\text{i}} + {1\over\infty} = {1\over d_\text{i}} \\ \therefore f &= d_\text{i} \end{align} $$
In your case, mounting the lens any closer than ~39 mm to the sensor won't yield a useful system at all.
General answer: Yes, image distance affects resolution
For a particular lens make/model, ignoring copy variation (put another way, assuming a particular lens is perfectly manfuactured), that lens's performance can be completely characterized by its angular resolution.
That is, given an incoming ray of light entering the lens at a given angle \$\theta\$ from the optical axis, with a small differential region \$\Delta\theta_\text{tangential}\$ by \$\Delta\theta_\text{sagittal}\$ the lens will have a certain resolution. We don't normally describe and report angular resolution though; it's easier to think about the lateral linear resolution of the lens when its angular resolution subtends a given linear distance at the image plane.
MTF (Modulation Transfer Function) charts for lenses describe the spatial resolution of the lens in "\$x\$ line pairs/mm" at a given linear distance \$d_\text{i}\$ from the center of the imaging plane. But that's really just two legs of a right triangle that could equally describe angular "line pairs/degree" at a given angular deflection from the center of the imaging plane. If the imaging plane moves further from the lens by a small factor of \$S > 1\$ (which according to the thin lens formula, equates to focusing on a closer subject), the lens's angular resolution doesn't change. Instead, the projected image plane would be larger, and the linear resolution would be reduced by the same factor S: "\$x/S\$ line pairs/mm" at a distance of \$S\times d_\text{i}\$.
Originally by user11924. Source · Licensed CC BY-SA 4.0
user11924
2y ago
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Not in the way you describe. The key issue is that a 39mm lens cannot form a real image on the sensor if the sensor is placed closer to the lens than its focal length. If image distance is less than 39mm, the thin-lens equation gives a virtual image, not one a camera sensor can record.
So your ~19.7mm option is not a valid imaging setup for a normal camera sensor.
For real-image use, the sensor must be at least about one focal length from the lens, with infinity focus occurring at the focal length and closer subjects requiring even greater image distance.
As for resolution: changing image distance changes magnification and field of view. The center of a lens image circle is often the sharpest part, so cropping the center with a small sensor does not usually reduce lens quality there. But a smaller image on the sensor means less magnification at the image plane, so the same subject detail occupies fewer sensor pixels unless you increase optical magnification.
So: image distance affects focus and magnification, and therefore recorded detail on the sensor, but your proposed “match the image circle by moving the lens inside its focal length” setup will not work as a real-image camera arrangement.
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