# Given aperture and focal length, what is the minimum distance for maximum resolution? [closed]

I'm going to describe a physics experiment involving imaging a proton beam using a CCD camera by reflecting it on a aluminium oxide surface, and we need some help about how close the camera should be to the surface (aluminium oxide) in order to get the best resolution.

Note that, from the point of view of the analyst of those images, the closer the camera to aluminium oxide surface, the less error that those data contains.

## Description of the setup:

Note: The angle of the surface with the horizontal plane is 45 degree.

### What we need exactly from those images:

The following image is one of the example that we are taking;

(The vertical size of the seen ellipse is 4cm only)

Although, we are taking RGB coloured images from the camera, we are transforming those images to gray scaled from because the only thing that we need from them is the brightness values that is contained in each pixels.

## Question:

Now the thing is, we would like to make the CCD camera close to the surface as much as possible; however, in this case, would we lose any resolution ?

To be clear, for a given focal length f and lens size of a CCD camera, what is the smallest distance that the camera should be located from the surface such that we get the maximum resolution that we can ?

Secondly, if we want to decrease that distance or increase the resolution, why type of cameras or lenses should we use ?

Note:

As I have mentioned in the "background" part of the question above, the only that we need need from those images are the brightness values, i.e the luminance, contained in each pixel. Moreover, we do not use the hue and saturation information contained in those images, only luminance.

Edit:

we just want to measure the size of the beam, and the aluminium oxide surface is used a way to reflect the beam to the camera.

• It would be interesting to know what kind of camera you need. – Rafael Jul 21 '18 at 1:11
• @Rafael As I have said, CCD camera – onurcanbektas Jul 21 '18 at 3:24
• @xiota Yeah, an actual proton beam like hydrojen ions. – onurcanbektas Jul 21 '18 at 6:55
• I'm voting to close this question as off-topic because it is not about producing photographs as defined in this group. It is about using camera sensors to measure the size of reflected proton beams. – Michael C Jul 21 '18 at 8:57
• @MichaelClark Umm.. have you even read the question ? – onurcanbektas Jul 21 '18 at 9:49

The closest focal distance is parameter of the lens, not the camera/sensor. So you should check for this parameter in to the camera manual. For example you see example one and example two. Both are 100mm lens, both Canon EF lens. But first have 90cm closest focus distance, second have 31cm closest focus distance.

To make change (decrease) in this distance you need to change lens with specific one. You can also try by adding magnification glass in front of the lens. Or (if it's possible) deattach the lens from camera and move it in direction of the mirror

• Thanks a lot sir. In case the manual or some specs of the lens is missing, is there a way to calculate the closest focal distance from other parameters of the lens ? – onurcanbektas Jul 21 '18 at 3:32
• @onurcanbektas, for sure. But I am afraid you need to know the entire optical schema of the lens to do this. What I will recommend you is to try to move the entire camera and refocus. And check the resulting image – Romeo Ninov Jul 21 '18 at 3:40

Just to supplement what has already been said, you can dramatically decrease the minimum focusing distance of any lens with the use of extension tube(s). Since extension tubes do not have any lens elements (it's a hollow tube), this could be beneficial to you regardless of the lens that you eventually use. You should be aware that an extension tube will prevent the camera from focusing beyond about 6" to 1' depending upon the focal length being used.

• Thanks you for your answer @Frank; however, I couldn't understand how the extension tube is going to help to reduce the min. focus distance ? – onurcanbektas Jul 21 '18 at 4:15
• Plus, I'm not familiar with the notation: what does 6'' to 1' means ? – onurcanbektas Jul 21 '18 at 4:15
• @onurcanbektas 6"=6 inches, 15 cm (avg), 1'= 1 feet, 30 cm (avg). And extension tubes do what I propose in answer: move the lens away from sensor :) – Romeo Ninov Jul 21 '18 at 4:45
• I do not think that our lens can move in our case from the camera. (I haven't seen the camera, but is it a property of all CCD cameras ?) – onurcanbektas Jul 21 '18 at 4:50
• @onurcanbektas, deattachability of lens is characteristic of particular construction of the camera. CCD is only the technology of the sensor. – Romeo Ninov Jul 21 '18 at 4:53

It is not clear if your target is the surface of the aluminum oxide or the "virtual" image of whatever is reflected by it. In the latter case you need to focus rather far (infinity?) and any lens will do.

For the other case, assuming that the closer, the better, the real limit is the ability for the lens to focus at a short distance.

• "Macro" lenses (the true ones) can get close enough to reach a 1:1 magnification (i.e., the image on the sensor is the same physical size as the object).

• You can also shorten the focus distance of a lens with either:

• a close-up lens, which is a converging lens of, usually, 1 to 4 dioptries, that screws like a filter in front of the lens. Since this is an additional optical device they can introduce distortions or softness. The good ones are mildly expensive.
• one or more extension rings, that go between the body and the lens. They don't introduce distortion of their own (if they are of good quality and keep the lens centered). But they "eat" a lot of light (which may no be a problem if you are shooting something bright).

Both devices make the camera "short-sighted": it can no longer focus on infinity.

But, if you want to shoot the AlOx surface, you have a problem of depth of field, since it has a 45Â° angle. The closer you get, the shallower the depth of field. For close-ups, the depth of field is driven by the magnification so if you want a decent focus you may not be able to use the whole sensor area (and if you try to use small apertures, you get diffraction problems).

So if your quality criterion is the pixels of image per millimeter of subject in the final result, a camera with a "dense" sensor may give better results than a camera with larger pixels. And the smaller the sensor, the denser it is, so maybe you should look into a compact or bridge camera, or even a smartphone (as long as you can get RAW from it).

• No, we just want to measure the size of the beam, and the aluminium oxide surface is used a way to reflect the beam to the camera. – onurcanbektas Jul 21 '18 at 3:29
• So, you fall in the "macro" case, and the DOF is very important because the edge blurriness is going to impact your accuracy. Unless you only want to measure the height at the center, along the only line which will be in perfect focus. – xenoid Jul 21 '18 at 9:00
• @xenoid Not really. The angle of the mirror with reference to the optical axis of the camera's lens will mostly be offset by the angle of the mirror to the optical axis of the beam emitter. The distance will be based on the total length of the optical path of the rays both before and after they are reflected by the mirror. Otherwise the reflex mirrors in SLRs wouldn't work the way they do. – Michael C Jul 22 '18 at 2:10
• The question is a bit ambiguous about what is really the target.... If the photo is the virtual reflection image why would it matter to be close to the mirror? – xenoid Jul 22 '18 at 13:05

Are you actually using a lens?

If you are not, and only need the scattered rays of the reflection of a narrow beam, In my opinion simply the distance is given by the overall size of the light smudge.

• Red is too close.
• Purple is too far.
• Green is maximizing your sensor's size.

Obviously, you need some exposure times, so additional scattered beams outside this "main" beams are also captured or not.

This can vary, in the case you are moving the plate or you are moving the beam, rasterizing, like scanning the plate.

If it is the second case you need to see what are the boundaries of those movements.

• Your drawing violates one of the fundamental principles of optics: The angle of incidence equals the angle of reflection. – Michael C Jul 22 '18 at 2:13
• I imagine two scenarios. But As it is not my research I do not have an idea what those measurements are for. 1. You could try to measure scattered beams, not mirrored beams. 2. You could try to measure bumpiness of an actual 100% glossy surface. The result the user is looking for is a smudge. Both could render a smudged image. – Rafael Jul 22 '18 at 18:17