Diffraction Limit Discussion Continuation

Started Feb 21, 2014 | Discussions thread
bobn2 Forum Pro • Posts: 55,673
Re: Cutting to the chase.

Jonny Boyd wrote:

Just another Canon shooter wrote:

Great Bustard wrote:

Jonny Boyd wrote:

The universal claim I'm disproving is that peak visible resolution always occurs at the same aperture.

You didn't disprove that the peak resolution always occurs at the same aperture, all else equal. At best, you've said that if the resolution is low enough and/or the photo is displayed small enough, there will be a large range of apertures where the loss of resolution either due to lens aberrations for apertures wider than the peak aperture or due to diffraction for apertures more narrow than the peak aperture, will not be noticed, none of which has anything, whatsoever, to do with being "diffraction limited".

+1. The formula Johnny uses can be justified under some assumptions, like a Gaussian blur. But it proves him wrong, as the plots illustrate (the formula, too, but I guess formulas are out of fashion nowadays).

If you really want a formula, then here we go. Bob, you should red this too.

We're going to consider two camera systems with sensors with resolution b and s when b > s.

We're also going to consider a lens with resolution at peak aperture of l_p.

We'll us units such that l_p = 1. If the resolution at ant aperture is l then 1 => l > 0.

If we want to compare the resolution of the two systems then we can use the formula

1/system resolution^2 = 1/lens resolution^2 + 1/sensor resolution^2

The resolution for sensor b is therefore r_bl = sqrt [(l^2 * b^2) / (l^2 + b^2)] which at peak aperture will be r_bp = sqrt [b^2 / (1 + b^2)]

Similarly for sensor s, r_sl = sqrt [(l^2 * s^2) / (l^2 + s^2)] which at peak aperture will be r_sp = sqrt [s^2 / (1 + s^2)]

When stopping down the lens, the relative drop in resolution for sensor b will be: Drop_rel_b = 1 - (r_bl / r_bp)

Similarly for sensor s, the drop will be: Drop_rel_s = 1 - (r_sl / r_sp)

The difference between these two relative drops is:

Drop_rel_b - Drop_rel_s = 1 - (r_bl / r_bp) - 1 + (r_sl / r_sp) = (r_sl / r_sp) - (r_bl / r_bp)

If this is equal to zero, then when stopping down to the same aperture then resolution drops at the same relative rate e.g. by 5% of peak resolution. If it is greater than zero, then the higher resolution sensor (b) experiences a greater drop relative to its peak resolution. Since its peak resolution is also great than for the lower resolution sensor (s), this will also equate to a great absolute drop.

So is there a greater relative drop i..e is Drop_rel_b - Drop_rel_s > 0?

Drop_rel_b - Drop_rel_s = (r_sl / r_sp) - (r_bl / r_bp)

= sqrt ([(l^2 * s^2) / (l^2 + s^2)] / [s^2 / (1 + s^2)]) - sqrt ([(l^2 * b^2) / (l^2 + b^2)] / [b^2 / (1 + b^2)])

= sqrt [(l^2 + l^2 * s^2] / (l^2 + s^2)] - sqrt [(l^2 + l^2 * b^2] / (l^2 + b^2)]

= l * ( sqrt [(1 + s^2] / (l^2 + s^2)] - sqrt [(1 + b^2] / (l^2 + b^2)])

= (l / sqrt [(l^2 + s^2) * (l^2 + b^2]) * (sqrt [l^2 + b^2 + l^2 * s^2 + b^s * s^2] - sqrt [l^2 + s^2 + b^2 * l^2 + b^2 * s^2])

If sqrt [l^2 + b^2 + l^2 * s^2 + b^s * s^2] > sqrt [l^2 + s^2 + b^2 * l^2 + b^2 * s^2] then Drop_rel_b - Drop_rel_s > 0

l^2 and b^2 * s^2 appear on both sides, so taking them out, if b^2 + l^2 * s^2 > s^2 + l^2 * b^2 then Drop_rel_b - Drop_rel_s > 0.

X = b^2 + l^2 * s^2 - s^2 - l^2 * b^2

If X > 0 then Drop_rel_b - Drop_rel_s > 0.

X = b^2 + l^2 * s^2 - s^2 - l^2 * b^2

= b^2 * (1 - l^2) - s_2 (1 - l^2)

= (b^2 - s^2) * (1 - l^2)

As established earlier, l <=1, therefore (1 - l^2) > 0.

Also as established earlier, b > s, therefore (b^2 - s^2) > 0.

Therefore X > 0 and Drop_rel_b - Drop_rel_s > 0.

Therefore the relative drop in resolution is great for a higher resolution sensor than a lower resolution one.

You do like wasting your own time. I don't think anyone has disputed that would be the case.

Therefore the absolute drop is also bigger.

Also probably true.

The final implication of this is that as sensor size decreases, the drop in resolution when stepping down constantly decreases.

Pixel count, you probably mean. Best not to complicate things with different sensor sizes at this point.

Given that human eyes have finite ability to distinguish resolution,

That statement is very ill formed, so much so as to be meaningless. What do we mean by 'resolution' and how does one 'distinguish' it? Best to stick with measurable things rather than make unevidenced and undoubtedly simplistic assumptions about human perception. What, for instance, is the role of acuity in all this? Is MTF50 a good model for perceived resolution?

there will come a point where the resolution at an aperture lower than the peak aperture will be indistinguishable from the resolution at peak aperture to the human eye.

However, I would hazard a guess that is probably true. So what? When the image is degraded so much that you can't distinguish between one f-number and another, are we really interested at all? Remember, the decision we're trying to inform here is the trade between DOF and resolution. Just thinking about that trade presupposes a concern for image quality which certainly won't be satisfied by a system so un-diffraction-limited that diffraction has ceased to be a player in resolution at any f-number (and interestingly, this 'diffraction limited' you're talking about has become precisely the reverse of what it means in an astronomical or microscope context). Let's just be clear, as camera with those properties is a dreadful camera, one no serious photographer would want.

That's exactly what I've said with words, illustrated with numbers and charts and have proved with numbers.

You've 'proved' nothing with numbers - working fictitious numbers cannot prove anything. Working fictitious anything can't prove anything.

-- hide signature --


Post (hide subjects) Posted by
Keyboard shortcuts:
FForum PPrevious NNext WNext unread UUpvote SSubscribe RReply QQuote BBookmark MMy threads
Color scheme? Blue / Yellow