Andyx01 said:
We are talking about comparable resolution capabilities, the bit you sum up as "If your lens has the resolving power to see the added resolution", that is the bit we are talking about.
I fail to see why crop v.s. FF is part of this topic when you clearly and accurately suggested it is about sensor density v.s. lens resolving power. The original author might as well have posted "Estimating extra reach (resolving power) of Canon v.s. Sony. - Makes as much sense.
Which brings us full circle - There is no extra 'reach' on Crop v.s. Full Frame, the diffraction limits are the same either way.
The Subject should read:"Estimating extra reach (resolving power) increased sensor densities provide for a given lens.
It doesn't really matter what the title of the thread is. The simple FACT of the matter is that smaller sensors nearly universally have smaller pixels than larger sensors. ONE large sensor has a lot of pixels, the 36.3mp Exmor, however even it's pixels are still larger than the pixels on the vast majority of modern day crop cameras. It doesn't matter if we call it "Estimating the extra reach of crop vs FF" or "Estimating the extra reach of small vs. large pixels". That's quibbling over semantics. Since smaller sensors are nearly guaranteed to have smaller pixels, the current title works fine.
Next, however, the whole notion that either a lens or sensor "sees" the resolving power of the other is a fallacy. I've explained this countless times on these forums, but here it goes again.
OUTPUT RESOLUTION (the measurable resolution of the image produced by a CAMERA, for the purposes of this post, defined as Lens+Sensor), is the CONVOLUTION of the resolution of the real scene as it's light passes through the lens and is recorded by the sensor. That's the key word here, convolution. Cameras convolve information. While it is possible for a lens to resolve 86lp/mm at f/8, and a sensor to have the ability to separate 116lp/mm, the notion that the sensor "outresolves" the lens at f/8 is a misnomer. What is really happening is the lens and sensor are working together to produce a BLUR SPOT. The size of that blur spot is what determines the resolution of the OUTPUT IMAGE.
We can very closely approximate the resolution of lenses and sensors by using the following formula:
Code:
blurSpot = SQRT(lensSpot^2 + sensorSpot^2)
The spot size of a lens can be computed by multiplying the resolving power in line pairs per millimeter, multiplying by two, and taking the reciprocal:
We can further convert the blur spot into spatial resolution by using the following formula:
Code:
spatRes = (1/blurSpot) / 2
We can combine these formulas into one single formula to take :
Code:
spatRes = (1/SQRT(lensSpot^2 + sensorSpot^2)) / 2
If we have a 1D X, 5D III, D800, 70D, and D5300 then (let's just assume they are monochrome sensors, for the sake of simplicity) each of those has a sensor spot of:
Code:
1DX: 6.92µm
5DIII: 6.25µm
D800: 4.9µm
70D: 4.16µm
D5300: 3.9µm
If we use the same theoretical lens, one which performs ideally at all apertures, on all five of these cameras, at apertures of f/2, f/4, and f/8, then the lenses DIFFRACTION LIMITED resolving powers are:
Code:
f/2: 346lp/mm
f/4: 173lp/mm
f/8: 86lp/mm
Converting these to spot sizes:
Code:
f/2: 1/(346*2) = 0.0014mm (1.4µm)
f/4: 1/(173*2) = 0.0029mm (2.9µm)
f/8: 1/(86*2) = 0.0058mm (5.8µm)
Running the numbers, we get the following:
Code:
1DX f/2: (1/SQRT(0.0014^2 + 0.00692^2)) / 2 = 71lp/mm
1DX f/4: (1/SQRT(0.0029^2 + 0.00692^2)) / 2 = 66.8lp/mm
1DX f/8: (1/SQRT(0.0058^2 + 0.00692^2)) / 2 = 55.5lp/mm
5DIII f/2: (1/SQRT(0.0014^2 + 0.00625^2)) / 2 = 107lp/mm
5DIII f/4: (1/SQRT(0.0029^2 + 0.00625^2)) / 2 = 94.6lp/mm
5DIII f/8: (1/SQRT(0.0058^2 + 0.00625^2)) / 2 = 68.6lp/mm
D800 f/2: (1/SQRT(0.0014^2 + 0.0049^2)) / 2 = 134lp/mm
D800 f/4: (1/SQRT(0.0029^2 + 0.0049^2)) / 2 = 111lp/mm
D800 f/8: (1/SQRT(0.0058^2 + 0.0049^2)) / 2 = 74lp/mm
70D f/2: (1/SQRT(0.0014^2 + 0.00416^2)) / 2 = 153.5lp/mm
70D f/4: (1/SQRT(0.0029^2 + 0.00416^2)) / 2 = 121lp/mm
70D f/8: (1/SQRT(0.0058^2 + 0.00416^2)) / 2 = 76lp/mm
D5300 f/2: (1/SQRT(0.0014^2 + 0.0039^2)) / 2 = 161.6lp/mm
D5300 f/4: (1/SQRT(0.0029^2 + 0.0039^2)) / 2 = 125lp/mm
D5300 f/8: (1/SQRT(0.0058^2 + 0.0039^2)) / 2 = 77lp/mm
Again, these are all theoretically diffraction limited apertures. Assuming such a case, small pixels, even the very small pixels of the D5300 are STILL resolving more detail at a diffraction limited f/8, which has a maximum theoretical resolution of 86lp/mm, than any of the full frame cameras with bigger pixels. There are diminishing returns, however the D5300 still enjoys over a 4% OUTPUT IMAGE resolution lead over the D800, and it enjoys a very large lead of 12.3% over the 5D III and a whopping 38.7% lead over the 1D X.
This is a DIFFRACTION LIMITED sensor. The notion that a higher resolution sensor cannot benefit at fully diffraction limited, narrow apertures like f/8, is patently false. The notion that a lens that is not resolving more than the sensor "cannot see" the resolution of the sensor is patently false. The two, lens and sensor, WORK TOGETHER to produce the final output resolution. In actuality, the specifics are certainly more complicated. Lenses tend NOT to be diffraction limited at wider apertures, and optical aberrations, of which there are many that affect the convolution of the incoming wavefront in different ways, will limit resolution at wide apertures on many lenses. However the same rules apply...for any given lens spot, regardless of whether it is limited by diffraction or aberrations, is going to CONVOLVE with the sensor. Higher resolution sensors, while they will eventually reach a point of diminishing returns, are STILL going to resolve more detail than lower resolution sensors.
)
;D
