Is my thinking about equivalence right?

Started 3 months ago | Questions thread
bobn2 Forum Pro • Posts: 69,811
Re: Is my thinking about equivalence right?

rogerstpierre wrote:

bobn2 wrote:

rogerstpierre wrote:

bobn2 wrote:

Olympus has to produce double the resolution of the Sony just to match it in a final image, due to being magnified twice as much, then there's not much of a comparison. You have to compare the Olympus lines with the lined for half the lp/mm on the Sony. On FF 10 lp/mm corresponds to 215 lp/ph, whilst on FF that final resolution is given by 20 lp/mm, so if we compare the two at the same final resolution, we find that the Sony gives an MTF of 0.93 in the centre and 0.85 at the edge, whilst the Olympus gives 0.77 in the centre and 0.69 at the edge.

You can't talk "magnification" with a digital sensor like you do with a film substrate. On a digital sensor the number of pixels, or data points you may want to call it, defines the size of the image, not the size of the sensor itself. Hence a sensor 1/2 the size with twice the number of pixels will produce an image twice larger when viewed at the same "magnification". This is where I don't understand when magnification is used as a variable to compare DoF between different sensor size. It has no relevance in a digital world, rather the total number of pixels that makes up an image does.

Call it 'enlargement', or what you want. Whether the mechanism is chemical or electronic. the image that the lens projects has to be enlarged to the size that you want ti view it. If your image frame is 17.3x13mm it needs to be enlarged twice as much to get any given size output as it does if it is 36x24mm. Your point has no impact on what I was saying, that the image the lens projects must be enlarged twice as much on mFT as on FF. It also has no impact on DoF, where the important factor is the saze of the blur in the viewed image. To take into account the doubled enlargement DoF calculations for mFT use a CoC that is half the size as they do for FF, resulting in the same size CoC in the viewed image.

You obviously don't understand how a digital image can be viewed.

With respect, I don't think that the problem here is with my understanding.

The size of the photosite that makes up the data point has no relevance when it come to the representation of a single RGB value. 1pixel is 1 pixel regardless of how big the photosite that captured the data is.

Erm, yes...and?

A 20mpix image viewed by any means will always be the same size regardless of how large the sensor that captured the image is.

ah.. You're assuming that images never get resampled on the way to being viewed? That's incorrect. They almost invariably get resampled.

The only difference from the sensor's pov will be in the signal to noise ratio.

We aren't talking about the sensor's point of view. The only points of view that matter with respect to what I was talking about is how large is the camera's image frame and how large is the size at which you view the photo. As I said, whatever is the mechanism used, and image that is tiny has to be enlarged to one that is big enough for you to view. The size of the pixels (or grains) is irrelevant to this. The only thing that they affect is the reproduction quality that you will get after enlargement.

The larger the photosite, the stronger the signal.

First, that's simply not true. The 'strength' of the signal depends on the amount of light energy collected by the photosite, which is only loosely coupled to its size. Second, even were it true, it's completely irrelevant to this discussion. The 'strength' of the 'signal' has no bearing at all on how much the image that the lens projects onto the sensor needs to be enlarged before viewing.

The quality of the lens optical components however, will dictate the resolution of the projected image, hence M43 lenses have to be optically superior to equal the same resolution with an image circle that is 1/2 the size indeed but that has nothing to do with the visualization of the image captured once it has been captured.

It has everything to do with that. It is precisely caused by the fact that to view the mFT image the same size it needs to be enlarged twice as much, where 'enlargement' means the ratio of the linear sizes of the camera's image frame and the image display frame.

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