determining magnification for use in DOF calculation

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bs0604
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Location: Sarasota FL

determining magnification for use in DOF calculation

Post by bs0604 »

I know that magnification is determined by "sensor size/subject size"
I have an APS-C sensor with size of 23.6 x 15.6 mm. Do I use an average of these 2 dimensions to plug in to the equation or just the smaller of the two sides?

hero
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Post by hero »

I think perhaps you might have misunderstood the formula. It isn't the actual physical dimensions of the sensor that is in the numerator, but rather, the size of the image of the subject projected onto the sensor.

Let's walk through a thought experiment. Suppose you take a photograph of a metric ruler whose markings are in millimeters, in such a way that those markings are in focus. When you are taking that photograph, what you are doing is you are exposing the sensor of the camera to light that projects an image of that ruler onto the sensor via a system of lenses. Now, the idea of magnification is to answer the question, how large is that projected image relative to the actual size of the ruler?

At 1:1 magnification, the projected size of the ruler equals the actual size. In other words, the spacing of the millimeter markings of the image projected onto the sensor is the same as the spacing on the actual ruler.

Now, at 2:1 magnification, the projected image onto the sensor is twice the size of the actual ruler, so the spacing of the millimeter markings of the projected image is 2 mm. And at 10:1 magnification, each millimeter marking would be separated by a full centimeter (10 mm).

As you can see, the physical dimensions of the sensor do not come into play in this calculation. Think of the sensor as the screen we might use when showing a movie, or looking a slideshow. The image that we project onto the screen has been enlarged, typically to fill the screen, but the size of that image would not be any different if the screen's dimensions were changed, so long as the projection or the screen's distance from the projector has remained the same.

So, for our purposes, magnification is purely a function of the relative size of an subject's projection to the actual size of the subject. The size of the projection may depend on the optical setup (what lenses are used, and their positions relative to the subject and the sensor), but the size of the sensor does not matter.

If we were to make a print of the sensor's output, typically there is a further enlargement that occurs in this step (unless it's a contact print). This step is independent of the aforementioned magnification and has no relationship to the optical setup. To use an analogy from the days of film, I might choose to create an 8x12" print from my 35 mm negative, but someone else might choose to print it at 24x36". The amount of enlargement here is at the discretion of the person making the print, and is the ratio of the print size to the sensor size; but it is not "magnification" as we use the word here.

Pau
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Post by Pau »

This is just a linear relationship: if you take a picture of a ruler placed following the long side, 23.6mm is the value you must enter, and the image of the ruler tells you the subject factor.
(of course if you do it at the short side the result will be the same)

edit: Hero was faster and made a more elaborated explanation
Pau

bs0604
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Joined: Tue Feb 19, 2013 8:09 am
Location: Sarasota FL

Post by bs0604 »

So if I have a flower petal which is 10 mm x 5 mm and it just fills my sensor in its longitudinal dimension--then 23.6 would be my numerator and 10 mm my denominator?
If correct, then if the subject does not completely fill the sensor, you have to do a bit of estimating regarding how many mm of the sensor are filled?
Also we are assuming that my camera's LCD screen represent the full size of the sensor?

ChrisR
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Post by ChrisR »

You got it ;)
LCD screens usually clip a fraction off the sensor image, but they often don't say.
Remember if it's "95%" by area that's 97.5% linear, so it'll hardly wreck your calculation.

Experiments with a ruler clear things up, of course.
Chris R

bs0604
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Location: Sarasota FL

Post by bs0604 »

I am missing something. I am determining magnification to put in to a formula or calculator to determine # of steps required for my stackshot. But the method outlined throughout this post is taking in to account the two dimensional accept of the subject and not the depth of the subject which is the important issue. I you have a 1 mm x 1mm subject that is 10 mm deep it is going to require more steps that a 1x1 mm subject which is only 2 mm deep. The determination of magnification is only taking in to account the 1x1 aspect and not the depth???

hero
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Post by hero »

Typically, magnification determines a step size, not number of steps.

The step size tells you how much you need to change the camera-to-subject distance between each shot, but in itself doesn't determine how many shots you need to take. If you determine that for your particular magnification and your subject that you need a step size of 0.1 mm, then if you want a total depth of 1 mm to be in focus on your subject, you'd need 1 mm / 0.1 mm = 10 shots. But if you want a depth of 5 mm to be in focus on your subject, you'd need 5 mm / 0.1 mm = 50 shots. The step size is the same, but the number of shots needed is also proportional to how much of the subject you need to be in focus once the images are stacked.

In summary, (number of steps) x (step size) = total depth you want subject to be in focus.

bs0604
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Joined: Tue Feb 19, 2013 8:09 am
Location: Sarasota FL

Post by bs0604 »

thanks. I realized this an hour after I placed my post. I need to refrain from posting at 3:00 am :P

ChrisR
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Post by ChrisR »

Posting can can risk a thinko ; but I try to make a rule to keep off ebay at 3am.
Chris R

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