[beaumont]

Right. I really botched that sentence. The beam coming out of the lens is still the same size but the target is smaller so part of the beam never hits the target (gets cropped).


I guess this is where the physics perhaps get muddled in my head a bit?

As I understand things, the captured image at the sensor is of fixed resolution, based on pixel density of the sensor. And whether I make a print at 4x6 or 12x18 the optical characteristics of the image remain constant. Bokeh, DOF, etc all happens between the leading edge of the forward lens element and the trailing edge of the rear lens element (or teleconverter).
Take an image from a 10MP APS-C (15x23mm) sensor and an image from a 10MP FF (23x35mm) sensor cropped to APS-C size. The APS-C image would simply enlarge to 4"x6" better than the cropped FF image due to pixel density (the cropped FF image would be only 6.5MP).
Teleconverters actually physically alter the optics and project back a magnified image of 'infinite' resolution. The teleconverter physically alters the DOF, bokeh, focal length and aperture characteristics of the "beamed" image itself.


Because of pixel density as I outlined above, wouldn't the same sized image (talking field of view here) from a FF image need to be enlarged more than a APS-C image of the same FOV?
Take 2 cameras with identical 50mm prime lenses; one camera is FF the other APC-S. Now take a picture with each of the same scene (the FF will capture have a wider FOV of course). Now crop the FF image to the same FOV as the APS-C shot. The FF sensor having fewer pixels per square inch would be lower resolution therefore would need to be stretched by 1.53x in order to make the same sized print.

I may be picking nits here, but I really do want to fully understand this subject matter.

Thanks

Wow, lot's of feedback. Hence the original comment..."I have no knowledge of the focal difference between using on a DSLR vs film nor do I probably need to understand it at this point".
Got my answer though..."is the focal length of the 77 truly only 7 longer than the 70?" The answer seems to be a unanamous "yes"...
Thx all.

[Marc Sabatella]

As I understand things, the captured image at the sensor is of fixed resolution, based on pixel density of the sensor. And whether I make a print at 4x6 or 12x18 the optical characteristics of the image remain constant. Bokeh, DOF, etc all happens between the leading edge of the forward lens element and the trailing edge of the rear lens element (or teleconverter).
Take an image from a 10MP APS-C (15x23mm) sensor and an image from a 10MP FF (23x35mm) sensor cropped to APS-C size. The APS-C image would simply enlarge to 4"x6" better than the cropped FF image due to pixel density (the cropped FF image would be only 6.5MP).

Sure, but who said anything abut cropping a FF image? I was talking about printing the entire image from both cameras, not cropping it after the fact from the FF sensor. The point being, the one from the APS-C sensor would look more "magnified" in *exactly* - and I mean *exactly* the same way the image looks more magnified with a TC. That is, image from APS-C camera would show face only, image from FF camera would show head-and-shoulders. Stick a longer lens or TDC on the FF camera - or, indeed, crop it - and it too would show face only, and hence also look more "magnified" when printed, in *exactly* the same manner as the APS-C camera.

Teleconverters actually physically alter the optics and project back a magnified image of 'infinite' resolution. The teleconverter physically alters the DOF, bokeh, focal length and aperture characteristics of the "beamed" image itself.

True, but I'm not (mostly) talking about DOF or bokeh - just field the field of view. And this is going to *identical*.

Because of pixel density as I outlined above, wouldn't the same sized image (talking field of view here) from a FF image need to be enlarged more than a APS-C image of the same FOV?

Only if you crop the image before enlarging it. I was never talking about that. And when i spoke of "magnification", I meant this in the naive sense of how big a pictured object looks on screen, not any actual calculation involving pixel density. That is, if you use a telephoto lens on a FF camera to picture a distant object - a bird, say - than print that picture (*without* first cropping it) at a given size - say, 12x18" - that bird will look bigger than it does to the naked eye. Now take that same lens and put it on an APS-C camera. Print the resulting image at 12x18", and the bird will look physically bigger on the print - more "magnified". And the difference in apparent "magnification" will be *exactly* the same as if one had used a TC or a longer lens on the FF camera, or had cropped the image first (although bokeh and DOF may indeed differ)

Take 2 cameras with identical 50mm prime lenses; one camera is FF the other APC-S. Now take a picture with each of the same scene (the FF will capture have a wider FOV of course). Now crop the FF image to the same FOV as the APS-C shot. The FF sensor having fewer pixels per square inch would be lower resolution therefore would need to be stretched by 1.53x in order to make the same sized print.

Sure, but once again, we were not talking about taking an FF image and then cropping it.

[Tirec]

A question on the "Crop Factor" with a twist.

My understanding is that using a 100mm lens on a DSLR (K100D, in my case) results in the the equivalent of using a 150mm on a 35mm film camera.

If I put a 2x teleconverter in the mix, would the teleconverter also be multiplied by 1.5 for an equivalent of a 3x teleconverter?

So would my 100mm lens with the 2x teleconverter return the equivalent of 450mm (150 crop factor of the 100mm x 3 crop factor of the 2x teleconverter)?

Or would it be 300mm ((100mm x 2x teleconverter) x 1.5 crop factor).

Curious how that all works together.

Thanks.

Tirec.

[Marc Sabatella]

If I put a 2x teleconverter in the mix, would the teleconverter also be multiplied by 1.5 for an equivalent of a 3x teleconverter?

No. Still 2X.

[pingflood]

A question on the "Crop Factor" with a twist.

My understanding is that using a 100mm lens on a DSLR (K100D, in my case) results in the the equivalent of using a 150mm on a 35mm film camera.

If I put a 2x teleconverter in the mix, would the teleconverter also be multiplied by 1.5 for an equivalent of a 3x teleconverter?

So would my 100mm lens with the 2x teleconverter return the equivalent of 450mm (150 crop factor of the 100mm x 3 crop factor of the 2x teleconverter)?

Or would it be 300mm ((100mm x 2x teleconverter) x 1.5 crop factor).

Curious how that all works together.

Thanks.

Tirec.

It's pretty simple. The crop factor applies to whatever the 35mm equivalent would be.

So a 100mm without converter = 100 * 1.5 = 150mm equivalent (I hate using this word but you get the point)

100 with 2x converter = 100 * 2 * 1.5 = 300mm equivalent

As for those claiming depth of field is the same as 35mm for a given focal length, please look up 'circle of confusion' and ponder how it applies when output at the same size.

[Marc Sabatella]

As for those claiming depth of field is the same as 35mm for a given focal length, please look up 'circle of confusion' and ponder how it applies when output at the same size.

I've tried to ponder this, but remain confused... I can see how the crop factor would preserve DOF at a given focal length or decrease it, and have no real way or measuring the effect.

[sewebster]

I've tried to ponder this, but remain confused... I can see how the crop factor would preserve DOF at a given focal length or decrease it, and have no real way or measuring the effect.

The DOF is defined as the distance between areas that have some (small) defined degree of "out of focus-ness." If you enlarge the image from a smaller sensor to the same output size as an image from a larger sensor then the perceived depth of field will be smaller (for the smaller sensor) because you "magnified" the image and therefore the "out of focus-ness" more.

The actual "image" that is projected on the sensor plane is the same in both cases, but if you have to blow up the image from the smaller sensor to a greater degree then it magnifies the imperfections.

[Marc Sabatella]

The DOF is defined as the distance between areas that have some (small) defined degree of "out of focus-ness." If you enlarge the image from a smaller sensor to the same output size as an image from a larger sensor then the perceived depth of field will be smaller (for the smaller sensor) because you "magnified" the image and therefore the "out of focus-ness" more.

Right, I do understand that as a general concept. What I don't understand is how this correlates with crop factor numerically. That is, when shooting, say, a 50mm lens at f/2.0, you get the field of view of 75mm. Some say you still have the DOF of the 50, and at *some* level this is true - in focus areas don't suddenly become OOF. But as we know, the greater enlargement does mean a little unsharpness becomes a lot. What I don't know is whether that effects we really do have the DOF typical of a 75mm lens on FF, or more, or less, or whether it depends on aperture, subject distance or anything else.

[Lowell Goudge]

Having jumped into the photo game after the introduction of DSLR, I have no knowledge of the focal difference between using on a DSLR vs film (nor do I probably need to understand it at this point.) However the question I have is when comparing the FA's to the DA's, are we comaparing apples to apples in the focal length? For example when using both FA 77 LTD and the DA 70 LTD on a DSLR is the focal length of the 77 truly only 7 longer than the 70?
Thx in advance...

to be clear, focal length, and this is not a function at all of sensor or film format, is the distance that a simple lens is from the focusing plane, for an object at infinity.

the only difference between film and digital SLRs is that a film SLR has a frame on the film that is 24mm high and 36mm long.

A DSLR has a sensor that is 16mm high, 24mm long, i.e. it is the center 2/3 of a film frame, nothing more nothing less.

[Marc Sabatella]

As another data point, consider a typical P&S camera. Very small sensor, so their focal lengths are actually quite small. Eg, the equivalent of a 50mm lens for a typical P&S might be 12.5mm (that is, the "crop factor" might be 4X). The whole "circle of confusion" business convinces me that this 12mm lens will produce pictures with somewhat less DOF than a 12mm lens would on FF. But by all accounts, it's a heck of a lot more DOF than a 50mm lens does on FF I've seen formulas for DOF that involve lots of different variables, but even as someone with a degree in math, it was more than I could bear to work through them to really answer the question: just how much *does* DOF change for a given crop factor.

[pingflood]

Right, I do understand that as a general concept. What I don't understand is how this correlates with crop factor numerically. That is, when shooting, say, a 50mm lens at f/2.0, you get the field of view of 75mm. Some say you still have the DOF of the 50, and at *some* level this is true - in focus areas don't suddenly become OOF. But as we know, the greater enlargement does mean a little unsharpness becomes a lot. What I don't know is whether that effects we really do have the DOF typical of a 75mm lens on FF, or more, or less, or whether it depends on aperture, subject distance or anything else.

The thing is, there is no such thing as 'in focus' and 'out of focus'. Nothing suddenly becomes OOF, it just gradually becomes less subjectively sharp. Since you have to magnify the same 50mm shot more on an APS-C (crop) size sensor than on a 35mm film sized one to get to say a 4x6 or 8x10 print, you are turning more of the 'subjectively sharp' part into 'out of focus', and thus end up with what is apparently less depth of field.

But I don't think it's as simple as 50mm DOF on APS-C = 75mm DOF on 35mm either...