I'm sure there are numerous sighs out there with the subject title as this has probably been answered numerous times. I have an APS-C camera. I understand the crop factor and its effect on photos in relation to full frame with angle of view. There are two things that are still unclear to me. 1. Let's say I have a 50mm EF lens mounted, what am I seeing through the viewfinder? Is it he full frame size or the cropped size? If it were a 50 mm EF-S lens mounted, would I be looking at the same scene through the viewfinder? With the EF lens, would I have to make sure my shot is composed to account for my APS-C sensor? 2. I read conflicting information on aperature with different sensor sizes (or at least confusing in my mind). I've read that f8 is f8. I've also read that the crop factor must also be applied to the aperature which in my mind means f8 is actually f12.8. If I buy an EF f1.4 lens for 4x the cost, does this mean the widest aperature I can get is f2.2 regarding light to the APS-C sensor? Thanks for being patient with the newbie.

The cropped size. Yes. No. 50mm is 50mm. When setting exposure f/8 is f/8 and f/1.4 is f/1.4 the sensor size/crop factor has no bearing. When you see people apply crop factors to f/stop values what they're doing is comparing DOF (dept of field) between the different format sizes -- a topic I assure you that you want to avoid. Joe No problem.

F8 is always f8. Because of physics you'll generally be placed different to get same field of view from a crop sensor than a fullframe, so you see approx a 75 to 80mm filled of view looking through a crop camera with a 50mm lens. If you want the same field of view from a 50mm on a crop you'll stand further away than you would on a fullframe, so your depth of field is different, you'll have less dof on the fullframe

Through the Viewfinder is actually, when you get to the details a bit more confusing. With an APS-C camera you also have a smaller mirror relevant to it's size. It's that mirror the reflects the image to the viewfinder. Thus you see the 'crop' size in the viewfinder. BUT the "actual view" through the viewfinder (pentamirror, pentaprism) may not be 1.0x the visual size compared to the image at the sensor. It varies, for instance: Nikon D5600 & D3400 Viewfinder Frame Coverage 95% horizontal 95% vertical (Approx.) Viewfinder Magnification 0.82x (Approx.) Nikon D7500 Viewfinder Frame Coverage DX (24x16): 100% horizontal and 100% vertical (Approx.) 1.3x (18x12): 97% horizontal and 97% vertical (Approx.) Viewfinder Magnification 0.94x (50 mm f/1.4 lens at infinity, –1.0 m–1) (Approx.) Canon 80D Magnification Approx. 0.95x (with 50mm lens at infinity, -1m-1) / 28.3° angle of view So you'll have to check your specific camera specifications for the exact view you get from your viewfinder or LCD.

With a "through-the-lens" camera (like a DSLR) what you see through the viewfinder is what has already passed through the lens, so essentially the same thing. Some viewfinders are a little off, so if you habitually add a bit of extra space in the viewfinder, you can crop to the best crop later. No. That instruction was incorrect.

I shoot with a crop sensor camera, so here are my thoughts on the deal. Forget trying to correlate between FX and DX. You can drive yourself crazy for no reason. There is no magic to Full Frame, it is just used as a "least common denominator" when comparing field of view of different sensor size cameras. You could just as well use medium format, DX or 1" sensors as the base. Just learn how your DX camera works and forget how it compares to a full frame. After a while you will figure out how depth of field (or anything else) is affected by different settings on YOUR camera. I don't even worry about how field of view differs between FX and DX lenses. It doesn't matter. I just know what field of view I am going to get with a certain focal length on my DX. I don't care what the field of view would be on an FX.

You see the size in the viewfinder which approximates what the camera sensor will capture regardless of camera model. Many camera's viewfinders show an angle of view which is fractionally smaller than what the sensor will capture (so it's not 100% accurate, but it is close). Ok, this one is a bit more mixed.... and to be fair... you sort of asked three questions: a) is f/8 still f/8? b) do you apply the crop factor to the aperture? c) what about focal length multipliers such as a 1.4x teleconverter? So we'll take these in parts. First... part (a) which is really about exposure calculations If you use a lens directly attached to the camera body (any camera body - regardless of full-frame vs. crop frame) then... (and here's the qualifier) for purposes of calculating the exposure, f/8 is f/8 is f/8. This is the simplest answer you're going to get today. It gets more complicated from here. Second... part (b) though the exposure calculations don't change, that doesn't mean nothing changes... it turns out to have an impact on Depth of Field. For purposes of calculating the depth of field ... (and there are lots of caveats here) then an APS-C sensor camera (I'll assuming a crop factor of 1.6x because you're using a Canon camera) you have to divide the focal ratio of the lens by the crop factor of the camera, to get the equivalent f-stop for purposes of calculating the depth of field. But I really have to caveat this because nearly all depth-of-field calculators ask you what camera model you are using and the the reason they ask you is because they're using that info to do this part of the math for you. Example: Suppose I want to take a photo using a 50mm lens at f/2... and I want to know what the "depth of field" will be if I focus my camera on a subject located 10' away. If I am using a "full frame" camera then the answer is: Focus distance: 10' Near-limit at f/2 is: 9.33' Far-limit at f/2 is 10.8' Total DoF for 50mm f/2 at 10' focus distance is: 1.45'If I am using an APS-C "crop frame" camera (1.6x crop factor) then the answer is: Focus distance: 10' Near-limit at f/2 is: 9.56' Far-limit at f/2 is: 10.5' Total DoF for 50mm f/2 at 10' focus distance is: .91' Notice here... that if you divide 1.45 (the total DoF for a full frame camera) by 1.6 (the crop factor) you get: .91' (when rounded to 2 significant digits). But there are more caveats... because the angle of view for the APS-C camera will be narrower. If you were taking a portrait of a subject, you wouldn't get the same framing & composition if you use the same focus distance. To do that, you need to back away from the subject to 16'. When you do that (APS-C at 16') you get: Focus distance: 16' Near-limit at f/2 is: 14.9' Far-limit at f/2 is: 17.3' Total DoF for 50mm f/2 at 16' focus distance is: 2.36'So whereas in the first case (when we didn't re-compose ... just change the camera body) the APS-C camera got a shallower DoF. But when we re-compose to get equivalent framing, suddenly the APS-C camera is getting broader DoF. When you do a full-frame camera but adjust the focal ratio by the 1.6x multiplier of the crop frame camera, you get f/3.2. And when you do the math for f/3.2 you get: Focus distance: 10' Near-limit at f/3.2 is: 8.97' Far-limit at f/3.2 is: 11.3' Total DoF for 50mm f/3.2 at 10' focus distance is: 2.31'Notice how the total DoF of the last two examples compare... they aren't absolutely identical, but nearly identical. BTW, you can go to DOFmaster.com to plug in the values and experiment. But this is why for any given shot with a lens & focal ratio, composing an equivalent frame of view on full-frame vs. crop-frame... the full-frame camera will be capable of a shallower depth of field ... and more to the point, if you multiply the focal ratio you are using by the crop-factor of the camera, you'll get a value that can be used to determine what the equivalent DoF would be as compared to a full-frame camera. So there are some caveats to this (it's not entirely straight-forward) ... but there are clearly some different with respect to how changing the sensor size will impact the depth of field. Third... part (c) what about focal length multipliers (e.g telecompressors (teleconverters), and focal reducers)? It turns out these have a real change to both the focal length and the focal ratio. And this really makes sense when you think about what a focal ratio actually is. First... the clear point of the teleconverter is to change the effective focal length of the lens. But notice that these things commonly come in a 1.4x and a 2x size. Why? Why not have a 1.5x size? It turns out the reason is because they are based on the powers of the square root of 2 and the square root of 2 is approximately 1.4 (it is really an irrational number like Pi that just goes on and on... it's usually shortened to just 1.414 or even 1.41. But for purposes of photography, you're not going to notice any real change in the difference from 1.4 vs. 1.41 so they typically just shorten it to 1.4 (for purposes of exposure calculations it's close enough). Each "full" f-stop in the sequence is simply the (rounded) result of taking the current f-stop and multiplying it by the square root of 2 (which we rounded to simply 1.4). So f/1.0 times the square root of 2 is 1.4. And 1.4 times the square root of 2 is "2" (the square root of any number... times itself... is simply the original number again). We could keep going... 2 x √2 = 2.8. 2.8 x √2 = 4. 4 x √2 = 5.6. 5.6 x √2 = 8 ... and so on to get f/11, f/16, f/22, f/32, etc. The reason for this is because if you change the diameter of a circle by √2 then you either exactly double its area or exactly halve its area depending on whether you were increase (multiply by √2) or decreasing (divide by √2). If I have a circle which is 10mm across, then the area is Pi x r^2. r=5mm (half of 10mm). So 5^2 = 25 and 25 x Pi = 78.54 mm^2 -- that's our total area. Now if we multiply that 10mm by √2 we get 14.14... and the number goes on but half of that is the radius (approximately 7.07mm) and if we square that we get 50 (actually we get exactly 50... not just approximately 50) and notice how this is EXACTLY double the 25 value we used earlier. So now we just multiply 50 x Pi and we get a total area of 157.08 mm^2 (which is PRECISELY double the area). This math works both ways ... increasing or decreasing (just divide by √2 instead of multiplying by √2). So that's maybe a little more math than you wanted, but... What's a "focal ratio"? It's simply the focal length of the lens... divided by the diameter of the clear aperture opening in the lens. So if I have a 100mm lens and it has an aperture opening which is 25mm across... then that's an f/4 lens because 100 ÷ 25 = 4. Now suppose we attach the 1.4x teleconverter... The new effective focal length just increased by 1.4x so our 100mm lens is now an effective 140mm lens. But what did we do to the size of the aperture opening? Nothing. It's still 25mm across. So when we divide 140mm by 25 we get: 140 ÷ 25 = 5.6 Ok, so without the teleconverter we were at f/4 and with the teleconverter we're at f/5.6. Notice how f/4 and f/5.6 are EXACTLY one full f-stop apart? It's just as if we had decreased the size of our circle by √2. And there you have it. There's a lot to photography which is all based in math. You don't necessarily need to memorize all the math to be a good photographer, but you should be aware that it is based on math. BTW, the same √2 is heavily used as part of the "inverse square" law which, among other things, determines how rapidly light intensities decrease as the distance from the light source increases and this is the basis for how flash exposures are calculated.

on the aperture question.... f8 is f8 in terms of exposure. however, in terms of DOF you do need to apply the crop factor.

Photography or Technology??????????????????????????????????????????????????????? I'd prefer PHOTOGRAPHY.