EF-S 17-55mm f/2.8 IS or EF 24-70mm f/2.8L

[Daniel Browning]

If you absolutely cannot ever sell any lenses, and you will never use a backup body (i.e. keep your 50D and 17-55), then buying the 17-55 would be a mistake, because after you upgrade it will sit in the closet. On the other hand, if you are open to the idea of selling your lens (most often within 20% of what you paid for it), or cna utilize a second body in the future, then it would be far wiser to buy the 17-55.

That said, they're really not comparable at all. The closest thing to a "full frame compatible" 17-55 is the 16-35 f/2.8 or 17-40 f/4. The 24-70 is just a completely different field of view.

The 17-55 on 50D is equivalent to a 28-90 f/4.5 on full frame.
The 24-70 on 50D is equivalent to a 38-110 f/4.5 on full frame.
Going from 28mm to 38mm is very different. And going from 38mm to 24mm is even more different.

[Mark Elberson]

The 17-55 on 50D is equivalent to a 28-90 f/4.5 on full frame.
The 24-70 on 50D is equivalent to a 38-110 f/4.5 on full frame.
Going from 28mm to 38mm is very different. And going from 38mm to 24mm is even more different.
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<div><span style="font-size: small;"]I found this insteresting...

The difference in focal length at the short end (17mm vs. 24mm) is only 7mm but the difference in FOV is about 18 degrees.
The difference in focal length at the long end (55mm vs. 70mm) is 15mm but the difference in FOV is under 6 degrees.</div>

[George Slusher]

I found this insteresting...
<div style="clear: both;"]<span style="font-size: small;"]
The difference in focal length at the short end (17mm vs. 24mm) is only 7mm but the difference in FOV is about 18 degrees.
The difference in focal length at the long end (55mm vs. 70mm) is 15mm but the difference in FOV is under 6 degrees.

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It's not the numerical difference but the ratio of the focal lengths
that is important, though that doesn't directly translate into FOV in the same ratio. It
does translate into magnification, which can be related to field of
view.



Here's an example: 10mm to 20mm is a factor of 2, so, if an object at infinity has an image 1mm long on the sensor at 10mm focal length, the image will be 2mm long on the sensor at 20mm focal length. (The magnification doubles--but that's strictly true only for objects at infinity, i.e., pretty far away.)


Now start at 100mm. If the image of an object at infinity is 1mm long on the sensor at 100mm, at 110mm (same numerical difference as between 10mm and 20mm), the image will be 1.1mm long. To get the image to be 2mm long, you'd have to go to 200mm.


To put this another way, an image that fills the frame in one dimension at 20mm will be only half as wide/long at 10mm. The same ratio would hold for 200mm and 100mm or any other pair of focal lengths that are in the ratio of 2:1.


Field of view can be related to magnification with trigonometry. You can look this up on Google, if you like.


The same sort of relationship can be found between f-stops. f/1.4 and f/2 are only 0.6 apart, but that's the same ratio as between f/11 and f/16, which are 5 units apart. Both are "one stop" apart. An exposure that requires 1/30 sec at f/1.4 would require 1/15 sec at f/2. An exposure at f/11 that requires 1/30 sec would require 1/15 sec at f/16. In the case of f-stops, the operative comparison is the square of the ratio of the stop numbers. (They go up in steps of the square root of 2. The "standard" f-stops are: 1.0, 1.4, 2.0, 2.8, 4.0, 5.6, 8, 11, 16, 22, 32. Each of those is one stop from its neighbors. (One stop = a difference of 1 EV.)


The square root comes in because f-stop is defined as the ratio of the focal length to the effective aperture diameter. Thus, an f-stop of 2 would mean that the effective aperture is half the lens focal length. An f-stop of 4 would mean an effective aperture 1/4 the focal length. Thus, the ratio of the effective aperture diameters for f/4 vs f/2 would be 2. However, it's the AREA of the aperture that affects exposure. The amount of light that reaches the sensor or film is proportional to the time the shutter is open and the area of the aperture, which is proportional to the SQUARE of the diameter. Each successive "stop" represents a ratio of DIAMETERS by the square root of two and thus a ratio of AREAS by 2.


For example, the effective area of an lens's aperture at f/2 would be *4* times the area of the same lens at f/4--two stops = 4x the light. For the same exposure, you could make the shutter speed 1/4 as long (e.g., from 1/15 sec to 1/60 sec).


Then, there are Guide Numbers for strobes ...