fredrick_rege

Thanks, Alan and C.G. The words that got me to understand this are "the same amount of light reaches the <b>sensor</b>". I guess that even though the aperture has twice the diameter, the physics of the light reaching the sensor makes it a wash. I'll dig into the math a little more to figure out why this is so. I believe you, I just need to prove it to myself:)

Thanks, C.G. But let's hold f-stop and exposure constant. In that case, shutter speed would need to be halved, right?

 

Basically this is a practical question... I'm have a subject for which there is one correct exposure. I'm indoors and need to shoot with a wide open lens. Does the 100mm f/2 get me twice as fast a shutter?

The title should be "longer is faster than shorter, right?"

Bare with me as I set up this question:

 

Imagine I have two lenses: both 50mm primes. Lens "A" has a maximum aperture of

f/2. Lens "B", when wide open, is at f/2.8 (one stop down from "A").

 

If I'm shooting both lenses wide open (in aperture priority mode), it makes

sense to me that "A" would be twice as fast as "B". For the former, given the

same composition and exposure, the shutter would need to be open for half as

long as the latter to capture the same image. This makes sense to me. (correct

me if I'm wrong, please... I'm new to this)

 

But what if I stop both lenses down to say f/4? My assumption is that they

would perform equally as fast (but the optics within the lens may vary image

quality). Correct?

 

Question:

 

Now let's make things a little more interesting... I trade in Lens "B" for a

100mm f/2 prime lens. Let's call it lens "C". If I shoot both "A" and "C" at

f/2, which one is faster?

 

From what I've read, the f-number(N) is equal to focal length(f) divided by

aperture diameter(d). In symbolic notation, N=f/d. So in this example,

consider lens "A":

 

2=50/d --> d=50/2 --> d=25 --> the aperture's actual size is 25mm.

 

The algebra for Lens "C":

 

2=100/d --> d=100/2 --> d=50 --> the aperture's actual size is 50mm.

 

So by my algebra, "C" has 4 times as much light gathering area as "A"

(a=pi*r^2... remember:)

 

In other words, the light gathering capability (at f/2) of "C" is 2 stops up

from "A" (halve the area of "C", then halve it again, and you get the area of

"A"). Correct?

 

So for a given composition, "C" would be 4 times as fast, even though they're

both at f/2? I've gotta be missing something here, right? Maybe I'm not...

maybe this is why long, fast lenses are desirable.

 

By the way, the DOF in "C" would be shallower than "A" in this example too, correct?

I have a recently purchased a Cannon XTi SLR. I found that opening jpg images directly in Photoshop CS2 (version 9.0.2) defaults to 72 ppi (and yes, I set the camera to Large/Fine - 3888X2592).

 

I wanted to create 6" X 4" prints... when I changed the Document Size to 6" wide X 4" high, I did not get the image quality I expected. This was a consequence of (as Will Hammond mentioned) the total number of pixels used in the image: only 432 X 288 pixels.

 

I found that if:

 

- I open the original jpeg in "Digital Photo Professional, version 3.0.1.5" (it came on a CD with my XTi)

-then go to "Tools|Transfer to Photoshop"

 

the same image opens at 350 ppi! After doing my 6" X 4" resize, the Pixel Dimensions were 2100 X 1400 - almost 5 times better than the 72 ppi version.

 

Don't ask me to explain why or how Photoshop opens the same image at different resolutions. All I know is that (when printing) I got better quality with the higher resolution.

 

To Will's point, I probably should be somewhere between 350 ppi and 72ppi. My point though is that the 72 ppi default was not sufficient.

 

If you're intent is to use your images on the web, this need not be a concern. In fact, you should consider the "Save for web" option in Photoshop to extract metadata that you may not want out there.

 

FORege