What the heck is an f-stop?

Started 7 months ago | Discussions thread
Great Bustard
Forum ProPosts: 24,681
Like?
Re: Sensor size, f-stop, exposure, total light, and noise.
In reply to Paul Anderegg, 7 months ago

Paul Anderegg wrote:

That does help...

I'm pleased to hear it! I made an error in my post above, which I'd like to correct here. One of the paragraphs should have read (correction in bold):

So, an f/1.4 lens on a 1/3" sensor is equivalent to an f/2 f/2.8 lens on a 2/3" sensor, as it will project the same total amount of light on the sensor for a given shutter speed (given equal transmission), thus resulting in the same noise for equally efficiency sensors.

And, of course, substitute f/2.8 for the rest of the post.

...which brings me to another question. Assuming the two f10@2000 1/3" and 2/3" models considered above..........that f10 isn't really the same is it?

Well, f/10 will result in the same exposure for the same scene luminance, same shutter speed, and same transmissivity of the lens, regardless of the format.

I will give a hypothetical example based on two small camcorders I own. Sony CX900 and Panasonic x920. Both provide the same picture brightness (luminance, zebras set to 80IRE would zig and zag in the same spots) at wide open aperture and no gain (0db). Both also provide the same amount of sensitivity boost, and provide comparable ability to discern (brightness-wise) a line of trees at night against a lit skyline.

The Sony has a 1" sensor, the Panasonic a 1/2.3".....it's actually probably 1/4", but that isn't completely relevant.

Actually, it's quite relevant, which I'll discuss in a moment.

The Panasonic has a maximum aperture of f1.5, the Sony an f2.8. Due to the larger chip in the Sony, that f2.8 "rating" is enough to provide enough light on the sensor to provide a brighter picture than the f1.5 lightfall on the tinier Panasonic sensor. At all gain settings, the Sony is also MUCH cleaner.

The f/1.5 lens will be 1.8 stops (3.5x) brighter than the f/2.8 lens, but a 1" sensor has 4.1x (2 stops more) area than a 1/2.3" sensor. This means that f/2.8 on a 1" sensor puts 0.2 stops (1.2x) more light on the sensor than f/1.5 on a 1/2.3" sensor, so the noise would be about the same for equally efficient sensors.

However, a 1" sensor has 15.1x the area of a 1/4" sensor, which means that f/2.8 on a 1" sensor puts 2.1 stops (4.3x) more light on the sensor than f/1.5 on a 1/4" sensor. This means the f/2.8 lens on the 1" sensor would have half the noise as the f/1.4 lens on a 1/4" sensor for equally efficient sensors.

Noise aside, please take the following hypothetical into consideration when answering the follow-up question. Given as I said they produce the same luminance wide open at no gain/ISO boost, if the Sony was rated at f2.8@2000, and the Panasonic was rated at f1.5@2000, then those f10@2000 ratings for the professional camcorders would be similarly skewed? If both cameras, 1/3" and 2/3" provide that critical measured 100IRE at f10@2000, then would not the 1/3" camera be able to provide a brighter image at wide open iris, because it could open past f1.8?

I just need to know if the 1/3" camera/lens combo first proposed would have a brighter image (more IRE's!) at 0db, no gain, lowest settings, boost set to off etc, at full wide, and then again at full telephoto, given the lenses listed in the B&H links. I actually use the 18x4.2BERm lens every night for my job, and it doesn't get darker when you zoom in through to full telephoto. All cheap 1/3" mount "pro" lenses do ramp the aperture down, as do all 2/3" mount lenses. this little secret nugget of high end 1/3" lenses got me thinking they could actually beat out the larger sensor better low light sensitive 2/3" models simply do to the wins they see in the math game.

You need to distinguish between the brightness of the image projected on the sensor and the total amount of light that makes up the photo. Here's my write-up on it:

http://www.josephjamesphotography.com/equivalence/index.htm#exposure

and here's gollywop's excellent article on it:

http://www.dpreview.com/articles/8148042898/exposure-vs-brightening

In short, assuming equally transmissive lenses and equally efficient sensors, then all you have to know is the aperture diameters, where the diameter of the [virtual] aperture (entrance pupil) is the quotient of the focal length and the f-ratio.  The system using the lens with the wider aperture, all else equal, will have less noise, in proportion to the ratio of the aperture diameters.

For example, let's say we used 12mm f/1.5 on a 1/4" sensor.  The aperture diameter is 12mm / 1.5 = 8mm.  For the same perspective and framing on a 1" sensor, you would use 47mm where 47mm f/2.8 has an aperture diameter of 47mm / 2.8 = 17mm.

Assuming the same scene luminance, same shutter speed, same lens transmissivity, and equally efficient sensors, the 1" system would have 8mm/17mm ~ half the noise as the 1/4" system.

Hope this all helps!

Reply   Reply with quote   Complain
Post (hide subjects)Posted by
Keyboard shortcuts:
FForum PPrevious NNext WNext unread UUpvote SSubscribe RReply QQuote BBookmark post MMy threads
Color scheme? Blue / Yellow