About three weeks ago, I started teaching photography at The Avon Old Farms School. We are just getting around to learning about the exposure meters and the interplay between shutter speeds, apertures and of course film speeds.
I think I can safely say that most people, even students of photography, fail to understand this dance. The newer, fancier, digital cameras apparently do all the “creating.” Sigh!
Here is an easy one – If you are at f/5.6 and your shutter speed is 1/125 of second, what should your shutter speed be set to for a similar exposure at f/4? Answer: 1/250 of second. Right?
Well, as I was trolling on Twitter today, I noticed a challenge posted by my friend Jonathan Canlas. He is a terrific photographer out of Utah and yes, he still shoots film. (Follow him on Twitter like I do – http://twitter.com/jonathancanlas/)
Here is his challenge: 400 iso f6.3 @ 30 sec = 800 iso f(X) @ 8 sec.
If you have no idea what those numbers mean, I recommend taking a course in basic photography. Could those of you seasoned pros out there perhaps solve this riddle? Give it a shot below in the comments section.
Keep in mind that we are trying to determine the aperture value here, or f-stop number. I'll see if Jonathan will chime in to give us an explanation of how he came up with his correct answer.
I'll send the first person to post the correct answer a little surprise, so don't post anonymously.
Photography should still be fun, so if you feel stressed out about this little puzzler, take a deep breath and then tackle it. Go!
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Well i am the first one? My knowledge being so low … if the iso would hqve been the same i would have say X=5.6 but the iso is double and the shutter preed as well so X=6.3 ;)
Stephane
f/5.6
just kidding. it’s f/4.5
I believe it’s f/3.2 since you gain two stops from the shutter (approximately) and one stop from the ISO.
Looks like Larry got it first . it should be X=f4.5. ISO gives u + 1stop, Shutter speed give -2 stops so fstop should be 1 stop higher
ISO increases 3 3rds – shutter increases 3 3rds fstop needs to open up 3 3rds – am under the weather
I had to look in my camera and go down a stop in aperture to figure out it was 4.5. Didn’t know it off the top of my head.
I agree with Brian (3.2), but with different reasoning. Can’t put it into words per se, but, to make the equation on both sides equal, you’d dial in at an f-stop of 3.2
400 iso f6.3 @ 30 sec = 800 iso f(X) @ 8 sec. is
iso 400 to iso 800 = +1Ev
30sec to 8 sec = -2Ev (approx)
f:6.3 – f:4.5 = +1Ev
4.5
hahaha….I completely screwed that up…you LOSE two stops on the shutter and gain one stop on the ISO. 1-2 = -1, so f/6.3 becomes f/4.5. Basic math fail. :-)
I could be wrong, but I believe it goes like this. Since you are increasing the ISO and slowing the shutter speed to make the image brighter, you need to close up the aperture to compensate. Since the ISO was increased 2 stops (400 to 640, then 800) and the shutter speed was halved twice (30 to 15 then 8) I believe you need to go up 4 f stops to f/10. Lets in less light to compensate so you get correct exposure.
400 iso f6.3 @ 30 sec = 800 iso f(X) @ 8 sec
(note: I pasted the original equation in my post just to reference it.)
The answer should be f/18
I’m sticking to my answer…f3.2 it is in my book :)
I realized I should probably explain how I got to this answer. As we are trying to get an equivalent exposure, the changes made in the other settings must be offset by the aperture change. ISO went up 1 stop (makes image brighter), shutter speed went down 2 stops (makes image brighter), so the image is now 3 stops brighter if we keep f/6.3. In order to bring it back you closed the aperture 3 stops => f/18 (had to look up the scale as I don’t have the 1/3rd stop scale memorized.
Decreasing the shutter speed makes the image darker not brighter! There is less time to collect photons :)
Going from 1/30th to 1/8th of a second most certainly makes the image brighter.
Usually when you say the speed of something went down, it means it is moving slower. But I see what you mean in terms of this problem, I was assuming the standard convention of not denoting the fraction for shutter speeds (30 = 1/30, 8 = 1/8), if that is not the case then my calculation is incorrect and the shutter did in fact move faster providing 2 stops less light giving f/4.5
640 is not a full stop of ISO. A full stop for ISO or shutter speed is a factor of 2 so 400 to 800 is one factor of 2 and one full stop.
For aperture the factor is square root of 2, which leads to all even multiples of 2 being full stops: f/2, f/4, f/8 etc…
An easy way to remember the full stops is to realize that there are 2 stops between these multiples of 2, for example going from f/2 to f/4 is 2 full stops. Usually people are familiar enough with the numbers to remember the number between 2 and 4 that corresponds to a full stop, in this example it’s f/2.8.
Another example would be the full stop between f/8 and f/16 which is f/11.3
It’s simple to recreate the aperture scale with a hand held calculator, start with the square root of 2 which is roughly 1.4142 this gives us our first aperture f/1.4. Then we multiply by 1.4142 each time to get the next aperture.
f/1.4
f/2 (1.4142 * 1.4142)
f/2.8 (2 * 1.4142)
f/4 (2.8 * 1.4142)
f/5.6 (4 * 1.4142)
Note however that the above has some rounding error and to get the correct values you must consecutively multiply by sqrt(2) without rounding (so instead of using 2.8 you would use 2*1.4142=2.8284.
My apologies, I was looking at this in fractions of a second, doh!
I had it wrong anyway, I was looking at this in fractions of a second.
Good point about using decrease to mean shorter exposure time. Using increase or decrease can be ambiguous and should probably be avoided. These things are often confusing because the language used by people is inconsistent. I should have said decreasing the exposure time, or switching to a faster shutter speed.
I think using the terms fast and slow makes things easier to understand, the ISO changed to a 1 stop faster value, and the shutter speed changed to a 2 stop slower value, so the aperture must be made 1 stop slower.
What slow and fast means in terms of manipulating numbers in each case can be different. Slower ISO is smaller numbers, slower shutter speed is bigger numbers, and slower aperture is smaller numbers like ISO. But slower always means gathering more light, and faster less light.
Since I decided to read the equation PROPERLY this time, I also came up with 4.5. ~sigh~ Used to dealing in fractions of a second, photographing weddings is different than landscapes!
The shutter speed actually went up two stops. It went from 30 seconds (not 1/30) to 8 seconds (not 1/8).
Jason, This is exactly how I calculated it.
I’m not sure about fractional f-stops because my manual camera never had 1/3 stops, and I only use full-stops in aperture priority mode, but this would be my answer – 100%. f/4.5
BTW, Big thanks to Seshu for putting up this challenge.
Of course, when you’re with a camera, it’s easier to calculate… One Full ‘click’ for the ISO, and three 1/3 ‘clicks’ for the aperture and you’re at your target shutter speed. That’s how my mind works in the field. Yes, things can go wrong, but when you’re looking at composition, lighting etc, I find that it works.
PS: 1 ‘click’ is a notch on the camera’s adjustment wheel, in case you didn’t figure it out. ;)
Actually, in terms of lenses conventionally slower mean gathering less light (it takes more time to gather) and faster means gathering more light. An f/2 lens is “faster” than an f/4 lens.
Yikes, yes, thanks again for correcting my mistake. I meant to say slower aperture is *bigger* numbers! The damn thing is a ratio of the focal length over the diameter they could have just defined it the other way around and had larger numbers mean larger openings…
Gee whiz! I turn the other way and I see 29 comments here! Wow. I’ve asked Jonathan for an explanation. He said he had posted it on his Facebook account. Now unless you and he are buddies there, it’s unlikely you will know the right answer (though as I see it, it’s been mentioned here several times already).
does anyone use square root of 2 when it comes to the multiplication factor of f/?
4.5btw
Heh, good point =)
Yes that is how the f/ scale works (see my comment below). The reason is that the area of a circle goes like the diameter squared, so to half the amount of light entering you must change the diameter by a factor of sqrt(2). The aperture is simply the focal length divided by the diameter, so it also changes by a factor of sqrt(2).
“Now unless you and he are buddies there, it’s unlikely you will know the right answer.”
There is only one right answer and it doesn’t depend on being friends with Jonathan on facebook :P
Start with:
400 iso f/6.3 @ 30 sec
First change the ISO a factor of 2 (i.e. one full stop) from 400 to 800. This increases the sensitivity of the sensor so now less light is needed. To maintain a constant exposure the aperture must let in one stop less light. So f/6.3 (which is f/5.6 + 1/3 stop) becomes f/9 (f/8 +1/3 stop).
400 iso f/6.3 @ 30 sec = 800 iso f/9 @ 30 sec
Now consider the shutter speed which changes by a factor of 4, or 2 full stops. The exposure time is 2 stops less at 8 seconds than it was at 30 seconds. The aperture must allow more light in if the exposure is to remain constant, so two stops more light than f/9 (f/8 +1/3 stop) is f/4.5 (f/4 +1/3 stop).
400 iso f/6.3 @ 30 sec = 800 iso f/9 @ 30 sec = 800 iso f/4.5 @ 8 sec
Easy peasy lemon-squeezy!
I get closer to 4.6, because
sqrt(800/400) x sqrt(8/30) x 6.3 = sqrt(8/15) x 6.3 ~ 4.6