F3.5 to F2.8 lens, is that going to make a big difference ?

Mountainsandh

Just starting to figure out lenses, so could do with bit of help here.
I want a lens to take pics of people in a pub.
It's a small place, at least 3 or 4 people playing music, the usual small old pub setting. Ideally I would like to shoot without flash.
So I figure I need as wide an angle as I can, and as fast a lens as possible.
All I have are my kit lenses (Pentax), and the widest angle is the 18-55mm F3.5, and a very very tight budget.

I'm not after a super quality lens, simply can't afford it, and not experienced enough to spend that kind of money yet.
So I was looking at the 24mm F2.8 lenses on Ebay, and there are some right cheap ones.

Would the 2 stops (is it 2 stops ??) make much difference ? Do you guys/girls think it's worth it ? or should I just make do with the kit lens ?
Thanks !

magicbastarder

it's not even close to two stops difference - it's less than a stop in difference, actually.

to work out the stop difference, square the max aperture value. aperture value is a function of the diameter of the aperture, but the amount of light is a function of the area of the aperture, hence the squaring.

joolsveer

I think it is half a stop so it will only make a slight difference.

Mountainsandh

Magicbastarder haven't a clue what you said, but I gather it won't make a difference anyway ?:o :D

thks a mill for replies btw, saving me money so

5uspect

f-number refers to the diameter of the aperture pupil but it's the area of the pupil that matters when it comes to f-stops. A pupil area of twice the size lets in twice as much light.

So that's why f-stops are measured in the sequence f/1.4, f/2, f/2.8, f/4 f/5.6, etc. each stop let in half as much light (or doubles it if you're opening the iris)

A quick calculation to determine the light gathering ability of two different f-numbers is to raise the f-numbers to the power of -2 (square it and invert it) and divide the two.

so:
(2.8 ^(-2)) / 3.2 ^(-2)
=
1.3

So you get 30% more light with an f/2.8 lens

If you were to get a 50mm f/1.8 you would get three times as much light.

Heebie

It won't make a big difference.. but you might find that the quality of the lens itself could be much higher on the 2.8.. you might also find that it's "sweet spot" where it really looks the best is in a different.. much more desirable place in the range. (the f3.5 might have it's sweet spot at 8 or 16 and the 2.8 at 5.6 or even 3.5)

I would probably see if I could bring the f3.5 somewhere, along with the f2.8 and shoot the same interesting thing at around the same settings on both lenses. (pick something with lots of details, and lots of different levels to it.. shoot wide-open, then 1-2 stops in, then around f8, 16 and 32) Just to be able to take a good look & see how the two lenses stack up against each other.

You could find that there's almost no difference between the two, or you could find that one blows the other out of the water. If the latter is true.. then it's a good buy.. if the former is true.. then it's a no-go.

Thecageyone

The difference will be in the IQ, as pointed out. If you use any post-processing program at all, you can brighten the images a little later on anyhow.

The 18-55 is a solid little lens, underestimated. Only thing is, you'll end up zooming in, it's just too hard to resist on a zoom of any range, and then you'll lose that f/3.5 and need to use longer exposures.

Up the ISO, use as slow a shutter speed as you can comfortably hand-hold and move your fet instead of zooming, try keep the lens wide - good luck

Mountainsandh

Thks all.
Heebie don't know anyone who could give me a loan of a 2.8, so it's a pity, but I take your point and if the occasion ever arises again and I can compare 2 lenses this way I will.

TheCageyOne, thanks a lot, brilliant advice, you're right about the zoom but thanks to you I will be zooming with my feet on the night .

Thankfully my Pentax k-x is really good in low light, and I discovered recently I was much better at hand holding than I thought... and the final shoot is the perfect excuse for me to go out and shoot other sessions beforehand so I can practice .
There will be some light on in the pub, but I'm also thinking of planting a few nice and pub like candles on the table, not intrusive, but maybe something to help light up those faces and improve the lightbulb glow.

The owners are friends and fine with me experimenting at leisure and as often as I like so it should be fun.

charybdis

5uspect wrote: »

f-number refers to the diameter of the aperture pupil but it's the area of the pupil that matters when it comes to f-stops. A pupil area of twice the size lets in twice as much light.

So that's why f-stops are measured in the sequence f/1.4, f/2, f/2.8, f/4 f/5.6, etc. each stop let in half as much light (or doubles it if you're opening the iris)

A quick calculation to determine the light gathering ability of two different f-numbers is to raise the f-numbers to the power of -2 (square it and invert it) and divide the two.

so:
(2.8 ^(-2)) / 3.2 ^(-2)
=
1.3

So you get 30% more light with an f/2.8 lens

If you were to get a 50mm f/1.8 you would get three times as much light.

I think you might want to re-check that math. Firstly, the two f-numbers in question are f/2.8 and f/3.5. not f/2.8 and f/3.2.

A more accurate way to calculate the difference in stops between two f-numbers is to convert them to their value in stops by finding their log to the base root two as all "integer" f-numbers are the integer powers of root two.

So:

  log[sqrt(2)](3.5) - log[sqrt(2)](2.8)
= (ln(3.5)/ln(sqrt(2))) - (ln(2.8)/ln(sqrt(2)))
= ~3.6 - ~3
= ~0.6

i.e.: two thirds of a stop, not one third.

Also, an f/1.8 lens is two stops faster than an f/3.5 lens, so you'd actually get four times as much light.

Mountainsandh

charybdis wrote: »

I think you might want to re-check that math...
So:
...

  log[sqrt(2)](3.5) - log[sqrt(2)](2.8)
= (ln(3.5)/ln(sqrt(2))) - (ln(2.8)/ln(sqrt(2)))
= ~3.6 - ~3
= ~0.6

i.e....

Also...QUOTE]
:eek::eek: glups. I love photography and goodness knows I'm willing to learn, but I hope I don't ever have to figure all those maths because there is no way me littl' brain could handle all that.:o

but I appreciate the help

Kintarō Hattori

I hate the maths in photography. I know it's worthwhile but I hate it.

Thecageyone

I can't get with it either. I was always useless at maths, no change there.

Atm I'm using a manual lens that requires you to set the f-stops by rotating a clicking ring - I just guess what might work best in the current lighting, and I'm rarely far off, the odd re-adjustment maybe. It's got constant f/3.5 if needed, but so far that's been too wide as I've only really shot with it outdoors in strong lighting. It's been used at f/8 - f/14 most. Indoors I'll just use the 50mm.

I just say it like it is ... the lower the f-stop number, the more light you'll get, but the danger of unwanted softness. Simple :P

Mountainsandh

Kintaro H. I'm desperate at maths, a right disgrace, glad to know I'm not the only one. I'm sure eventually I'll have a better idea, like, in 20 years...;)

Thks TheCageyOne, that softness/sharpness/aperture thing is something I haven't really looked into yet, I'm self teaching and happy enough to understand aperture/shutter speed basics, iso, and even focal length a bit better. That sharpness/aperture relationship (or maybe lens performance like Heebie was hinting at, rather than basic softness/sharpness ?) is something that has been niggling at me to explore of late, has to be the next thing on my list. Although metering is a big issue too... so many things to learn.

charybdis

Kintarō Hattori wrote: »

I hate the maths in photography. I know it's worthwhile but I hate it.

Nobody really needs to know how to do this stuff for the purposes of most photography. In the same way that when asked "What is 2 + 2?" you don't actually add two and two in your head, you just remember the answer, when thinking about exposure and its related properties it's helpful to just think in stops so you don't have to try and work things out in your head when you want to adjust something. Everyone should understand what a stop is and everyone should be familiar with the basic "integer" stop scales for aperture (... f/1, f/1.4, f/2, f/2.8, f/4, f/5.6, f/8, f/11, f/16, f/22, f/32 ...) shutter speed (... 1, 1/2, 1/4, 1/8, 1/15, 1/30, 1/60, 1/125, 1/250, 1/500, 1/1000 ...), and ISO (... ISO 50, ISO 100, ISO 200, ISO 400, ISO 800, ISO 1600, ISO 3200 ...). If you understand what a stop is and somewhat memorise the basic stop scales, you'll understand exposure as much as you'll realistically need.

Mountainsandh

charybdis wrote: »

f/1, f/1.4, f/2, f/2.8, f/4, f/5.6, f/8, f/11, f/16, f/22, f/32 ...

Charybdis do you mind me asking, how come my camera comes up with numbers like 3.5, and 5, etc... for aperture, that don't necessarily come up in scales like that ? That's why I thought the 2.8 would be several stops lower than 3.5 !

edit : was just looking at one of my pics at 5.8 ?!?
http://www.flickr.com/photos/38994151@N03/4902281622/meta/in/photostream/

charybdis

Mountainsandh wrote: »

Charybdis do you mind me asking, how come my camera comes up with numbers like 3.5, and 5, etc... for aperture, that don't necessarily come up in scales like that ? That's why I thought the 2.8 would be several stops lower than 3.5 !

edit : was just looking at one of my pics at 5.8 ?!?

The scales in my post are the scales most traditionally used with each entry one stop different from the ones beside it. DSLRs allow you to set shutter speed and aperture value in halves or thirds of stops. You can tell what yours is set to by counting the number of "clicks" required to get from one number to the next on the scales I posed (or just check your manual/settings). If you can set your lens to f/3.5 or f/5 I'm fairly sure your camera allows you to set your f-number in thirds of stops. This means you have to "click" the relevant wheel three times in one direction to change up or down a full stop.

If you want to look at the f-number scale with half, third, or quarter stops included, this Wikipedia page has them.

You can also think purely in terms of "clicks" if you find it easier, e.g.: if you increase your shutter speed two "clicks" you have to open up your aperture by two "clicks" to keep the exposure the same as what it was before (assuming your camera is set to work in thirds-of-a-stop mode, you're effectively decreasing your exposure by two-thirds of a stop by using a faster shutter speed and increasing your exposure by two-thirds of a stop to balance it out by using a wider aperture).

5uspect

charybdis wrote: »

I think you might want to re-check that math. Firstly, the two f-numbers in question are f/2.8 and f/3.5. not f/2.8 and f/3.2.

A more accurate way to calculate the difference in stops between two f-numbers is to convert them to their value in stops by finding their log to the base root two as all "integer" f-numbers are the integer powers of root two.

So:

  log[sqrt(2)](3.5) - log[sqrt(2)](2.8)
= (ln(3.5)/ln(sqrt(2))) - (ln(2.8)/ln(sqrt(2)))
= ~3.6 - ~3
= ~0.6

i.e.: two thirds of a stop, not one third.

Also, an f/1.8 lens is two stops faster than an f/3.5 lens, so you'd actually get four times as much light.

Assuming a circular aperture at two different sizes:

[LATEX]A_1 = \pi r_1^2[/LATEX]
and
[LATEX]A_2 = \pi r_2^2[/LATEX]

so
[LATEX]
A_1 = nA_2
[/LATEX]
where n is the area ratio between two f-stops and hence the difference in light available.

then
[LATEX]\pi r_1^2 = n\pi r_2^2[/LATEX]

or
[LATEX]r_1^2 = nr_2^2[/LATEX]

rearranging
[LATEX]n = \frac{r_1^2}{r_2^2}[/LATEX]

or even
[LATEX]n = \frac{d_1^2}{d_2^2}[/LATEX]
where d is the diameter

since n = 2 between stops then
[LATEX]2 = (\frac{d_1}{d_2})^2[/LATEX]
or
[LATEX]\frac{d_1}{d_2} = \sqrt{2}[/LATEX]

Depending on which you put on top or bottom you can use a 2 or a -2 exponent.

There's no need to complicate things with natural logs.

charybdis

5uspect wrote: »

Assuming a circular aperture at two different sizes:

[LATEX]A_1 = \pi r_1^2[/LATEX]
and
[LATEX]A_2 = \pi r_2^2[/LATEX]

so
[LATEX]
A_1 = nA_2
[/LATEX]
where n is the area ratio between two f-stops and hence the difference in light available.

then
[LATEX]\pi r_1^2 = n\pi r_2^2[/LATEX]

or
[LATEX]r_1^2 = nr_2^2[/LATEX]

rearranging
[LATEX]n = \frac{r_1^2}{r_2^2}[/LATEX]

or even
[LATEX]n = \frac{d_1^2}{d_2^2}[/LATEX]
where d is the diameter

Depending on which you put on top or bottom you can use a 2 or a -2 exponent.

There's no need to complicate things with natural logs.

I used natural logs as a handy way to convert base root two logs to a handy format so people could play along at home.

The point was that the actual f-number scale is derived from the integer powers of root two and talking about light in terms of percentage increase or decrease isn't really useful for the purposes of photography. So, when one wants to calculate the difference between two f-numbers in stops, I think it's best to do it the way I did (or just check the scale).

5uspect

charybdis wrote: »

I used natural logs as a handy way to convert base root two logs to a handy format so people could play along at home.

The point was that the actual f-number scale is derived from the integer powers of root two and talking about light in terms of percentage increase or decrease isn't really useful for the purposes of photography. So, when one wants to calculate the difference between two f-numbers in stops, I think it's best to do it the way I did (or just check the scale).

But your math hurt the OPs brain.

Anyway you can get the f-stop scale from a percentage increase/decrease just as easily.

A = 2.^(-0-10);

= 1.0000 0.5000 0.2500 0.1250 0.0625 0.0313 0.0156 0.0078 0.0039 0.0020 0.0010

Each stop gives you half as much light as the previous one because of the reduction in area

d = sqrt(A/(4*pi));

Convert to diameter, normalize and calculate your f-stop

f = d./d(1)

f =

1.0000 0.7071 0.5000 0.3536 0.2500 0.1768 0.1250 0.0884 0.0625 0.0442 0.0313

1./f ' =

1.0000
1.4142
2.0000
2.8284
4.0000
5.6569
8.0000
11.3137
16.0000
22.6274
32.0000

so conversely the difference between f/1.4 and f/4 is:

(1.4 ^(-2)) / 4 ^(-2) = 8.1633 times as much light collected.

Since each stop doubles the light you see its basically a geometric progression as you crank the aperture: 1, 2, 4, 8, 16, 32 ...

or simply [LATEX]2^n[/LATEX]

DaireQuinlan

Gentlemen ! We're all aduults here, I'd say most people are as comfortable expressing themselves in either natural logarithms or integral powers of root two as the fancy takes them. Indeed, I often convert between the two on the fly as I'm shooting just to keep the old noggin agile, often converted into base 5, which is my favourite base because it means I can count up to 10 with only one hand.

Thecageyone

5uspect wrote: »

Assuming a circular aperture at two different sizes:

[LATEX]A_1 = \pi r_1^2[/LATEX]
and
[LATEX]A_2 = \pi r_2^2[/LATEX]

so
[LATEX]
A_1 = nA_2
[/LATEX]
where n is the area ratio between two f-stops and hence the difference in light available.

then
[LATEX]\pi r_1^2 = n\pi r_2^2[/LATEX]

or
[LATEX]r_1^2 = nr_2^2[/LATEX]

rearranging
[LATEX]n = \frac{r_1^2}{r_2^2}[/LATEX]

or even
[LATEX]n = \frac{d_1^2}{d_2^2}[/LATEX]
where d is the diameter

since n = 2 between stops then
[LATEX]2 = (\frac{d_1}{d_2})^2[/LATEX]
or
[LATEX]\frac{d_1}{d_2} = \sqrt{2}[/LATEX]

Depending on which you put on top or bottom you can use a 2 or a -2 exponent.

There's no need to complicate things with natural logs.

Wait, what!? That looks hell-a-more complimicated :eek:

magicbastarder

charybdis wrote: »

I used natural logs as a handy way to convert base root two logs to a handy format so people could play along at home.

i use natural logs to heat my house.

5uspect

Thecageyone wrote: »

Wait, what!? That looks hell-a-more complimicated :eek:

Ah not really, its just the difference in area between two circles.
We know that that the area doubles for an f-stop and pi is constant so we can write that difference in area in terms of the diameter of the circle.

Since the area, A, of a circle is [LATEX]A = \pi r^2[/LATEX]

And the diameter, d, is twice the radius, r, we can write the difference in terms of the diameter squared.

So the f-number is a measure of the diameter of your aperture. But it's the area of the aperture that determines the light gathering ability. But since you can write the area in terms of the diameter you can just use that instead.

Perhaps there would be less confusion if f-number was measured in area rather than diameter. The numbers wouldn't seem so odd.

Area 1/Diameter
1.0000 1.0000
0.5000 1.4142
0.2500 2.0000
0.1250 2.8284
0.0625 4.0000
0.0313 5.6569
0.0156 8.0000
0.0078 11.3137
0.0039 16.0000
0.0020 22.6274
0.0010 32.0000

AnCatDubh

Maths, and computational theory aside - Yes, 2.8 will make a big difference, when you know what you are doing with.

Listen to the buzz of the wedding shooters for example who deal with low light scenarios constantly. 2.8, 2.8, 2.8, is mostly what you will hear - never, 3.5, 3.5, 3.5...

There is a reason for this. If you are struggling with light, you will be infinitely grateful for anything that gives you a little extra scope - that allows you a shutter speed of 1/60 rather than 1/30, that allows you ISO of 800 rather than 1600. You will be trying to squeeze the last bit of performance from your gear and an extra stop of light availability will be significant.

charybdis

5uspect wrote: »

But your math hurt the OPs brain.

Anyway you can get the f-stop scale from a percentage increase/decrease just as easily.

A = 2.^(-0-10);

= 1.0000 0.5000 0.2500 0.1250 0.0625 0.0313 0.0156 0.0078 0.0039 0.0020 0.0010

Each stop gives you half as much light as the previous one because of the reduction in area

d = sqrt(A/(4*pi));

Convert to diameter, normalize and calculate your f-stop

f = d./d(1)

f =

1.0000 0.7071 0.5000 0.3536 0.2500 0.1768 0.1250 0.0884 0.0625 0.0442 0.0313

1./f ' =

1.0000
1.4142
2.0000
2.8284
4.0000
5.6569
8.0000
11.3137
16.0000
22.6274
32.0000

so conversely the difference between f/1.4 and f/4 is:

(1.4 ^(-2)) / 4 ^(-2) = 8.1633 times as much light collected.

Since each stop doubles the light you see its basically a geometric progression as you crank the aperture: 1, 2, 4, 8, 16, 32 ...

or simply [LATEX]2^n[/LATEX]

The reality is you can do it either way, I just think it's more elegant to do it by converting the f-number into a value in stops through base root two logarithms as it's the series from which the scale was derived.

• [latex]\sqrt{2}^{0} = 1[/latex]
• [latex]\sqrt{2}^{1} = 1.4142[/latex]
• [latex]\sqrt{2}^{2} = 2[/latex]
• [latex]\sqrt{2}^{3} = 2.8284[/latex]
• [latex]\sqrt{2}^{4} = 4[/latex]
• [latex]\sqrt{2}^{5} = 5.6569[/latex]
• [latex]\sqrt{2}^{6} = 8[/latex]
• [latex]\sqrt{2}^{7} = 11.3137[/latex]
• [latex]\sqrt{2}^{8} = 16[/latex]
• [latex]\sqrt{2}^{9} = 22.6274[/latex]
• [latex]\sqrt{2}^{10} = 32[/latex]

BONUS ROUND: The "A" paper system, which all of you are familiar with (at least in terms of familiarity with what an "A4" piece of paper is), is derived from the same series. The ratio of the long side to the short side of an A4 sheet of paper is [latex]\sqrt{2}:1[/latex]. This ratio means that a doubling or halving in the area of a piece of paper allows you to maintain the same ratio throughout. So, "A0" paper is of the aspect ratio [latex]\sqrt{2}:1[/latex] and has an area of [latex]1m^{2}[/latex], and each successive "A" paper size (A1, A2, A3, A4, A5, etc.) has the exact same aspect ratio but precisely half the area. This means anything designed for the aspect ratio of "A" paper is completely portable between all the sizes without modification.

5uspect wrote: »

Ah not really, its just the difference in area between two circles.
We know that that the area doubles for an f-stop and pi is constant so we can write that difference in area in terms of the diameter of the circle.

Since the area, A, of a circle is [LATEX]A = \pi r^2[/LATEX]

And the diameter, d, is twice the radius, r, we can write the difference in terms of the diameter squared.

So the f-number is a measure of the diameter of your aperture. But it's the area of the aperture that determines the light gathering ability. But since you can write the area in terms of the diameter you can just use that instead.

Perhaps there would be less confusion if f-number was measured in area rather than diameter. The numbers wouldn't seem so odd.

Area 1/Diameter
1.0000 1.0000
0.5000 1.4142
0.2500 2.0000
0.1250 2.8284
0.0625 4.0000
0.0313 5.6569
0.0156 8.0000
0.0078 11.3137
0.0039 16.0000
0.0020 22.6274
0.0010 32.0000

I think using area to express aperture value becomes mentally unwieldy sooner than the current f/x method. The current way is adequate as soon as people can somewhat remember the scale and understand that there is one stop between each value.

DaireQuinlan wrote: »

Gentlemen ! We're all aduults here, I'd say most people are as comfortable expressing themselves in either natural logarithms or integral powers of root two as the fancy takes them. Indeed, I often convert between the two on the fly as I'm shooting just to keep the old noggin agile, often converted into base 5, which is my favourite base because it means I can count up to 10 with only one hand.

You can count to 31 on one hand if you switch to base 2. Counting to 1023 on two hands is a good party trick, assuming everyone stays awake.

AnCatDubh wrote: »

Maths, and computational theory aside - Yes, 2.8 will make a big difference, when you know what you are doing with.

Listen to the buzz of the wedding shooters for example who deal with low light scenarios constantly. 2.8, 2.8, 2.8, is mostly what you will hear - never, 3.5, 3.5, 3.5...

There is a reason for this. If you are struggling with light, you will be infinitely grateful for anything that gives you a little extra scope - that allows you a shutter speed of 1/60 rather than 1/30, that allows you ISO of 800 rather than 1600. You will be trying to squeeze the last bit of performance from your gear and an extra stop of light availability will be significant.

If you want to see the difference between f/3.5 and f/2.8, set your camera to manual exposure mode, dial in f/3.5 and an appropriate shutter speed, take a photo, then slow your shutter speed by two-thirds of a stop by "clicking" the wheel twice in the direction that decreases it (assuming your camera controls shutter speed in thirds of a stop), take another photo; in terms of exposure, the second photograph will be what an f/2.8 lens would've provided at the shutter speed used in the first photograph. I suspect you'll find the difference isn't all that great.

Don't be afraid to crank the ISO as high as necessary, a well exposed photograph taken at ISO 1600 will look a lot less noisy than an underexposed photograph taken at ISO 400.

Mountainsandh

Thanks again everyone, very useful answers here, except I regretfully have to inform Charybdis and 5uspect that "all that maths" is still a mystery to me, but then I did mention before I was an absolute disgrace :D. I'm sure it will help others less mathematically challenged people who come across this thread some time in the future though.

It's a pity there was a 24mm F2 lens for cheap on Ebay that got swiped under my nose, that would probably have made a little bit more of a difference. Like AnCatDubh said, even that little bit of shutter speed might help, especially so since I'm going to shoot musicians, with fiddle bows flying etc... and I'm not really looking for the slow shutter speed impression this time anyway. The actual shoot won't be until October so I do have a bit of time to think and practice.