Why is aperture a ratio?

[Hertz van Rental]

I'm just confused as to why aperture is expressed as a fraction. I understand how the stop system works to increase/decrease the area by 2, but for different focal lengths, I don't understand why the same f number corresponds to the same amount of light (assume the same shutter speed).

It's to do with Physics and Maths. But mostly Maths.
In particular, the Inverse Square Law and the area of a circle.

Firstly, the Inverse Square Law:
Say you have a light shining on to a piece of paper. The amount of light falling on to the paper at, for example, 1 foot from the light is 'x'.
If you move the paper 2 feet from the light source the amount of light falling on the paper is now only a quarter as bright.
This comes from Maths.
Think of a pyramid.
At a given height you can work out the surface area of the base of the pyramid.
If you double the height of the pyramid then the surface area of the base increases by a factor of four.
With light, if you double the distance from the light source then the same amount of light has to illuminate four times the area.
This means the illumination level is 1/4 of what it was.
With me so far?
Now if we look at the diameter of the opening letting light into a lens:
With a 50mm lens the light reaching the film/sensor for an opening of a certain diameter is 'x'.
For the same diameter opening on a 100mm lens, the light source is now effectively twice the distance away from the film/sensor so the illumination is only 1/4 of what it was for the same diameter opening on the 50mm lens.
For a 25mm lens the illumination would be 4x brighter.
And so on.
Now, if you go back to Maths, increasing the diameter of a circle by 2x makes the surface area of the circle 4x larger.
Decreasing the diameter by a factor of 2 reduces the surface area to 1/4.
See a co-incidence?
If, for a 50mm lens, you make the diameter of the opening that lets the light in = 'a' then increasing the diameter of this opening by 2x allows 4x the amount of light in.
So, for a 50mm lens an opening diameter of 'a' gives a light level of 'x' at the film/sensor.
On a 100mm lens an opening diameter of 'a' gives a light level of 1/4 times x.
If you now increase the diameter of the opening on the 100mm lens by 2 (2 x a) you are letting 4x the amount of light in so the illumination level on the film/sensor increases by 4.
The decrease of illumination by increasing the distance of light from the film is cancelled out by increasing the amount of light you let in.
Still with me?
Now, if on the 50mm lens you divide the diameter of 'a' into the focal length you get a number.
If, on the 100mm lens, you divide the larger diameter opening that gives the same level of illumination into the 100mm lens you get... the same number.
For example:
50mm divided by 2mm = 25
100mm divided by 4mm = 25
The result you get is a ratio and that is what the f number is.
It's just a convenience that means that f2 on a 50mm lens gives the same level of illumination on the film/sensor as f2 on a 100mm lens and a 30mm lens and a 25mm lens and a 1000mm lens and...
Basically, as you change the focal length of a lens the hole letting the light in gets bigger or smaller to keep the amount of light coming in the same and the relationship is reflected in the f-number so you don't have to do the calculations yourself.
Isn't Maths wonderful?

PS And that was all done without the aid of Google. :mrgreen:
I spent many years trying to explain this to students.
It's easier with a blackboard to help. But if you are still confused or not clear on anything just ask.
I'm happy to go over stuff as many times as you like - and I don't have an ego (like some people) so I don't get arsy.

 

[tirediron]

Uhm...I don't think I've been here long enough to know what that means

Helen_B, a member famous for her detailed for her wealth of knowledge and detailed explanations of the technical aspects of photography

 

[Hertz van Rental]

Helen_B, a member famous for her detailed for her wealth of knowledge and detailed explanations of the technical aspects of photography

You forgot to add her bad habit of re-writing other people's perfectly good posts for no real reason. :lmao:

 

[ksmattfish]

You forgot to add her bad habit of re-writing other people's perfectly good posts for no real reason. :lmao:

Usually she just removes the condescending attitude and taste of ass, Hertz, which makes it more pleasant and helpful for everyone.

Ohhh! And she's also not afraid to provide links to her work, so we can look and judge for ourselves if she knows what she's talking about. Have a good day!

 

[Josh66]

Because:

Isn't that pretty much exactly what epp_b said in his post using the paper tube analogy?

I don't know what you're talking about...

Are you giving me **** for not agreeing with (or for saying the same thing as) a post that was made after me, and then edited to remove the thing you're talking about?

The reason I restated the question was because it didn't seem like you really understood what the OP was asking. The OP clearly knew the effects that the focal length had on the physical size of the aperture, since he said (in his original post) essentially the same thing you did.

 

[Antarctican]

Ooo, lousy attitude ksmattfish. Looks to me like Hertz went through a lot of effort to provide a detailed and helpful explanation to the question asked. As he has done for many others who have asked for help/answers over the years. So why would you flame him?

 

[DeadEye]

Hertz ~ Thank you for a very good detailed explanation. It seems to me that just about everything in exposure is easier to think about in terms of ratio. Shutter speeds tend to double or half 15,30.60,125,250 also ASA 50, 100, 200, 400, 800 . In example if I were to take a light reading and found that at ASA 100 and f2.8 I needed 1/125 T for correct exposure. Well what I wanted was a faster shutter so then by bumping ASA to 200 and also doubling the T to 250 the exposure would be the same and not require another metering , or if I wanted to drag the shutter at 60 then easy ratio says stop down 1. It gets quite intuitive were you are not even thinking about it as it gets done.

Same for setting up a lighting scenario, Could you imagine a meter that read in LUX only

 

[Helen B]

Thanks for the kind words guys.

I hope that when I offer an alternative version or an expansion of that which has already been explained I am using my own words rather than rewriting previous posts. If learning from texts, I believe that it is easier to learn things from texts written by more than one author. Sometimes I try to correct things that I believe that previous posters have got wrong, such as some of the attempts at the definition of aperture in this thread: link.

Here's my version of this one.

A 100 mm lens does let more light in from an object at f/2 than a 25 mm at f/2. The area of the aperture is greater, so more light gets through it. The entrance pupil has 16 times the area, so 16 times the amount of light can be collected from an object.

It doesn't much matter exactly how far the light travels within the lens, but it does matter how large the image of the object is - ie how much the light is spread out in the image. If the object is effectively a point source (eg a star), and the image of it is a true point, the 100 mm lens will produce an image that is 16 times as bright as that produced by the 25 mm lens with both lenses set at the same f-number (aperture).

If, however, the object is not a point source, and the image is not a point but spread over an area, the area of the image will matter. The light from an object that comes through the lens from gets spread out over different areas. This is the normal case in photography, and the difference in the area of the entrance pupil and the difference in the area of the image exactly match as you change focal length and hence the larger image produced by the 100 mm lens will have the same intensity ('brightness') as the smaller image produced by the 25 mm lens.

The conclusion is that for constant brightness the area of the image should be in proportion to the are of the entrance pupil. As the square of the focal length is in proportion to the area of the image, and the square of the diameter is in proportion to the area of the entrance pupil you can deduce that for constant brightness the focal length should be in proportion to the diameter of the entrance pupil. Written another way, the ratio of focal length to the diameter of the entrance pupil (which is the f-number) should be constant.

Best,
Helen

PS

Ooo, lousy attitude ksmattfish. Looks to me like Hertz went through a lot of effort to provide a detailed and helpful explanation to the question asked. As he has done for many others who have asked for help/answers over the years. So why would you flame him?

I'm wondering why Herz started the flaming? What was his purpose?
"You forgot to add her bad habit of re-writing other people's perfectly good posts for no real reason."
Do you think that was justified?

 

[tirediron]

I don't know what you're talking about...

Are you giving me **** for not agreeing with (or for saying the same thing as) a post that was made after me, and then edited to remove the thing you're talking about?

The reason I restated the question was because it didn't seem like you really understood what the OP was asking. The OP clearly knew the effects that the focal length had on the physical size of the aperture, since he said (in his original post) essentially the same thing you did.

Apologies, you are correct - I did misunderstand. I read the relevant posts probably three times, and read them incorrectly all three times. Sorry for any confusion caused to the OP or anyone else. :blushing:

 

[epp_b]

A 100 mm lens does let more light in from an object at f/2 than a 25 mm at f/2. The area of the aperture is greater, so more light gets through it. The entrance pupil has 16 times the area, so 16 times the amount of light can be collected from an object.

OK, but, ignoring the subject and speaking in terms of perfect physics (which is what we do in physics), the same f/number has the potential to let in the same amount of light, regardless of the focal length, does it not?

 

[Helen B]

OK, but, ignoring the subject and speaking in terms of perfect physics (which is what we do in physics), the same f/number has the potential to let in the same amount of light, regardless of the focal length, does it not?

It depends on what is meant by 'the same amount of light'. Depending on how that is defined the answer can be either 'yes' or 'no'. You could say that you need more light through a longer lens than through a shorter lens to achieve the same image brightness.

Best,
Helen

 

[epp_b]

It depends on what is meant by 'the same amount of light'. Depending on how that is defined the answer can be either 'yes' or 'no'. You could say that you need more light through a longer lens than through a shorter lens to achieve the same image brightness.

I suppose so, but what I mean is: the whole point of expressing aperture using a ratio is so that we can figure out how much light to let through the lens without having to calculate the exact absolute size in millimeters for each focal length.

Say you're in a huge, completely dark room with a single light source sitting on a table. You point the camera at the light source using 50mm at f/5.6 and take a shot. You then use 200mm at f/5.6 and move the light source further back (or take several steps backwards) so that the light source (the subject) fills the exact same amount of the frame as the first picture.

Assuming an absence of human error, the pictures will be identical, yes?

 

[tirediron]

No, the whole point of expressing aperture as a ratio is so that we have a standard by which to calculate our exposures and gauge DoF. We don't really care how many lumens strike the sensor/film, nor for practical purposes do we care about the actual size of the aperture.

To answer the question in your second paragraph, assuming an abscence of human error, then yes, the two images would be identical.

 

[compur]

I'm just confused as to why aperture is expressed as a fraction.

Lenses of shorter focal length have a wider field of view (they "see" more
of the scene) therefore more light is reflected into them than a lens of
longer focal length pointed at the same scene under the same conditions.
Usually, anyway.

But, all lenses of the same format produce approximately the same size
image circle "out the back end" onto the film or sensor.

Therefore a shorter focal length lens is compressing a larger scene into
that image circle than the longer lens is.

If the physical dimensions of both apertures were the same (and all else
being theoretically equal), the shorter (wider angle) lens would produce a
brighter image (more light focused into the same area) than the longer
lens.

This is why f/stops or apertures are expressed as a ratio of focal length.

For Example
If you had f/1.8 on a 55mm lens, wouldn't that equate to more light into the sensor than f/1.8 on a 24mm lens? The actual diameter of the opening for the 55mm lens is 55/1.8=30.5mm, while for the 24mm lens it is 24/1.8=13.3mm, right? Now the area just scales linearly, so the areas are different... maybe I'm missing a concept here??

You'd be right if the same amount of light entered both lenses but that is
usually not the case. More light will enter the 24mm lens (in most average
scenes) than the 55mm lens.

This can break down if you are shooting light sources like the "candle in a
dark room" example, but that's why we have light meters.

 

[Joves]

Well if you want you can look at f-ratio as being f-percentage, if it will make it easier for you. It being a pecentage of the available light to the focal plane from the lenses diameter.