Welcome to Scientography, a series I'm writing on the scientific aspects of photography. Today, I'm discussing aperture. I'll be explaining what aperture is, what an f-number means, as well as identifying why longer focal length "fast" lenses are so expensive, so read on to get your dose of knowledge for the day...

If you found this article, chances are that you have some idea as to what aperture controls, or even what the settings you set on your camera represent. You might say, "Hey, IForgotMyBurnerKeyOnce, isn't aperture what affects depth of field?" and you'd be right!

**Aperture**

Aperture controls the amount of light that enters the lens by narrowing (or widening) the aperture "blades." These five to nine curved or straight blades roughly form a circle who's diameter is adjusted based on the f-number that is set in the camera (or on the lens itself, for you who have fully manual lenses). The f-number is an easy way to relay to the photographer a sense of how thin or wide the depth of field will be, typically by the 'f/' notation.

You probably already knew that, but I want to get a concrete idea of what aperture is engraved in your mind before we delve deeper.

F-numbers range from f/0.7 (Carl Zeiss 50mm f/0.7 Planar - only 10 ever made) to f/32, typically. A lens with a lower f-number will have a thinner depth of field and will collect light more quickly, requiring faster shutter speeds or a lower ISO setting (or slower film).

**F-Number**

But what exactly does the f-number represent? A quick search landed me on a Wikipedia page, which states "the f-number of an optical system is the ratio of the lens's *[sic]* focal length to the diameter of the entrance pupil." We can logically deduce that, since the aperture of a lens is adjustable by the aperture blades in correlation to the f-number, the aperture is the entrance pupil referenced in the definition of f-number.

Through the definition of what an f-number is, we get the following equation:

Where:**N** is the f-number (notated *f*/**N**),* f* is the focal length of the lens (in mm), and

**D**is the diameter of the aperture (in mm).

So, now we have the essentials to calculate the f-number, focal length, and diameter of the maximum aperture of a lens.

For example, the Canon 50mm *f*/1L USM in the lead picture has a maximum aperture of *f*/1. Given this information, that means that the diameter of the aperture, at its widest setting, is:

*N=f/D1=50mm/DD=50mm*

You can use basic algebra to calculate using the original formula, or you can remember the other two formulas derived from the f-number formula, *f=N*D* and *D=f/N*.

**Examples**

That one was easy, but what's the diameter of the aperture of the Canon 85mm *f/*1.2L II USM at *f/*1.2? Easy.

*N=f/D1.2=85mm/D1.2D=85mmD=85mm/1.2=70.8333... mm*

**WAIT!** You're saying that the aperture on the 85mm *f*/1.2L II USM has a *wider* aperture than the 50mm *f*/1L USM, even though the f-number is 20% narrower? Yes. As focal length increases, the size of the aperture must be increased.

If you plot the formula for aperture diameter, *D=f/N*, as *z=x/y*, you can see just how the parameters are interconnected:

Going left to right along the Y axis increases maximum aperture from *f*/8 to *f*/0.95 (Leica 50mm f/0.95 Noctilux).

The higher a point is on the Z axis, the larger the diameter of the aperture at max, and

The depth (X-axis) is focal length, with the furthest back points being 100mm, closest being 10mm.

As you can see, the longer the focal length, the wider the diameter of the aperture must be, especially for longer focal lengths.

**Stops**

Since you're such a wise person and knew what aperture was, then tell me, what a "stop" of light is? Surely, you could tell me a stop of light is the difference between EV 0 and EV +/- 1 on your light meter, but what does it mean for you?

"EV" is short for exposure value. Technically, your camera does all the work for you, what with fancy built-in light meters and such. The output you see on your light meter in your viewfinder is typically not the absolute EV value. More so, it is the relative EV value. To get this value, the camera meters the light coming in through the lens, checks what exposure settings you have, then reports to you whether you'd overexpose (EV > 0), underexpose (EV < 0), or perfectly expose (EV = 0) the shot.

Conveniently, a change of -1 EV (-1 stop) is equal to:

1/2 * Shutter Speed or

1/2 * ISO

The aperture also affects the EV, but not in the way that you'd think. You'd think a difference of +/-*f*/1 would be equal to a difference of +/- 1 stop, but that's not the case at all.

**It's All Relative**

See, EV is calculated by the following:

Where:**EV** is Exposure Value,**N** is f-number, and **t** is shutter speed (in seconds).

Fortunately, I'm going to skip the long explanation for why this works, but essentially if 1 stop is halving the shutter time or halving the ISO sensitivity, then to do a stop for aperture, you must halve the *area* of the aperture.

Since I was an electrical engineering student, it's easiest to explain it in terms of electricity. If EV is the power of a circuit (Watts), then Voltage is the shutter speed, ISO is Resistance, and aperture is Amperage.

In order to halve the power of a circuit (EV -1), you can halve the voltage (shutter speed /2, double the resistance (ISO /2), or halve the amperage (Aperture *area* /2).

**You're Making My Head Hurt**

It's ok if your head hurts, mine does too when I think about this, but it makes sense. By halving the area of the aperture, we halve the amount of light let through the aperture, thus underexposing by one stop.

Since we know the area of a circle is:

How do we halve it and figure out a way to easily do this repeatedly? Easy:

Let sub-o be the original parameters and let sub-n be the new parameters:

*(That should be a over two, not over pi)*

And so,

So, in order to get the stopped down radius, you must multiply the original radius by *sqrt(2)^-1*.

If you wanted to get diameter, you'd multiply both radii by two, which would bring you to the same exact equation. So we cay safely say:

In order to halve the area of the aperture, or to reduce exposure by 1 EV, or "stop", the diameter of the aperture must be multiplied by 1 over the square root of 2.

Which means that one stop difference in f-number is as follows:

So, to find the next stop down of f-number, multiply the f-number you're at by square root of two.

The area issue is why f-numbers are typically an interval of sqrt(2)^x, such as (from *f/1*):

*f/1, f/1.4, f/2, f/2.8, f/4, f/5.6, f/8, f/11*, *etc...*

**What's Your Point?**

The point of all this is that I was reading an article about the Canon 200mm f/1.8L. It's touted as the fastest 200mm Canon has ever built. Then, this picture caught my eye:

From the picture, you can obviously see the 200mm *f*/1.8L is huge! But what size aperture would be required?

Using the information above, I found that the aperture must be **111.11mm** across! For comparison, the *f*/2.8L would have to have a diameter of just 71.4mm! That means that the diameter of the aperture on the *f/*1.8L is 60-ish percent bigger than the *f/*2.8!

**How About in English?**

One thing to take away from this is that "faster" lenses are more expensive for a reason. They've got to incorporate a large aperture, especially when working with longer focal lengths. Every time you increase the size of the aperture, you bring in a host of issues with aberrations that I can't get into right now.

The other thing you should take away is that a lens rated at *f/*4 is only 1 stop slower than an *f/*2.8. When buying a lens, is the extra stop really worth the extra jump in price? Though, it does allow you to double your shutter speed, or halve your ISO.

More Scientography articles as I can write them!