The Aperture Value or the F Stop is an amazing tool to understand the relation between the lens you are using and the brightness of the image you perceive when viewing it. It says that, for a given “F Stop”, all lenses focus an image of equal brightness on the image sensor, irrespective of their focal lengths. How does this happen?

Let us denote the amount of light in an image by ‘I’.

The amount of light captured by an Image Plane (the digital image sensor of your camera) is directly proportional to the area of the aperture of the lens. Let us denote area of the aperture area as ‘A’. More the area, more light is allowed in to form the image. Thus I ∝ A

The amount of light transmitted is inversely proportional to the Square of the distance from the center of the lens to the image-plane. Quite simply, if the distance between the lens and the image plane increases, the light’s intensity falls geometrically. If ‘d’ represents the distance, then I ∝ (1/d)^{2}. (square of inverse of d)

The distance from the center of the lens to the image plane is also called the ‘Focal length’ of the lens. Let us denote it by ‘f’.

Note that by definition f is same as d.

Hence interchangeably, I ∝ (1/f)^{2}.

After joining the relation of ‘A’ and ‘f’, we get I ∝ (A/f^{2}).

Assuming that the aperture of the lens is a circle, the area of the aperture can be calculated as A = π r^{2} where ‘r’ is the radius of the circle.

Since radius is exactly half of diameter, it can be calculated by knowing the diameter of the circle.

The diameter of of an image formed by a lens is calculated as Focal Length / Aperture value. (focal length is already defined by us as ‘f’).

Let Aperture value be represented by ‘F’. Hence diameter of the image can be represented as f / F.

Since we need the radius, we divide this by 2 and we get (f/F)/2, which is the same as r = (f/2F).

Hence area of the circle is π r^{2}, which is same as π (f/2F)^{2}.

Filling these in the light amount equation (I ∝ (A/f^{2})), we get I ∝ ( (π (f/(2F)^{2}) / f^{2}). Expanding the equation, it can be represented as:

I ∝ (π f^{2})/(2F)^{2} / f^{2}. Cancelling the f^{2} from top and bottom, we are left with I ∝ π /(2F^{2}).

You will notice that we are left with only constants (π is a constant, F is value determined in advance). Thus, for a given F value, I is always the same.

This makes us conclude two things:

1) Irrespective of the focal length of a lens, at a given aperture, the amount of light falling on the image sensor is the same.

For example, this means that F/8, a 35mm lens and a 500mm lens, both allow the same amount of light into the sensor. This helps you quickly estimate what your image will look like depending on the available light.

For example, if you shoot a sunset with a 35 mm lens and immediately afterwards shoot the same scenery with a 500mm lens (keeping the aperture, shutter speed and ISO constant), the images may represent more or less of the scenery (35mm will fit in more, while 500 mm will fit in less), you will notice that both the images are actually **equally** lit. Just because the 35mm fits in more of the scenery, does not mean it’s image gets noticeably brighter or the 500mm gets darker (unless you point the 500mm to a darker part of the scenery).

So if you take pictures of the same scenery with different apertures (keeping other parameters the same), you will develop a mental idea of how much light gets into the image sensor. Some images will be brighter than normal and some will be darker. On another occasion, under similar lighting circumstances if you are required to shoot with a different lens, you can switch lenses without worrying about which aperture to use. Just recall your memory and select the aperture value that appeared the best in your experiment with the other lens.

2) Its all fine to stare at a mathematical equation realizing the absolute truth. But how did it come about? Since the focal length is varying, the lens’s initial diameter (the front of the lens that allows the light in) is varying, the quality of the glass is varying, the construction of the lens is varying, how did we land up with a Constant?

The fallacy lies in the pre-determined F factor. During engineering, exhaustive tests are carried out (backed by very complex mathematics) to determine the actual diameter of the aperture so that light passing through the lens will equal that of a standard test lens.

If diameter (D) = f/F, even F = f/D is true.

Are the images obtained from same aperture of different lenses absolutely the same? Definitely not. Some differences always crop in because of manufacturing and technology issues and the circumstances under which the photography is being done. However, realizing that aperture value has predetermined the amount of light entering the camera, we can adjust the shutter speed and ISO sensitivity to improve the image.

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## 2 responses to “Aperture Value – The Great Light Equalizer”

Nice article. Just wanted to add that the aperture value also effects the depth-of-field. For example the same shot with f 8 will have a much shallower depth-of-field at 500mm than at 35mm, even though they will be equally lit. Depth-of-field is in itself a pretty big topic to discuss, may be another article.

You bet DOF is a pretty big topic to discuss. I have a few articles planned; mostly on topics that I think will make it easier for digital photography enthusiasts to understand the photography basics and get better images.

In fact, the Aperture value keeping the light same was a terrific revelation to me and I cannot wait to try out some standard scenes with a range of apertures to develop the mental idea of what aperture works best for a specific scene.

Of course there are other factors – ISO, Shutter speed, DOF that will affect the composition and image quality, but now there is at-least a starting point – One constant, rest being variables.