# Why do lower f-stop numbers mean larger openings?

Because the lens works much like a funnel, since it gathers light, the larger the working diameter of the lens, the brighter the projected image. The brightness of this image will depend on the brightness of the scene and the magnification made by the lens. The longer the focal length, the more the lens is magnified. Enlargement writing to produce an image has its effect on the brightness of the image. In other words, the greater the focal length, the more the lens is magnified. This higher magnification result is a larger but darker image of the objects.

Another way of saying this, the brightness of the image intertwines the working diameter and the focal length. Because these two factors are so interwoven, it is difficult to measure the brightness of the image. We are forced to retreat in a mathematical proportion that will eliminate the chaos of determining the brightness of the image. This is true because a relationship is dimensionless. If I say that the proportion of boys and girls in a sixth grade class is 3 boys for every 4 girls, I have given a proportion that works regardless of the number of students. For example, if the class consists of 28 children, then 12 children and 16 girls is the breakdown (the relationship is dimensionless).

For the camera lens: if the working diameter is 4 inches and the focal length is 4 inches, then the focal ratio (number f) = 4 ÷ 4 = 1 (written as f / 1 (f / 1 produces a very high brightness) image). If the working diameter is 2 inches and the focal length is 4 inches, then the number f is 4 2 2 = 2 (written as f / 2.).

The splendor of using a ratio is that any lens that works with the same number f as another lens, provides the same image brightness regardless of the dimensions (diameter or focal length), for an identical scene. It's complicated; but the f-number system really takes away the chaos.