Because the lens operates much like a funnel in that it gathers light, the greater the working diameter of the lens, the brighter the projected image. How bright this image will be is dependent on the brightness of the scene and the magnification realized by the lens. The longer the focal length, the more the lens magnifies. The deed of magnifying to produce an image, takes its toll on image brightness. In other words, the longer the focal length, the more the lens magnifies. This higher magnification result is a larger but dimmer image of objects.
Another way to say this, image brightness intertwines the working diameter and the focal length. Because these two factors are so interwoven, gauging image brightness is demanding. We are forced to fall back on a mathematical ratio that will take the chaos out of figuring out image brightness. This is true because a ratio is dimensionless. If I tell you the ratio of boys to girls in a 6th grade class is 3 boys for every 4 girls, I have given you a ratio that works independent of the number of students. For example if the class consist of 28 kids, then 12 boys to 16 girls is the breakdown (ratio is dimensionless).
For the camera lens: if the working diameter is 4 inches and the focal length is 4 inches, then the focal ratio (f-number) = 4 ÷ 4 = 1 (written as f/1 (f/1 is produces a very bright image). If the working diameter is 2 inches and the focal length is 4 inches, then the f-number is 4 ÷ 2 = 2 (written as f/2.).
The splendor of using a ratio is, any lens operating at the same f-number as another lens, yields the same image brightness regardless of the dimensions (diameter or focal length), for an identical scene. It’s complicated; but the f-number system actually takes away the chaos.