The shape of most camera lenses is a circle. Thus the answer to your question revolves around the math used to find the area of a circle. This is because the amount of light passed to film or digital sensor is chiefly a function of the capturer area of the lens (working diameter).
You know the formula to figure out the area of a circle, it is: Area = Radius squared multiplied by Pi. Thus the area of a 10mm circle = 10 ÷ 2 X 10÷2 X 3.1416 = 75.5398 sq. millimeters. Thus the area of 20mm circle is 20 ÷ 2 X 20 ÷ 2 X 3.1416 = 314.1593 sq. millimeters. Thus a 20mm dimeter lens passes 314.1593 ÷ 75.55398 = 4. In other words, a lens twice the diameter of another passes 4X more light.
Now the basic unit of exposure used in the jargon of photography is the f stop. This is an increment of exposure change = a doubling of halving of the exposing energy. In other words, to double the exposure you increase or decrease the area of the lens to achieve a 2X change. This can be accomplished using the lens aperture or by adjusting the shutter speed or a combination of both. As to the difference between a 10mm diameter lens and a 20mm lens, this change establishes a 4X change equal to 2 f-stops.
Thus the formula to figure out the area of a circular lens is the key to your question. But maybe even better is a factorial: Multiply the diameter of any lens by the square root of 2 and you compute a revised diameter that yields a 2X (1 f-stop) change. This key number is 1.4142. This value is also the factor used to compute the f-number set: 1 – 1.4 – 2 – 2.8 – 4 – 5.6 -8 – 11 – 16 – 22 – 32. Note, each number going right is its neighbor on the left multiplied by 1.4. Each number going left is its neighbor on the right divided by 1.4. Again, multiply or divide the diameter of a circle by 1.4 yields a revised diameter that has twice or half the surface area. As to the lens, this translates to a 2X change in light transmission.
Another factorial: The amount of light passed by a lens intertwines its diameter with its focal length. If the capture area of the lens is doubled or halved we get a 2X change. If the focal length is doubled or halved we get a 4X change. This is because the lens produces an image by projection. If we double the focal length, the image magnification changes 2X but the area of the projected images changes 4X. This intertwining creates a dilemma. To solve this simply we fall back to a ratio. This math divides the focal length of the lens with its working diameter and conceives a value called the focal ratio. This is the familiar f-number system we use. Again, the f-number intertwines the light loss or gain of focal length change with the light loss or gain of aperture change. We use the f-number to take the chaos away.
Now exposure has a math formula called the law of reciprocity. E = exposure I = intensity of the projected image T = dwell time of the exposure. Formula E=!T (exposure = intensity multiplied by time. This law holds for general photography however its accuracy often fails when film is caused to undergo a prolonged exposure (1 second or longer) or when film is exposure using super short exposure time (1/1000 of a second or quicker).
All that being said – If more than one lens is used to project an image; and the images are superimposed, the exposing energy will be increased. Thus if 4 otherwise equal lenses are deployed, each contributes 25% of the exposing energy.