I apologize for this one before hand, but this is going to take
some math... The amount of light transmitted by a camera lens is
determined by the area of the lens's apperature. Assuming that the lens
is a circle (which is not quite true, but a fair approximation), the
area of the apperature equals pi times the radius squared, or A=pi*r^2.
Turning this equation around gives us r=(A/pi)^1/2.
The F-number of a lens is also determined by an equation. The
F-number is equal to the focal length of the camera divided by the
diameter of its apperature, or F=l/d. Since the diameter is equal to
twice the radius, we can write this as F=l/2r. Now if we keep the focal
length constant, we can put these two equations together to solve for
the F-number in terms of the area of the apperature (the amount of
So now we just need to plug in your numbers. If F=4.5, then we can
solve the equation to give us A=0.0388. But we want the F-number for a
lens with twice as much light transmitted, so it will have A=0.0775
(=2*0.0388). Now we can plug this in to find the new F-number, F=3.2.
The smaller the F-number, the more light is transmitted.
In photography, going from an F-number to one that lets twice as
much light in is called a 'full-stop'. As a general expression, we can
say that F-numbers that are a full-stop apart are separated by a factor
of 1.414 (the square root of 2). In the example we just did, the simple
way of doing it is to divide 4.5 by 1.414. This would give us the same
result, F=3.2. If you take a closer look, you will see that this
general rule comes straight from the equations we just used.
To get a lens that transmits half as much light, you would just go
a full-stop in the other direction. Can you figure out what the
F-number is for a lens that transmits half as much light as an F-4.5
(published on 10/22/2007)