Q:

WHAT IS THE F-NUMBER OF A LENS THAT TRANSMITS TWICE AS MUCH LIGHT AS A F-4.5 LENS?

- JENNIFER IGLESIAS (age 18)

KILLEEN HIGH SCHOOL, KILLEEN, TX, BELL

- JENNIFER IGLESIAS (age 18)

KILLEEN HIGH SCHOOL, KILLEEN, TX, BELL

A:

Jennifer -

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 light transmitted):

F=1/2r

r=(A/pi)^1/2

F=1/[2(A/pi)^(1/2)]

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 lens?

-Tamara

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 light transmitted):

F=1/2r

r=(A/pi)^1/2

F=1/[2(A/pi)^(1/2)]

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 lens?

-Tamara

*(published on 10/22/2007)*