I'm trying to wrap my head around the math/physics of how illuminance on the film/sensor plane remains equal for a given f-number regardless of the focal length of the lens. Since a longer lens will have a larger entrance pupil at a given f-stop than a shorter lens at the same f-stop, how can the light falling on the film/sensor be equivalent?

Empirically, I know this to be true. Say I take an incident meter reading (not reflected/TTL) off my subject, and get f/5.6 at 1/100s, I'll get proper exposure at the point at which i took the meter reading with that setting whether I use a 50mm lens or a 200mm lens (obviously composition will be different).

Wikipedia's f-number article reads:

A 100 mm focal length f/4 lens has an entrance pupil diameter of 25 mm. A 200 mm focal length f/4 lens has an entrance pupil diameter of 50 mm. The 200 mm lens's entrance pupil has four times the area of the 100 mm lens's entrance pupil, and thus collects four times as much light from each object in the lens's field of view. But compared to the 100 mm lens, the 200 mm lens projects an image of each object twice as high and twice as wide, covering four times the area, and so both lenses produce the same illuminance at the focal plane when imaging a scene of a given luminance.

I'm not sure I agree with the explanation on wikipedia. The film/sensor doesn't care how big the image is that falls outside the bounds of the sensor; the film/sensor "sees" what it sees, and that's it.


4 Answers 4


Illuminance is "the amount of luminous flux per unit area."

Suppose the 100mm lens was aimed at a large wall, and suppose distance was so that the lens sees a 10 x 10 foot area of that wall, reflecting illuminance back to the camera. OK, so this is a square, but it is about the subject, not the sensor (area matters, shape does not).

Then the 200mm lens would see a 5 x 5 foot area of same wall, half as wide and 1/4 as much area, and so 1/4 as much illuminance.

However, then the 2x diameter (4x area) of f/4 aperture of 200 mm will let 4x the light though at same f/4, therefore f/4 is f/4 exposure, regardless of focal length. 4 x 1/4 = 1 (same).

This is why we use the system called f/stops with the funny numbers, so f/4 will be f/4 and have meaning to us.

FWIW, not asked yet, but about the same argument is the reason that camera distance from the subject does not affect the exposure. The mountain is the same Sunny 16 daylight exposure regardless if we are on it, or 25 miles away. The Sun is special (at 93 million miles), but this also applies to subjects not illuminated by the Sun).

When the subject is seen by the camera at greater distance, that lighted object area also appears smaller. When ten times more distant, the subject dimensions are only 1/10 size, which is 1/10 x 1/10 = 1/100 the area. Inverse square law says the light is 1/100 as bright at 10 times distance. So 1/100 the light in 1/100 the area is the same apparent intensity, per unit of area. It exactly balances out, same exposure. Distance from camera does not affect exposure. Distance from a flash does matter.

  • \$\begingroup\$ Thanks @WayneF. Than makes sense when considering a reflective light meter reading (TTL), but I'm not making the connection when I use an incident light meter at the subject. The incident light meter doesn't care how big the wall is; it simply measures how much light is falling on that point. Any help with that aspect of the question? \$\endgroup\$ Commented Aug 13, 2016 at 2:31
  • \$\begingroup\$ Not sure I can connect those different subjects. The lens discussed sees the wall "directly" as units of area. And the reflected meter sees "the reflection" from the wall, and puts exposure of its average at midrange (dependent on subject color reflectivity, etc). The incident meter doesn't see the wall, it simply measures the incident light, and adjusts exposure to put it midrange (independent of subjects colors or reflectivity, so dark and light things properly adjust accordingly). Incident is the way to bet for exposure, but the incident meter is Not about what the lens sees. \$\endgroup\$
    – WayneF
    Commented Aug 13, 2016 at 3:17
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    \$\begingroup\$ Simply because the direct light is what the incident light actually sees. What else could it read? But yes, instead looking through the lens is affected by other different factors, illuminance and area of subject and focal length and area of aperture, etc (which balances out for the same exposure). But that does not prevent the incident meter from simply metering the light directly. It has of course been designed and calibrated to agree, but the method and rules and factors are entirely different than what the lens sees. No need for it to be otherwise. \$\endgroup\$
    – WayneF
    Commented Aug 13, 2016 at 3:51
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    \$\begingroup\$ Don't get me wrong, I truly appreciate the feedback. I'm looking for the mathematical explanation, however, and not "because that's what an incident meter is supposed to do." The other posts (suggested duplicates) stop there. I'm trying to bring the conversation one step further. \$\endgroup\$ Commented Aug 13, 2016 at 4:03
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    \$\begingroup\$ @the_meter413 What you are not getting: The focal length divulges the power of the lens. Long lenses magnify more. This translates to a larger projected image. The price paid for higher magnification is reduced image brilliance. Consider a slide projector set-up. Increasing the projector to screen distance yields more magnification. This bigger image requires that the light that comprises the image, occupy a larger area (less photon per unit area). The result is a dimmer projected image. Same for the camera, we compensate by enlarging the aperture to mitigate. \$\endgroup\$ Commented Aug 13, 2016 at 16:36

A longer lens will have a larger entrance pupil for the same f-number. The f-number is a dimensionless ratio. It is the ratio of the entrance pupil to the focal length.

A 100mm lens with a 25mm diameter entrance pupil is 25mm/100mm = 1/4 = f/4 or 1:4

A 200mm lens with a 50mm diameter entrance pupil is 50mm/200mm = 1/4 = f/4 or 1:4

A 200mm lens with a 25mm diameter entrance pupil is 25mm/200mm = 1/8 = f/8 or 1:8

Remember that the entrance pupil is not the diameter of the physical aperture diaphragm. Rather, it is the diameter of the diaphragm as seen through the front of the lens. The lens elements in a telephoto lens will magnify the size of the physical diaphragm when it is viewed through the front of the lens. The EP is what matters, not the physical aperture. It is the portal through which the lens sees the world.

For how this works out with both constant aperture and variable aperture zoom lenses, please see: How do zoom lenses restrict their widest aperture at the telephoto end?


It is true that a longer focal length lens for a given f number does have a larger entrance pupil. it is also true that more photons will land in this entrance pupil emitted from any particular source than a smaller one.

However, for a longer focal length lens, the exit pupil will also be further from the image, all else equal. What is truly important to exposure (or "illuminance"), the watts per meter squared hitting the image, is how big the exit pupil appears to be, or its projection upon the image plane. This quantity is described by the numerical aperture of the lens, and the f number is 1/(2NA). In essence, the longer the focal length, the further away the base of the cone is. Even if it is larger, it appears to be the same size as a smaller cone placed closer, and thus the exposure is the same.


Focal length dictates image size. If you double the focal length, the height of the image of objects double. This is a 2X magnification change. Consider the projected image of a checkerboard square. Suppose the dimension of the projected square is 10mm by 10mm with a 50mm lens mounted. The area of the projected rectangle is 10 x 10 = 100 square millimeters. Now we mount a 100mm lens. The image of the square enlarges to 20 x 20 = 400 square millimeters. The image of the square now occupies 4X more area (400 ÷ 100 = 4). Stated differently, if we double the focal length, we magnify the object 2X and this enlarged image covers 4X more space. The result is a drop-off of image illumination to ¼ of the initial brightness (4X light loss). True if the diameter of the circular aperture remains unchanged for both set-ups.

Now assume the diameter of the aperture of the 50mm lens is set to 6.25mm. We calculate the focal ratio (f/number) thus 50 ÷ 6.25 = 8 written as f/8.

Now we mount a 100mm lens and preserve the same aperture diameter. The 100mm lens functions as 100 ÷ 6.25 = 16 written as f/16. This is a 2 f/stop reduction. Since each f/stop is a doubling or halving of light energy. The reduction in exposure energy is 4X = two f/stops.

So the question is: What must we do to maintain the same exposing energy? The answer is: we must increase the diameter of the aperture to compensate for the light loss induced by doubling the focal length. We will need 4X more surface area. To achieve, we turn to the geometry of circles.

Factorial: If we multiply the diameter of any circle by the square root of 2 = 1.4142, we calculate a revised diameter that yields 2X more surface area. Remember, we need 4X more light -- thus 4X more surface area. Here we go: 6.25 X 1.4142 = 8.8388mm. That gets us 2X more surface area. So we do this again: 8.8388 X 1.4142 = 12.5mm. This will be the diameter of the aperture of the 100mm lens functioning as 100 ÷ 12.5 = 8 written as f/8.

Conclusion: A 50mm lens functioning at f/8 has an aperture diameter of 6.25mm. A 100mm lens functioning at f/8 has an aperture diameter of 12.5mm. This 12.5mm entry has 4X more surface area thus allowing 4X more light to play on film or sensor.

We need a system to take the chaos away. Ratio to the rescue. A ratio is dimensionless. Any lens functioning at the same focal ratio passes the same exposing energy (with minor variation unimportant in this discussion).

The key to all this is the square root of 2 which we can round to 1.4. Using 1.4 as a multiplying factor we discover a set of numbers: 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. Going the other way, each number is its neighbor on the right divided by 1.4 This number set yields a sequence of circles, each with double or half their neighbor’s surface area.


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