Is there any mathematical proof to show that the exposure in a camera is independent of the focal length used, if the f-number (ratio of focal length to lens diameter) remains constant?

  • \$\begingroup\$ This answer to the previous question covers the math. (It's not complicated!) \$\endgroup\$
    – mattdm
    Commented Jul 17, 2019 at 13:51

1 Answer 1


That's why f-number is f-number, in other words F-number is dependent of the focal length in order to get away the focal length from the formula.

This is the theory. Actually, the real value is T-number which is slightly different from F-number.

  • \$\begingroup\$ I'm having a doubt for a while. The thing is: the aperture area is proportional to the square of the ratio of focal length to the f-number, as evidenced by: en.wikipedia.org/wiki/Aperture#Aperture_area Now, when focal length is increased, the f-ratio remaining constant, the aperture area still increases allowing more light in. Am I correct? \$\endgroup\$ Commented Jul 17, 2019 at 11:15
  • \$\begingroup\$ And yet, despite that, the exposure remains constant. How is that possible? \$\endgroup\$ Commented Jul 17, 2019 at 11:16
  • \$\begingroup\$ @LumosMaxima Does it help if you consider that the area of a scene recorded is proportional to the square of the focal length? \$\endgroup\$
    – mattdm
    Commented Jul 17, 2019 at 12:28
  • 1
    \$\begingroup\$ Longer focal length magnifies the objects in the image, makes them look larger. In fact, it magnifies the entire image, however the sensor size crops the frame back to the same size. So this smaller part of the image in the cropped frame gets only part of the light, not all of it. But the f/stop calculation increases aperture diameter due to longer focal length. This exact exposure compensation for focal length is the entire purpose of the f/stop calculation fstop = focal length / aperture. It creates constant exposure regardless of focal length zoom. f/4 is f/4, regardless of focal length. \$\endgroup\$
    – WayneF
    Commented Jul 17, 2019 at 15:29

Not the answer you're looking for? Browse other questions tagged or ask your own question.