I think I am quite good at math and understand basic geometry, but I have problems understanding the function of a diaphragm in cameras.

Let's say a camera is in a specific state. It captures the light from a specific "cone", going outside of the camera lens. Now, if the diaphragm moves (decreases the diameter of the hole), and nothing else moves in the camera, it can only result in a "black material" appearing around the image, right?

It will not change the size of objects in the final image, or sharpness of objects, or the "rate" (amount/time) of light, that is getting to the center of the image (which is not covered by a diaphragm). Am I right?

But if the diaphragm is never shown in the final image, and only covers rays around it, what is the purpose of it, as those rays would never make it to the image sensor anyway, right?

Edit: I think I am starting to get a clue.

My intuition was, that if there is an obstacle between two points A and B, A can not see B. The point A is on the object, and B is on the sensor, and the fact that the line between them is not straight (refracted by a lens) does not change anything.

I did not realise, that there are lots of rays (in with different directions) going from A, through the lens, and reaching B. The obstacle near the lens behaves very differently. By covering half of the lens, 2x less rays would get from A to B, but the sensor B would still receive light from A (B will still "see" A).

EDIT 2023: I have built this simulation of a camera. You can see three points emitting light on the right, a diaphragm, a lens, and a sensor on the left. I recommend trying this:

  • move or remove the lens (to "focus" or "unfocus")
  • change the "hole" of a diaphragm (to allow more or less light hit the sensor)
  • 1
    $\begingroup$ I think you need to ponder a bit more on ray tracing through an imaging system, and where you could put an aperture that would not interfere with the image, but change the amount of light making the image. $\endgroup$
    – Jon Custer
    Commented Jan 3, 2019 at 14:12
  • 1
    $\begingroup$ Briefly speaking: no, that's not right. You can check how such black ring never appears, and also how that apperture affects exposition and focusing. It's quite long to explain. Are you familiarized with geometric optics? What things do you already know? Also, I suggest you to make the distinction between aperture and field diaphragms. $\endgroup$
    – FGSUZ
    Commented Jan 3, 2019 at 14:25
  • $\begingroup$ A simple lens will send every light ray from the same single point on the subject, to a single point on the film plane, no matter where that light ray hits the lens. Likewise, light rays from a different point on the subject will all be referred to a different point on the film plane. An aperture placed close to the lens blocks some of those light rays, but it does not change the paths of the light rays that are let through. $\endgroup$ Commented Jan 3, 2019 at 15:26
  • $\begingroup$ ...meanwhile, if the aperture is very close to the lens, then the image of the aperture itself will be so far out of focus as to be invisible. $\endgroup$ Commented Jan 3, 2019 at 15:27
  • 1
    $\begingroup$ @SolomonSlow That seems like an answer rather than a comment. $\endgroup$
    – rob
    Commented Jan 4, 2019 at 3:19

1 Answer 1


The argument in your edit is essentially correct.

The diaphragm is introduced very close to the lens, where objects are maximally defocused, so that every ray from every object passes through the lens. The diaphragm removes some of those rays, but it still allows multiple rays to come through and form an image.

This means that the "shadow" of the diaphragm is completely defocused, so that it covers all of the image by darkening it as compared to what you'd have with an open diaphragm (with more light to go around in total), but it does not form an image on the sensor because it's at the plane of the lens.

As to why you have a diaphragm in the first place: this allows you to fiddle with the depth of field, i.e., with the range of distances at which objects will appear in focus. The wider your aperture is, the more rays you have, and that means that the rays that reach the focal plane encompass a wider cone of angles, which in turn implies that there is a reduced tolerance for moving the detector backwards or forwards while still keeping the object in focus.

Conversely, for a fixed detector plane, that reduced tolerance means that the wider the aperture, the shallower the range of lengths at which objects will appear in focus.

You use a diaphragm when you want to have a larger depth of field (i.e. where you want objects at many different distances to appear in focus) and you're OK with losing some light overall. Thus, you reduce the aperture, killing some rays from the outside of your lens, and in the process you reduce the cone of angles and you expand the depth of the focal plane.

  • $\begingroup$ Yes! The key for me was to realise, that there are many different rays between a point on an object and a point on a sensor. The rest is a simple geometry. The Diaphragm only "makes the lens smaller". Less light gets to the sensor, which is usually a bad thing (we may need to "emphasize" colors by software, which could add noise), but the blurriness of too near / too far objects is decreased, as the lens area decreases. $\endgroup$ Commented Jan 21, 2019 at 2:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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