My understanding is that a larger focal length corresponds to a larger working aperture diameter, since the numerator is larger.

Why doesn't a larger working diameter correspond to a brighter image? Shouldn't there be more light coming into the lens?

Wouldn't a subject framed with 42.5/1.2 lens be brighter than a 35/1.2 if the subject is framed the same?

  • \$\begingroup\$ "Constant aperture" means the F-stop stays the same, not the actual aperture diameter. \$\endgroup\$ Commented Jan 26, 2019 at 0:49
  • \$\begingroup\$ "Working aperture" is usually understood to mean the same thing as entrance pupil. \$\endgroup\$
    – Michael C
    Commented Jan 26, 2019 at 1:37

2 Answers 2


Shouldn't there be more light coming into the lens?

No, because the lens is collecting light from a narrower angle of view.

But beyond that, exposure is about brightness per unit area, not about total light collected by a lens, or even about total light spread over the entire surface of a sensor.


Ok, the only possible explanation is that similar to Gauss' law: the sub-section, sphere surface area cutout, smaller field of view actually produces less light than the large field of view did.

The field of view is proportional to the focal length AND the working aperture, since theres a cone from the center of working aperture, out the barrel, to the fringes of the field of view, and scaling horizontally also scales the cone vertically.

So -- even though there is more light gathered, the actual scene being focused is producing less light, and they happen to balance out because focal length is correlated identically with working aperture and the field of view.

In summary: only the F-number depicts the amount of light gathered per unit area of the scene, regardless of zoom/distance and framing


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