Consider the following two modifications to a camera lens:

  1. Obstructing the about 75% of the lens's front element with an opaque filter as shown here: https://www.youtube.com/watch?v=43W159gPpdQ.

  2. Severely damaging the front element as here: https://www.youtube.com/watch?v=FWzsXeXCwuc

In both cases, the lens continues to produce a complete and sharp image that is free from any obvious distortions or degradation.

My question is: how is it that a camera lens is able to function so well under these circumstances? More precisely,

  • why does obstructing the front element in case 1. not result in a significant reduction in the lens' coverage?
  • why does damaging the front element as in case 2. not seriously compromise the lens' ability to produce a sharp image on the focal plane?
  • \$\begingroup\$ Can't say I agree with the premise that these are sharp images from a well-performing, damaged lens. In the first video, everything looks soft. In a video (1920x1080), its not possible to inspect the lens performance at full resolution. Having hi-res images would help support the basis of the question and perhaps reveal more about the effects. That said, in the second video, I notice some shift in sharpness in the lower right. Consider separating into two questions, about cinemorph filter and about damaged lens. \$\endgroup\$
    – Aaron
    Commented Jun 2, 2015 at 19:54

1 Answer 1


Mainly because every point on the front surface of the lens has light from every point in the Field Of View striking it, and other than that which is blocked by the aperture diaphragm the lens refracts all of that light to every point on the image sensor.

It is the same reason reflecting telescopes (Newtonian, Maksutov, Cassegrain, Schmidt, etc.) don't have a blind spot in the center where some of the incoming light is blocked by the secondary mirror(s) and support(s).


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.