1
$\begingroup$

This question already has an answer here:

What happens to the image of an object kept before a lens if the upper half of lens is covered by an opaque coating?

I understand that the full image will be formed and only the intensity of the image is reduced. But is the intensity uniformly reduced? That is, will the upper and lower half of image have same intensity?

(The object is kept at a distance $1.25f$ before the lens. Half the object is above the principle axis. $f$ is the focal length of lens).

$\endgroup$

marked as duplicate by John Rennie, Kyle Kanos, user36790, AccidentalFourierTransform, Gert Apr 10 '16 at 0:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Possibly this answer will answer your question? physics.stackexchange.com/q/233603 $\endgroup$ – Farcher Apr 7 '16 at 10:00
  • 4
    $\begingroup$ Possible duplicate of A half covered lens $\endgroup$ – John Rennie Apr 7 '16 at 11:07
  • $\begingroup$ @JohnRennie I am asking about the distribution of intensity. The answer given says that upper half of image is brighter. Is it a mistake? $\endgroup$ – Aditya Dev Apr 7 '16 at 11:24
0
$\begingroup$

You are projecting a real image. It's quite easy to trace the rays. You are just reducing the "viewport", so in essence, it's no different than just using a smaller oddly shaped lens at some weird position. The regular knowledge about the projections still works... angles do matter (you have vignetting, becuase the lens covers less of a spherical angle when viewed from the side), but the brightness of the image is still proportional to the area of the lens, so at least approximately speaking, the image is uniformly illuminated.

$\endgroup$
  • $\begingroup$ Answer given says the upper half is more intense. $\endgroup$ – Aditya Dev Apr 7 '16 at 11:22
  • $\begingroup$ The brightest part is always the part that is the one for which the lens has the largest apparent angular size viewed from the screen. So yes, in that sense, that part is slightly brigher. But that's the case even for a circular lens (the middle portion of the image is the brightest). If the numerical aperture is small, that's a small angular correction (cosine of the angle the image makes with the lens normal). $\endgroup$ – orion Apr 7 '16 at 12:06
0
$\begingroup$

It is reduced uniformly.

Every small small part of lens can produce entire image,

Each part gets intensity for full image uniformly, so it refracts uniformly.

$\endgroup$

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