Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Apologies in advance for complete ignorance of optics (beyond conic sections)

How is the size, distance, and angle of the mirror used here calculated? The mirror is I believe the open flap in front of the projector.

enter image description here

share|cite|improve this question

That sort of thing is done by simple ray tracing. It is pretty easy to compute the path of the rays at the edge of the beam by tracing from the edge of the object. In a projector like that, the object is the LCD screen inside the projector. In practice one generally doesn't have to do the ray tracing by hand though -- information about the required size of the optics is readily available in the same software that is used to design the projector lens.

share|cite|improve this answer
I was referring to the mirror that reflects the image onto the whiteboard, not the lens – Joe Stavitsky Feb 22 '12 at 20:10
So was I. That sort of thing is easy to include in the optical design software. In fact, for that particular design it would be necessary. – Colin K Feb 22 '12 at 20:23
could you give an example of such software please? – Joe Stavitsky Feb 22 '12 at 20:30
CodeV, produced by Optical Research Associates; Oslo, produced by Sinclair Optics, Inc.; or Zemax, produced by Radiant Zemax. – Colin K Feb 22 '12 at 20:41
You wouldn't be too far off if you just imagine the beam of light expanding from the edge of the projector lens to the edge of the projected image. Then you just look at how big that beam will be at whatever distance from the projector you want to place the mirror. Be aware that if you just do this with an off-the-shelf projector, your image will be very distorted into a trapezoidal shape. Probably it will be partially out of focus as well. – Colin K Feb 23 '12 at 23:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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