Being very interested in how VR/AR headsets work, I want to figure out the physics of looking at a screen right in front of one's face, figuring out the limits and parameters to what one could potentially focus on.

In particular, let's suppose we have a simple system defined by d1, d2, and m where:

  • d1 is the distance from eye to lens,
  • d2 is the distance from eye to object (phone), and
  • m is the lens magnification; for example a typical fresnel lens sheet out of plastic, which are cheap and thin.

The question is, given the constraint of the human being able to comfortably see the screen without straining eyes (related to human focal length I'm assuming?), what is the shortest d2 we can make, and how does it depend on d1 and m? We can assume some standard screen size such as 7cm x 21cm.

I'm not sure how to approach the problem partly because I don't know how to define what a human can comfortably focus on. I've read online that approximate human eye focal length is anywhere from 17mm to 22mm, which may be a useful starting point.

enter image description here


1 Answer 1


The lens in VR systems, and in, e.g., digital cameras with a fake viewfinder, are placed at a bit less than approximately the focal length away from the object (the original image), thus making the image appear to be approximately at infinity from the point of view of your eyeball. As your diagram shows, you can trace the intermediate rays back for that system to a virtual object plane that's probably not at infinity but at a "comfortable reading distance" that is compatible with a relaxed lens in the eyeball.

  • $\begingroup$ Thanks for your answer, it's a little abstract but I think it's making some sense. Do you know in a more concrete way how d1, d2, and m may interact? Given the human eye focal length? $\endgroup$
    – JDS
    May 5 at 18:02
  • 2
    $\begingroup$ @JDS what really matters is p and q in the Lensmaker's Formula $\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$ for the lens m . you want to choose p,q,and f so that the diverging cone between m and the eye's lens is no larger than the lens (so no loss of light), and that the virtual image at distnance q from the lens m is a comfortable focus distance frm the eye's lens (thus d1 + q) $\endgroup$ May 6 at 11:24
  • $\begingroup$ Any chance you'd be interested in adding a diagram with some extra comments? I started a nice bounty :) cheers $\endgroup$
    – JDS
    May 8 at 6:44
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    $\begingroup$ @JDS I'll think about that, but it wouldn't be much more than taking your original image and extending the 3 lines between the eye-lens and the fresnell lens ("m") to the left until they intersect. That will show the location of the virtual object that your eye is focussing on. $\endgroup$ May 9 at 13:00

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