1
$\begingroup$

I've seen this solution which explains how head mounted displays work but I need to know where to place the lens. The previous solution used an objective lens to focus collimated light, I'm assuming that isn't at work in the HMD case because the screen is too close for us to assume light from points on the surface are parallel.

Effectively how do Oculus Rift type devices go from display screen to a single lens (in my case a single Fresnel lens) to my eye, all within the focal length of the lens while still resolving a clear image?

EX: If I have a 3x magnification Fresnel lens and I want a screen 15cm from my eyes, where in between should I place the lens so I can focus for a long time without a headache?

$d_{eye}=150mm$

$M_{lens} = 3$

$f=~?$

$d_{lens}=~?$

enter image description here

Thanks for the patience.

$\endgroup$
  • $\begingroup$ I think you should read a few introductory publications about the topic like "Head-Mounted Display Systems" by Jannick Rolland. The optical design of these systems is not quite as simple as a single lens system and just putting the image at infinity is not going to be sufficient to avoid the headache/nausea problem (if it can be avoided, at all, which so far nobody has managed to do, not even for the much more simple problem of 3d tv/movie projection). $\endgroup$ – CuriousOne Jun 17 '15 at 21:46
  • $\begingroup$ It just occurred to me that you could look at the design of ocular lenses (eyepieces), which do just what you want: magnify a small image and project it to virtual infinity. There are some very cheap and rather high performance wide angle (50 degrees and more) eyepieces out there for as little as $40-50 per pair. Just make sure that the object plane of the ones you select is outside of the barrel, or you are in for some machining... $\endgroup$ – CuriousOne Jun 17 '15 at 23:29
  • $\begingroup$ Thanks a lot for the information. Jannick Rolland's write up was a good read. I'll also try and pick up some eyepieces. Thanks! $\endgroup$ – Andrew Stewart Jun 18 '15 at 18:05
  • $\begingroup$ saburchill.com/physics/chapters3/0015.html This provided useful information. $\endgroup$ – Andrew Stewart Jul 10 '15 at 22:16
2
$\begingroup$

It's been a month, but maybe you are still curious. I've been struggling with the almost exact problem for.... too long. So here goes.

We basically want the effect of reading glasses. The screen is too close to your eyes (less than the LDDV Least Distance of Distinct Vision) so we need to create a virtual image that is at or beyond the LDDV. No need for infinity, we look at things about 10 inches away comfortably ( on average). A real image can be projected on a screen, or into a camera. A virtual image can't. It appears larger than the original and is oriented in the same direction (up). Side note, microscopes give a virtual image at about 10" away. So think about looking through a magnifying glass, you see a bigger version of the object. This is what we want. So how do you get that? You have to put the object closer than the focal length of the lens. The question asked for a 3X magnification lens. Usually fresnel lenses are sold by focal length, 90, 120 or 180mm are very common. If you need to convert from magnification power, divide the LDDV/magnification. So 10" or 254mm / 3 = 83mm or so.

Let's call the lens focal length f. Let's call the distance to the object p. Let's call the distance to the virtual image q. Let's call the distance from the lens to your eye de.

The virtual image size is a function of the distance from the lens to the object and is independent of the distance of the observer, for small changes of de anyway. Most of what you notice as changes are due to the convex lens being imperfect at forming images. Edges being blurry when you are far away etc. I should say I'm not a physicist so some this is mostly correct but I may miss use a term here and there... but it's certainly correct enough to answer the original question.

We need the object closer than f. f < p We need the distance to the virtual object plus the distance from the lens to your eye to be LDDV or greater. abs(q) + de >=LDDV

1/f = 1/p +1/q for a virtual image q is negative.

I could do substitutions with deye = 150-p etc. but I'm not sure the person asking the question picked good example values. f =80 with a decent sized screen and lens give horrible results even if you get the distances correct. Too much distortion. Try 180mm focal length.

I made an excel table to find a set of values so I could pick a good one. I found the best lens focal length by buying every focal length of lens I could find and trying them.

1/180 = 1/95 +1/q q=- 201 de = 250 - 201 = 49

So put the lens 100mm from the object, you put your eye 49mm from the lens (or a bit more, not every ones eyes are perfect, LDDV is an average not an absolute) and you get a nice picture. f=180, p=95, and the overall distance dEye becomes about 150mm. I would suggest p=100mm gives better results as the overall distance is 125mm. You can play around, but that's the relevant equation and how to apply it.

http://micro.magnet.fsu.edu/primer/java/lenses/converginglenses/index.html This page has a good java widget that allows you to see the effect of changing p on q. Note the object needs to be closer than the focal length.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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