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I am trying to make a single-eye head mounted display. It will be a prototype for displaying information while the person is doing other work. The important part is that the virtual image of the screen is around 1 meter away from the wearer so he will not have to adjust his focus much while working with his hands.

However I cant figure out the parameters of my lens. I have made a graphic to explain the situation:

hmd lens problem

I want to get my fixed screen size to cover all of the FOV, F,B and possibly C(adjustable) can be derived.

  • Is it possible to make a function out of this using the lens formula?
  • My field does not have to be perfectly flat, do i still have to worry about weird optical effect just using a biconvex lens?

Any tips on how to get my values are greatly appreciated.

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  • $\begingroup$ I would start by reading this article? vr-lens-lab.com/field-of-view-for-virtual-reality-headsets $\endgroup$ – Farcher Oct 20 '16 at 9:54
  • $\begingroup$ I got that far, but i cant get a handle on actually calculating the dimensions. This article shows only the general relations. Thanks for the suggestion though :) $\endgroup$ – Kapytanhook Oct 20 '16 at 10:06
  • $\begingroup$ The image does not show up by itself: The certificate there is expired, and in many contemporary browsers, the expired certificate is preventing browser from showing the image. It would be better to replace https with http. $\endgroup$ – StR Oct 28 '16 at 1:29
  • $\begingroup$ Weird, I just used the stackexhange image uploader. Maybe something on your side? $\endgroup$ – Kapytanhook Nov 1 '16 at 8:57
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It is indeed an interesting project.

I have a better idea. If you have the convex lens of focal length $C$, then the image will be at $\infty$ and your eye can see it easily. In this case the magnification will be $\frac{f_{eye}}{f_L}$. Usually the eye focal length is taken as 55 mm and the lens you will need is ~10 mm then the magnification of 5 will be there and one can see the small images clearly. You may also want to make them as reading glasses (top part with 0 power and bottom part with the high power convex lens).

Small changes in the focusing can be done by adjusting the position of the instruction paper relative to the lens using a screw or something.

I hope this will help

regards,

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I'm not sure where your actual, physical screen is, but let's assume that it is 1 cm to the right of the lens. That means you have a negative image distance.

The simple equation for this is 1/f=1/p+1/q
f=focal length
p=object distance
q=image distance

Then you would have
p=-1 cm
q=100 cm
f=-1.01 cm

Notice the focal length is negative. This is not a convex lens, it's concave.

Finding optics like this is going to be tricky, I think. Also I suspect you'll have to look into aspherical lenses, which will be even trickier.

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  • $\begingroup$ Hey David, thanks for responding. See the image I have provided, The screen is C distance away and D high and wide. This object and image formula does not seem to work for head mounted displays. I assume this is because we are talking about the human eye, not a screen to project an image on. Just take for example the oculus rift, the screen is only a few cm removed from the lens. Somehow it required a highly convex lens to make the virtual image appear 10m away. I cant apply this formula on existing solutions and come to the same focal point that their lenses have. Am I missing anyhting? $\endgroup$ – Kapytanhook Oct 27 '16 at 10:17

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