I went to my Optometrist yesterday, and I noticed that they wanted to test my vision at 20ft. However, because they had a small room that could only accommodate 10ft, they used a vision chart that was reversed behind me, and effectively simulated 20ft by asking me to read the chart via the mirror. Now, my question is: can I use mirrors to shorten that length even more, so I can measure my vision at 6 inches say? If I'm using mirrors to mimic 20ft then the vision test should still be testing my distance vision, and not my near vision, so my pupils and lens shouldn't be accommodating... I've tried drawing out the ray diagram for this with my brother, but we're both stumped. Can anyone enlighten us? I can't stop thinking about it now.


You're correct, the diagram would look something like this. A Snellen chart could be L distance from the mirror, with the observer sitting L distance from the mirror too. The virtual image would lie 2L from the observer.

enter image description here

For two parallel plane mirrors you'll end up with a ray diagram that looks something like:

enter image description here

Where $I_n$ are the $n^{\text{th}}$ virtual images of our initial object placed between mirror $M_1$ and $M_2$ e.g. images of images of images of images... I've only drawn rays bouncing from the left mirror $M_1$ here, but they'd of course bounce off $M_2$ in exactly the same way forming $I_n^{'}$ virtual images.

The kind of system you're looking for would be one where the size of each subsequent virtual image is reduced by a series of off-set parallel plane mirrors:

enter image description here

  • $\begingroup$ Thanks for that. What would a ray diagram look like though where the actual snellen chart was 6 inches from my face, but simulated with mirrors to appear to be 6 ft away? $\endgroup$ – Jennifer Kendricks Nov 10 '20 at 19:15
  • $\begingroup$ @JenniferKendricks in order to have a plane mirror setup where the virtual image is not $2L$ from the observer, where $L$ is the distance from the observer to the plane mirror, you would need to use more than one mirror... In that instance, the ray diagram becomes slightly more complicated to draw. $\endgroup$ – smollma Nov 10 '20 at 20:53
  • $\begingroup$ Would it be possible for you to draw out what sort of arrangement would be required for 20L, whereby L is the distance as per your ray diagram above? I would be so, so grateful. It's just bugging me too much... I've been trying! $\endgroup$ – Jennifer Kendricks Nov 10 '20 at 21:14
  • $\begingroup$ @JenniferKendricks check the answer. I think for $n$ mirrors separated horizontally by $L$ in the last arrangement you'd have an effective distance of $2n\times L$. So for the last setup, you'd have an effective distance of $10L$ if $L$ were the distance between the object and the bottom mirror. Probably only works for small angles between the mirrors such that $\cos(\theta_r)\approx 1$. If you start changing the separation distances of mirrors and objects, the geometry becomes harder... $\endgroup$ – smollma Nov 10 '20 at 23:14
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    $\begingroup$ Thank you so so much for this! I'm going to sit and try and work my way through it, but I think it makes sense to me now. I'll ask any more questions if I have them, but I can't thank you enough. I had never studied optics before but it's fascinating, and the more I look into it the more questions I have. Thank you!! $\endgroup$ – Jennifer Kendricks Nov 12 '20 at 12:28

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