I have thought of this question due to personal experience. I am short-sighted, and over the last three years my short-sightedness has worsened. Taking a lifeguard certificate again now that I did already three years ago, I noticed while diving that I have almost no vision anymore when trying to find and collect some training rings, that I had no problem to visually identify three years ago. In all cases, I am not wearing any goggles while diving.

So, I came to think whether short-sightedness has an effect that adds to the effect of the eyes being covered by water and the cornea having almost no refraction. This process is explained in other questions here, but I claim that my set of closely related questions is different, not only due to the inclusion of short-sightedness.

I then tried out my normal glasses under water and noticed no effect at all, not positive, not negative. I am now confused as to why the glasses have no effect, but the short-sightedness that they should correct does have a worsening effect. In this, I am assuming, without being able to justify it, that the glasses should have the same effect under water as above water, because, as the water touches them from both sides, the light should have a different path inside the length (less bending at entry and exit), but that the light rays exiting the glasses under water should be parallel to the light rays that would exit it in air. But what is instead the mechanism at work here that renders air-glasses useless under water?

Second, I notice that, when wearing swimming goggles, my short-sightedness is alleviated under water. This means that I can see better under water with swimming goggles and without any contact lenses or corrective glasses than I can above water without glasses or lenses. A quick search on the web found other questions of short-sighted people that noticed this effect, but I could not identify how this is working out for normal-sighted or far-sighted people. How does this work?

Then I pondered, as contact lenses are swimming on the eye, but, when underwater, have also contact with water instead of the air they are designed for, do contact lenses have an effect under water (ignoring the risk of them swimming away quickly)?

Finally, what would be the dimensions of a pair of glasses that actually work under water (be it for normal-sighted or short-sighted persons), when we, in turn, ignore their performance in air? How thick and large would they be?

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    $\begingroup$ Why doesn't the linked question/and answers also answer your question? Your glasses don't help because they also have close to the refractive index of water. The accepted answer there also deals with goggles and contact lenses. $\endgroup$ – Rob Jeffries Mar 12 '18 at 23:36
  • $\begingroup$ Does have a worsening effect , hmmm. Have you considered a change in the distance between the eye lens and the back of the eye caused by water pressure. A shortening of that distance would be consistent with worsening vision if short sighted. One way to tell is 3 foot depth vs 6 foot depth. $\endgroup$ – John Heath Mar 13 '18 at 0:33
  • $\begingroup$ True, I did not read the accepted answer closely enough. That reduces my question to the sub-questions 2 and 4. $\endgroup$ – Marie. P. Mar 13 '18 at 12:38
  • $\begingroup$ @John Heath I will try that out. When I went swimming yesterday (I can go only on Mondays), I did not explicitly test this, but when diving at 2-3m deep, I did have more problems spotting the rings than I thought I would have had based on being on the surface before, squinching my eyes together and looking at things that were just in front of me as close as the rings were later on. I will try to test it more carefully and replicably next week, e.g. trying to read a diver's watch dial an arm's length away at different depths! $\endgroup$ – Marie. P. Mar 13 '18 at 13:15

Glasses which correct for nearsightedness produce a virtual image nearer to the eye than the actual object:

diverging lens animation

I'm also nearsighted: with my glasses off, I can only focus about a foot in front of my face. My glasses take objects that are infinitely far away and diverge the light coming from them, so that there are virtual images less than a foot from my face; that's how I'm able to focus on objects through my glasses.

I have also observed that, underwater, I can see all the way across the pool with my swim goggles on. Most swim goggles are bubble-shaped, so we can model the swim goggles as a plano-convex lens:

lens shapes (source)

The curved surface of the "goggle-lens" is the water-air interface; the flat "surface" is the air-air interface, where there isn't any refraction. Now if you build a plano-convex lens out of glass and use it in the air, it's a converging lens. That's because the speed of light is slower in glass than in air, so the light bends towards the fat part of the lens. However your goggles are acting as a lens made of air and submerged in water. Since the relative indices of refraction are reversed, light moving from the water into the "air lens" is bent away from the fat part of the lens. The goggles, when underwater, therefore act like diverging lenses, which is the way to correct for nearsightedness.

I wonder if people who don't need glasses, or people who are farsighted, find things blurrier underwater with goggles on? Perhaps such a swimmer will comment.

  • $\begingroup$ what do you mean with "air-air interface"? Do you mean: "The curved surface of the "goggle-lens" is the water-plastic interface; the flat "surface" is the plastic-air interface, where there isn't any refraction"? - The goggles are a simple layer of see-through plastic. $\endgroup$ – Marie. P. Mar 13 '18 at 16:04
  • $\begingroup$ If the plastic of the goggles has the same thickness throughout, the goggles would fit in the second diagram as a "zero-meniscus" lens, neither converging nor diverging. That's why, looking through the goggles in the air, there is not very much change to the image. The refractive interface is between the water outside the goggles and the air inside the goggles. Proving that the thin plastic layer doesn't matter much is straightforward, but more than I can write today. $\endgroup$ – rob Mar 13 '18 at 16:35

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