Wiki fact-check: vertex distance

This gives a formula for a modified lens strength, based on distance moved.

Seems to me that “corrected” and “perceived” are not equivalent.

  • perceived means moving the original lens to distance ‘x’ will behave the same as the putting the calculated lens at the original distance
  • corrected means that if you move D to ‘x’ you must also adjust it to the calculated value to maintain the same effect.

(If the perceived lens is stronger, then you would have to replace it by a weaker lens at the new offset to give the same effect.)

The rules given here are that for a plus lens, moving away from the eye means it is perceived as being stronger, and vice versa. A minus lens does the opposite. (Note that they are using the opposite sign for distance : positive value is closer to the eye. Hence the opposite sign in the equation.) These are consistent with the woolly hand-wavey description I had planned to add to the page before I realised the answers I was getting were backwards.

Our wiki page gives a bigger value in the plus case. So I guess it is using the perceived story, rather than corrected, version.

  • I convinced myself, so boldly made the edit. But perhaps someone could double-check.

I’ve been struggling to see exactly how the lenses at the two positions can be considered equivalent. I’ve think I’ve finally twigged that they’re not… or at least they are only equivalent for objects a long way away (in that they both adjust the incoming parallel rays to be compatible with the far point of the eye). That bit is fairly trivial to demonstrate. Objects at other distances behave very slightly differently (requiring slightly different amounts of accommodation) in the two scenarios, and give different near points.

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Another thought occurred to me this morning… for differentials, infinity is not a sensible distance to choose to make the vertex equivalence correction. Would be much more sensible to make the two prescriptions equivalent at screen distance.

So I did some sums… to keep the numbers simple, I am supposing distance to blur is 10cm and the screen is 50cm away.

The contact-lens full correction is obviously -10dpt, and diffs are -8dpt (2dpt more plus for a screen 0.5m away).

Now we naively apply the vertex equation to these 8dpt differential contacts, and calculate that we need 1/(-1/8 + 0.025) = -10dpt glasses.

But then we do it from first principles, using the thin-lens formula. The lens is 2.5cm from the eye, and therefore 47.5cm from the screen. We want the virtual image 10cm from the eye, and therefore 7.5cm behind the lens. 100/47.5 - 100/7.5 = -11.23 so we actually need -11.25 dpt differntial glasses.

You do get the correct answer if you first apply the vertex correction to the full prescription : 1/(1/-10 + 0.025) = -13.33 dpt and then add 2dpt plus to reduce these glasses to differentials = -11.33dpt. (Actually, that should now be 1/0.475 = +2.1 dpt since the screen is slightly closer. => -11.23 dpt).

Well… I thought it was interesting…

(For an object at infinity, both -8dpt contacts and -10dpt glasses at 2.5cm will create a virtual image 12.5cm from the eye. So the naive approach is doing what was intended. But for an object 50cm from the eye, -8dpt contacts put the virtual image at 10cm, whereas -10dpt glasses at 2.5cm put the virtual image at 10.75cm)