diopter rule-of-thumb for moving in-focus-point fwd/bkwd?

Discussion in 'Optometry Archives' started by David Combs, Nov 25, 2008.

  1. David Combs

    David Combs Guest

    Suppose an N-diopter lens gives me a sharp focus at 20 inches,
    but I want it to be at, say, 28 inches (for viewing a wide computer-screen).

    Is there a pretty-good rule-of-thumb for, from that information alone,
    computing a "subtract"?


    David Combs, Nov 25, 2008
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  2. David Combs

    Dave Bell Guest

    Feels like there should be a simple rule, doesn't it?
    You'd be working with sums of reciprocals though, so no simple addition
    rule. Easy enough to derive it, though:

    In the example, 'N' is pretty close to 2D.
    39.25" (1 meter) divided by 20" = 1.9625D, if it gives you the effect of
    seeing the object at infinity.

    You need 39.25/28 = 1.4018D *total*, so -0.56D "add".
    Close enough for almost anyone, -0.50D.

    If you tried to use the 8" difference directly in a formula, you'd get
    something like 39.25/8 = 5D, definitely NOT what you want.

    (All that supposes you have or want to use no compensation, or focusing
    power of your own lens. Post-cataract surgery, e.g.)

    Dave Bell, Nov 25, 2008
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  3. David Combs

    David Combs Guest

    The above "pretty close to 2D" comes *from* the
    *following* sentence-- ie up above you're showing the result of
    the calculation, *before* showing where you got it from?
    By "effect of seeing the obj at infinity", do you mean
    that with the glasses on, when my eyes feel totally relaxed,
    things at infinity show up sharp -- in fact, sharper than
    at any other distance?

    Or do you mean "with 20-20 eyes, the feeling (totally relaxed,
    no strain at all) I'd get looking at something at infinity --
    that FEELING is what I'd feel, WITH the glasses, looking at a
    20-inch distant object"?

    (That sounds to me more like it. Correct?)
    Cool. Thanks!
    David Combs, Nov 26, 2008
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