Focal length by using sun (and trees)

Discussion in 'Optometry Archives' started by winke, Oct 11, 2007.

  1. winke

    winke Guest

    Can someone help refine the equation for Focal Length and Dioptre. I
    am looking for the optical equation to calculate a lens's Dioptre at
    home if the light source is not at infinity.

    I went through the info in a Sci.Med.Vision thread to calculate the
    focal length of several readings glasses.

    ---------------- QUOTE ----------------
    The relationship is: Diopters = 1 / Focal-length

    For instance, a focal length of 0.5 meters will have a power of 2

    The easy way, is to just take the lens outdoors, and hold it above a
    paper. If the lens forms an image of the sun at 0.5 meters -- the
    power is 2 diopters. If at 1 meter, the power is 1 diopter.
    -------------- END QUOTE --------------

    I have a dozen pairs of ready-made reading glasses. Most of them no
    longer have the strength of the lenses marked on them. I recall they
    were originally marked to within 0.25 dioptre.

    The sun is weak in London these days and I used the sun as it came
    through some tall trees which were very approximately 10 metres away.
    The trees seemed to help because it was harder to focus the sun
    behind thin cloud than to focus a silhouette image of twigs and

    However my calculations for my reading glasses seem to be about 0.2
    Disoptre LOWER than expected.

    In other words, a guessed 1 Dioptre lenses gave me a focal distance
    of 112cm which calculates to about 0.9 D. A guessed 2 D lens was
    calculated to be about 1.76 D. Etc.

    Are my results approximtely consistent with the light source not
    being at infinity but very roughly 10 metres away?
    winke, Oct 11, 2007
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  2. The light source - to wit,the sun - is over 200 million kilometres away -
    near enough to infinity for all practical purposes.

    However, that is not what you are focussing an image of, is it?
    Nicolaas Hawkins, Oct 11, 2007
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  3. winke

    winke Guest

    Yes, that's right, you're following my posting so far.

    I also mentioned that the trees which I focus are about 10 metres

    Are you able to help by posting the equation I am looking for?

    Does using my 10 metre value in the equation approximately account
    for the 0.2 Dioptre shortfall?
    winke, Oct 12, 2007
  4. VF = VI + P

    VF = final vergence, for a real image, VF = 1/distance in metres
    VI = initial vergence, for light from an object, VI = - 1/dist
    P = power in dioptres

    So, for your case, you have

    P = 1/(distance to image) + 1/(distance to object)

    and you get an error of 0.1 dioptres if you ignore the final term for an
    object 10 metres away.
    Timo Nieminen, Oct 12, 2007
  5. Gee whiz ....
    You also stated that the light source was about 10m away:That is what I was correcting.
    I'm not sure that such an equation exists outside of textbooks on optics.

    In my understanding (which I do not pretend is comprehensive) 10m is close
    enough to optical infinity as to make a negligible difference in
    estimations of dioptric power such as you are making.

    I suggest that the more likely source of the discrepancy is inaccuracy in
    your measurement of focal length. What is your tolerance in those
    measurements? An error of ±3mm would account for the discrepancy you
    mention and, absent a graduated optical bench (which by your own
    description you are not using), greater accuracy would be difficult to

    If you want pinpoint accuracy in regard to the powers of the lenses,
    perhaps you should take them to an optometrist and have them measured on a
    machine specially designed for the purpose, or be satisfied with the rough
    guesstimations that you have already done.
    Nicolaas Hawkins, Oct 12, 2007
  6. winke

    Dave Bell Guest

    1/fl = 1/Do + 1/Di

    fl = lens focal length,
    Do = distance from lens center to Object,
    Di = distance from lens center to Image.

    Then, Diopters = 1/fl in meters, which is the left-hand side, already.

    So, 1/fl = 1/10 + 1/1.12
    1/fl = 0.992857 which is pretty darned close to 1.0 !

    Working your other one backwards,
    1/1.76 = 0.562 meters, and
    1/fl = 1/10 + 1/0.562 (yes, = 1/10 + 1.76 ...)
    Diopters = 1.86, not so far from 2.0

    Dave Bell, Oct 12, 2007
  7. winke

    Dr Judy Guest

    Results are close enough given the inaccuracy inherent in the method,
    with dim light you will not get precise focus and likely not a precise
    measurement of distance either. Also, The ready made readers will
    come in 0.25D steps and an error of about 0.125D in manufacture is
    within tolerance.

    So the possible answers are +1, +1.25, +1.25, +1.75, +2, etc. Round
    your calculated answer up or down to the closest 0.25 interval.

    Dr Judy, Oct 12, 2007
  8. winke

    redbelly Guest

    An equation does exist, both inside and outside of textbooks:

    1/(focal length) = 1/(object distance) + 1/(image distance)

    It's just about the most basic formula used in optics. It's not just
    something found in textbooks, it really works in the real world. The
    upshot is, "winke" needs to add 0.1 diopters (1/10m) to his
    calculation to account for the non-infinite object distance. So:

    0.9 diopters become 1.0 diop.
    1.76 diopters become 1.86 diop.
    Actually, it accounts completely for the discrepancy in the first
    Actually, an error of 120 mm is necessary.
    The OP measured 112 cm. In order to get 1 diopter, one would need to
    get 1m = 100 cm. That's a difference of 12 cm or 120 mm.

    If the 112 cm were off by only 3 mm, that would change the result by
    just 0.002 diopters:
    1/1.12m = 0.893 diop.
    1/(1.12-0.003)m = 1/1.117m = 0.895 diop.


    redbelly, Oct 13, 2007
  9. winke

    winke Guest


    Excuse my delay in replying but I've been away for a week.

    I'm the OP and appreciate you working through the figures. It is
    exactly the info I was looking for.

    Thank you.
    winke, Oct 23, 2007
  10. winke

    winke Guest

    Tom, I am the OP and what you write is useful for me to know because it
    helps me understand some of the other variables involved which may have
    affected my readings.

    winke, Oct 23, 2007
  11. winke

    redbelly Guest

    You're welcome!


    redbelly, Oct 26, 2007
  12. winke

    lena102938 Guest

    Made to order, come in smaller steps, I bet
    It is very common prescriptions like -2.1 or + 1.7
    lena102938, Oct 31, 2007
  13. winke

    otisbrown Guest

    Dear Lena,

    Most measurements (Trial-Lens kit) can be made to
    about 1/8 th diopter.

    But typically, most commerical prescriptions are
    quantized to 1/4 diopter increments.


    otisbrown, Oct 31, 2007
  14. winke

    lena102938 Guest

    Yes, we do it in our physics labs , like that,
    instead of using laser as a source.
    lena102938, Oct 31, 2007
  15. winke

    otisbrown Guest

    otisbrown, Oct 31, 2007
  16. winke

    lena102938 Guest

    Mike, I know that.

    Why would some body needs eighths.eighth ?
    lena102938, Oct 31, 2007
  17. winke

    lena102938 Guest

    Otis, Thank you for that reference, very much!!!!!!

    PS I knew that most prescription step is 0.25.
    Simply Dr Judy pointed that it is step of the off the shelf readers,
    like prescription have another step
    lena102938, Oct 31, 2007
  18. winke

    lena102938 Guest

    PS : 1/8 ?
    I believe that person can distinguish 1/8, especially with ads or
    Taking into account all aberrations.
    Especially if -5, 1/8 will be very helpful.

    Is somebody ever was thinking that eyes constantly
    moving and changing shape? (forget accommodation)
    That we are blinking?
    That the eye - brain system has amazing
    shock-absorbing mechanism: we move head, arms,
    even walking: the picture is still stable.
    How much processing it takes ?
    Because of that OD can tell "you will use".
    The power of adaptation is almost enormous.
    lena102938, Oct 31, 2007
  19. winke

    lena102938 Guest

    True, and may be even more then 1/8

    I think it is their imagination, probably they think that they are
    Leonardo de Vinci
    lena102938, Oct 31, 2007
  20. winke

    lena102938 Guest

    lena102938, Oct 31, 2007
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