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Medical Forum / General / Vision / October 2007Focal length by using sun (and trees)
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. http://tinyurl.com/32fd97 ---------------- QUOTE ---------------- The relationship is: Diopters = 1 / Focal-length For instance, a focal length of 0.5 meters will have a power of 2 diotpers. 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 leaves. 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? > 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 [quoted text clipped - 33 lines] > Are my results approximtely consistent with the light source not > being at infinity but very roughly 10 metres away? 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?  SignatureNicolaas. >> Can someone help refine the equation for Focal Length and Dioptre. >> I am looking for the optical equation to calculate a lens's [quoted text clipped - 39 lines] > > However, that is not what you are focussing an image of, is it? Yes, that's right, you're following my posting so far. I also mentioned that the trees which I focus are about 10 metres away. 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? > >> 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. [cut] > >> 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 [quoted text clipped - 17 lines] > Does using my 10 metre value in the equation approximately account > for the 0.2 Dioptre shortfall? 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. SignatureTimo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/ E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html >>> Can someone help refine the equation for Focal Length and Dioptre. >>> I am looking for the optical equation to calculate a lens's [quoted text clipped - 41 lines] > > Yes, that's right, you're following my posting so far. Gee whiz .... > I also mentioned that the trees which I focus are about 10 metres > away. You also stated that the light source was about 10m away: >>> Are my results approximtely consistent with the light source not >>> being at infinity but very roughly 10 metres away? That is what I was correcting. > 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? 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 achieve. 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. SignatureNicolaas. ... Accuracy can never derive from rough aproximations. > >>> Can someone help refine the equation for Focal Length and Dioptre. > >>> I am looking for the optical equation to calculate a lens's [quoted text clipped - 57 lines] > > I'm not sure that such an equation exists outside of textbooks on optics. 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. > 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. Actually, it accounts completely for the discrepancy in the first measurement. > 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 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. Regards, Mark >and, absent a graduated optical bench (which by your own > description you are not using), greater accuracy would be difficult to [quoted text clipped - 9 lines] > > ... Accuracy can never derive from rough aproximations. > On Oct 11, 8:36 pm, Nicolaas Hawkins wrote: > its approximtely consistent with the light source not > An equation does exist, both inside and outside of textbooks: > [quoted text clipped - 25 lines] > Regards, > Mark Mark 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. W > Mark > [quoted text clipped - 5 lines] > Thank you. > W You're welcome! Cheers, Mark > > >>> Can someone help refine the equation for Focal Length and Dioptre. > > >>> I am looking for the optical equation to calculate a lens's [quoted text clipped - 108 lines] > > > ... Accuracy can never derive from rough aproximations. Yes, we do it in our physics labs , like that, instead of using laser as a source. Dear Lena, For example, here is a "classic" trial-lens kit to confirm the 1/4 diopter increments. http://premieremedical.safeshopper.com/442/5419.htm Enjoy, Otis > > > >>> Can someone help refine the equation for Focal Length and Dioptre. > > > >>> I am looking for the optical equation to calculate a lens's [quoted text clipped - 113 lines] > > - Show quoted text - On Oct 30, 8:18 pm, "otisbr...@pa.net" <otisbr...@pa.net> wrote: > Dear Lena, > [quoted text clipped - 124 lines] > > > - Show quoted text - 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 > 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 [quoted text clipped - 33 lines] > Are my results approximtely consistent with the light source not > being at infinity but very roughly 10 metres away? 1/fl = 1/Do + 1/Di where: 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 I'm sure any Opticians Practice will 'neutralise' (in other words 'measure') the power of the lenses for you if you ask them nicely. They use a machine called a focimeter, and as most of these machines are fully automatic these days they will give you a printout of the lens powers. You could also drop a few coins in the charity box on your way out (usually for Vision Aid Overseas or the Glaucoma Association) as means of a 'thank you' ;-) If you get up to West/South Yorkshire i'd do them for you. Cheers Jem >> 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 [quoted text clipped - 45 lines] > > Dave > However my calculations for my reading glasses seem to be about 0.2 > Disoptre LOWER than expected. [quoted text clipped - 5 lines] > Are my results approximtely consistent with the light source not > being at infinity but very roughly 10 metres away? 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. Judy > > However my calculations for my reading glasses seem to be about 0.2 > > Disoptre LOWER than expected. [quoted text clipped - 8 lines] > The ready made readers will > come in 0.25D steps Made to order, come in smaller steps, I bet It is very common prescriptions like -2.1 or + 1.7 > Made to order, come in smaller steps, I bet > It is very common prescriptions like -2.1 or + 1.7 Again, you rely on your imagination. I've never seen a prescription for anything other than quarters or eighths. -MT > > Made to order, come in smaller steps, I bet > > It is very common prescriptions like -2.1 or + 1.7 [quoted text clipped - 3 lines] > > -MT Mike, I know that. Why would some body needs eighths.eighth ? > Mike, I know that. Then why would you say this?: >> > It is very common prescriptions like -2.1 or + 1.7 Were you being sarcastic? > Why would some body needs eighths? Because sometimes we examine engineers and photographers. They often want maximum precision. And you can be that precise, with a dim room and a careful subject. But precision is wasted when refraction varies +/- 1/8 from morning to night and when real-world fabrication tolerances are +/-1/8. Each trial set and phoropter is marked in quarters, but they all include +/- 0.12 sphere lenses if you want to refine the endpoint. Antique trial sets often include +/-037, 062, and 087 but new trial sets don't. Eighth-diopter _cylinder_ trial lenses are pretty rare. -MT > > Mike, I know that. > [quoted text clipped - 12 lines] > precision is wasted when refraction varies +/- 1/8 from morning to night and > when real-world fabrication tolerances are +/-1/8. True, and may be even more then 1/8 I think it is their imagination, probably they think that they are Leonardo de Vinci > Each trial set and phoropter is marked in quarters, but they all include +/- > 0.12 sphere lenses if you want to refine the endpoint. [quoted text clipped - 5 lines] > > -MT > I think it is their imagination, probably they think that they are > Leonardo de Vinci They imagine the precision is more important than it really is. -MT > > I think it is their imagination, probably they think that they are > > Leonardo de Vinci > > They imagine the precision is more important than it really is. > > -MT True. > They imagine the precision is more important than it really is. As a final step I insist my optometrist let me fine tune the cylinder angle by me tweaking the trial lenses in the trial frame while I look at an astigmatism target. I can refine that within 1 degree of optimal, better than the 5 degree increments usually specified. Unlike power, the angle seems like something you can trust the (knowledgeable) patient to optimize. > > They imagine the precision is more important than it really is. > [quoted text clipped - 3 lines] > than the 5 degree increments usually specified. Unlike power, the angle > seems like something you can trust the (knowledgeable) patient to optimize. If you have absoluttely "regular" asrigmatism, which is almost impossible in real eyes. Also if you astigmatizm is large it is detectable. Astigmatism like cyl +1 or -1 and angle 1deg? Imagination. > If you have absoluttely "regular" asrigmatism, which is almost > impossible in real eyes. Matters not whether it is regular. There will still be one best angle to apply a cylindrical correction, even if the error being corrected is not itself perfectly cylindrical. > > If you have absoluttely "regular" asrigmatism, which is almost > > impossible in real eyes. > > Matters not whether it is regular. There will still be one best angle to > apply a cylindrical correction, even if the error being corrected is not > itself perfectly cylindrical. Best angle has reasonable precision. Beyond threshold "best angle " is absoluttely relative, in the brain Like people without astigmatism but wearing progressive lenses which (that lenses) have a lot of astigmatism even do not see that they wear astigmatic lenses. All depends of perception. Also it is a thing like adaptation. When trying lenses, after changing angle few times you will not be sensetive enough to the angle changing. You should start again next time. > astigmatism target. I can refine that within 1 degree of optimal, better > than the 5 degree increments usually specified. Unlike power, the angle > seems like something you can trust the (knowledgeable) patient to > optimize. The greater the cylinder power, the more it matters. I always specify to the degree, but with 050 cylinder you can repeat the test three times and get three different answers. I've guided many hands up to the axis wheel to let them confirm my cross-cylinder results. Some people are wildly unreliable or inconsistent on the crossed-cylinder tests. -MT > > astigmatism target. I can refine that within 1 degree of optimal, better > > than the 5 degree increments usually specified. Unlike power, the angle [quoted text clipped - 10 lines] > > -MT But they do not understand, probably, that after you make the even perfect prescription, then OP will make lenses which have spec. of pressision I do not now what is presison spec. of making lenses, But after that OP will insert lenses into frame and I am sure, that error in angle much more than 1 deg And then they wear glasses which slide on nose, shift, bend and so on, so on, so on > > astigmatism target. I can refine that within 1 degree of optimal, better > > than the 5 degree increments usually specified. Unlike power, the angle [quoted text clipped - 8 lines] > cross-cylinder results. Some people are wildly unreliable or inconsistent on > the crossed-cylinder tests. But do they grind them to specs? What about premade blanks? I've probably told this story before, but here it is again: I went to see my OD. He said I was a ten (degrees). He ordered me a six pack. I went in a few weeks later and he overrefracted me. Not good. He looked in my eye, and the contact was rotated. He moved it back. That was better. He told me to blink. It rotated right back to where it wanted to be. He tried it once more, with the same results. So he did a calculation and said I needed a 25, but they only come in 10 degree increments, so he tossed a coin (not literally) and said I was a 30. He ordered me another six pack, even though I still had five of the others in the box. > I've guided many hands up to the axis wheel to let them confirm my > cross-cylinder results. Some people are wildly unreliable or > inconsistent on the crossed-cylinder tests. I would expect that. When it comes to instrumentation (physical or musical), there are many who will just never "get it". > > Made to order, come in smaller steps, I bet > > It is very common prescriptions like -2.1 or + 1.7 [quoted text clipped - 3 lines] > > -MT PS : 1/8 ? I believe that person can distinguish 1/8, especially with ads or cylinder. 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. Dear Lena, Subject: The OD takes a refractive-STATE "snap-shot" once every two years. Lena> Is somebody ever was thinking that eyes constantly moving and changing shape? (forget accommodation) Otis> In an engineering-scientific sense? Yes, of course they even made a motion-picture of the eye's adaptiveness. Otis> A picture is worth 1,000 words. And animation of the natural eye's dynamic adjustments -- is worth 10,000 words. Watch the blue tint animation of the eye's change -- with applied minus and plus lens. > That we are blinking? > That the eye - brain system has amazing Otis> Yes it is. The performance is EXPECTED and confirmed. (But denyed by the powers-that-be.) > shock-absorbing mechanism: we move head, arms, > even walking: the picture is still stable. Otis> The natural eye is a high-performance control system -- in all respects. > How much processing it takes ? Otis> Profound -- modern technology only gets "close". > Because of that OD can tell "you will use". > The power of adaptation is almost enormous. Otis> With they are in total denial of Otis> Watch the natural and fundamental eye change it refractive STATE will applied plus and minus lenses. http://vision.berkeley.edu/wildsoet/myopiaprimer.html Otis> How much clearer can this be??? Otis> Why the denial of these objective scientific facts by the majority opinion ODs? > > > Made to order, come in smaller steps, I bet > > > It is very common prescriptions like -2.1 or + 1.7 [quoted text clipped - 19 lines] > Because of that OD can tell "you will use". > The power of adaptation is almost enormous. 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. Best, Otis > > > However my calculations for my reading glasses seem to be about 0.2 > > > Disoptre LOWER than expected. [quoted text clipped - 11 lines] > Made to order, come in smaller steps, I bet > It is very common prescriptions like -2.1 or + 1.7 Hello Winke, A small point regarding evaluation of eyeglass lenses. The focal length used in that business is actually the "back focal distance". That is, the distance from the back surface of the lens (surface nearest the eye) to the focal plane on that side. This is usually a few mm shorter than the actual focal length, often referred to as the EFL or effective focal length. So the distant object should be on the same side as the convex surface and the focal length measured from the concave surface to the image. You can also use an opticians lens measure, often called a lens clock. It looks like a pocket watch with three prongs. The lens clock is used to measure the depth of the curvature of spherical surface. The dial numbers are in diopters assuming 1.530 for the refractive index. You would use this to clock both surfaces and sum the results. The answer is not perfect because the index of refraction of the lenses is probably not 1.530. More likley it is crown glass with index 1.523 or CR39 plastic with index 1.49. Also, this simple method does not factor in the lens thickness. However, it is good enough for labeling a bunch of reading glasses to the nearest 0.25 Diopters for personal use. Lens clocks go for $60 to $160 new, depending on the vendor. I got one for half of retail on ebay a few years ago. I am not suggesting that you go to that expense. Just throwing it in for information as another way to get the job done. Tom Hubin thubin@earthlink.net ********************** > 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 [quoted text clipped - 33 lines] > Are my results approximtely consistent with the light source not > being at infinity but very roughly 10 metres away? > Hello Winke, > [quoted text clipped - 31 lines] > Tom Hubin > thubin@earthlink.net 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 |
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