View Full Version : Optics discussion at sci.med.vision
judyb
09-17-2000, 02:30 AM
Hi all
The following message was posted to sci.med.vision under the title "lens power via lens clock" I answered using my nom de plume there, but thought one of you optical types might have a better reply. My memory of thick lens optics is rusty!
i have been experimenting with a co worker's lens clock on pts lens and i love
the idea of having a portable lensometer, but i haven't been able to get
the knack of getting a reliable reading. sometimes i'm within +/- 050, next
time i'm off by 1.5 d. my optician buddies only use it to get base curves, not
power. and none of their books have this info (i've been told). anyone
know how to do this?
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Judy B
Jeff Trail
09-17-2000, 04:05 AM
Originally posted by judyb:
Hi all
The following message was posted to sci.med.vision under the title "lens power via lens clock"
i have been experimenting with a co worker's lens clock on pts lens and i love
the idea of having a portable lensometer, but i haven't been able to get
the knack of getting a reliable reading. sometimes i'm within +/- 050, next
time i'm off by 1.5 d. my optician buddies only use it to get base curves, not
power. and none of their books have this info (i've been told). anyone
know how to do this?
Judy,
The reason they are getting variable powers are that the lens clock is based on the index of glass (1.53) so when you sag another material you HAVE to allow for the difference in index of material :)
There is a way to convert it if you want to go to the trouble, to find the refractive "power" because of the index the sag gauge was based on but here is a formula to figure it out
using the example a +6^ curve on crown glass..compared to a sag curve of CR39 ..
if the front surface of the lens clocked at +6 then you need to find the radius of curvature to get to the next step...
(clocked curve)+6 = 1.53(index)-1 over r1
so..1.53-1 would give you .53/+6.00 (clocked curve) which would give you the radius of curvature.. .53 devided by +6 = .0883 m
now to find the conversion you just run the same formula but plug in the radius as the devider... 1.498 (cr39)-1 over .0883 m(radius from the first formula) so now you end up with .498 devided by .0883 = 5.64^
So the refractive power of 6^ on the clock on a glass lens is ^.. but if you clocked a CR 39 lens with that same clock, the refractive power of that 6^ is actually 5.64^ ... so you would be off that .50^ you mentioned Judy :)
I would prefer just popping it into my lensometer :).. but you can (if in a bind) use that formula to compensate for the index the lens clock is set on.. BUT that is the reason the power is off, you (or they) were trying to use a gauge based on one index to get the power of a lens with a DIFFERENT index.. :) simple huh?..hope this was easy to follow..
Jeff "can't we make life to complicated at times?"Trail
Darryl Meister
09-17-2000, 02:00 PM
Hi Judy,
As Jeff pointed out, the difference in refractive index is the biggest factor affecting the power difference in most instances. However, as you alluded to, lens thickness will also play a role for higher plus powers.
Jeff's formulas can be equated for the radius to arrive at a simpler expression:
Actual Power = (Index - 1)/0.530 * Clock Power
Once you have measured both surfaces and converted them using the correct index, you can simply add the two powers together to arrive at the focal power for thin lenses. To compensate for the gain in back vertex power caused by lens thickness (for 'thick' lenses), you can add the additional plus power produced by the following approximation:
Allowance = Thickness / Index * Base^2
where the Thickness is given in meters and Base is the actual front curve power of the lens.
Here are some general comments about measuring lenses precisely with a lens clock:
1. You need to know the index of refraction of the lens material. This isn't always the easiest thing in the world to determine.
2. You need an accurately calibrated lens clock. You also need to use the lens clock correctly (e.g., holding it perpendicular to the surface).
3. You need to caliper the lens thickness if the lens is plus-powered.
4. You need to determine the minimum and maximum surfaces powers on the toric back surface, if the lens has cylinder power. You may also need a protractor or something to establish the correct axis.
5. If the lens is aspheric or progressive, lens clock readings will not be accurate.
Anyway, between my posting and Jeff's I hope this is enough to describe some of the difficulties of the topic!
Best regards,
Darryl
Jeff Trail
09-17-2000, 07:10 PM
Originally posted by Darryl Meister:
1. You need to know the index of refraction of the lens material. This isn't always the easiest thing in the world to determine.
2. You need an accurately calibrated lens clock. You also need to use the lens clock correctly (e.g., holding it perpendicular to the surface).
3. You need to caliper the lens thickness if the lens is plus-powered.
4. You need to determine the minimum and maximum surfaces powers on the toric back surface, if the lens has cylinder power. You may also need a protractor or something to establish the correct axis.
5. If the lens is aspheric or progressive, lens clock readings will not be accurate.
Best regards,
Darryl[/I]
You know Darryl I went over to the Sci.Med. group yesterday and posted the formula (as well as the "short" version) even though I like knowing the long version as well :)
I even mentioned the "index" problem.. and really I can see no way of really doing it accurately any longer.. with all the indexes out there how are they going to determine if it's a 1.53, 1.54, 1.55, 1.56, 1.57, 1.59 etc. etc. ... and as for toric (cross curve) oh my God :rolleyes: most opticians I know do not even attempt trying to clock the ocular to determine cylinder and sphere power..
I guess for a quickie to get CR39 then that might be easy but as for other indexes..well I just do see it being done, Shoot I handle more lens then a lot of the opticians (being wholesale) and I couldn't begin to differentiate the indexes (from 1.54 to 1.59 just by looking at a lens..poly's are easy (got that "ting") but the rest.. :rolleyes:
Then toss in something like Seiko "luscious" with a spherical fitting button with aspherical edge..YIKES...
10 years ago, yea it coulda been done regularly, currently? I don't think it's really an option, to many factors to take into account to get a really dependable answer.
I guess if you wanted to get technical enough you could do a "reverse" formula and take the curve (ocular)base&cross, front curve and power and come up with the index. if it was in a lensometer..oh well, to much math for me to worry about it..I prefer just pulling the box tab and entering the index and grinding it!! :)
Jeff "sometimes I HATE math" Trail
judyb
09-17-2000, 11:39 PM
Jeff and Darryl
Thanks for the quick response. I can't imagine why that original sci.med.vision poster wanted to use a lens clock to measure power in the first place; lensometers are so easy to use!
I knew there were complicated calculations and that the index of the material was critical. Posting the math will either scare him away from a career in optics or convince him to take a course.
BTW, I use a different name there because Outlook Express sends mail to newsgroups using my son's e-mail account and I can't fiqure out how to change it to mine. If I post via a forum like this one or Deja I can use my own name but Outlook provides a much smoother way of reading posts.
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Judy B
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