Focal 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?

 

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?

 

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?

 

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.

 

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
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.

 

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

 

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

 

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
measurement.

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

 

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

 

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

 

You're welcome!

Cheers,

Mark

 

Made to order, come in smaller steps, I bet
It is very common prescriptions like -2.1 or + 1.7

 

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

 

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

 

Mike, I know that.

Why would some body needs eighths.eighth ?

 

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

 

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.

 

True, and may be even more then 1/8

I think it is their imagination, probably they think that they are
Leonardo de Vinci

 

True.