Re: Retinal image size calculation for human eye.

From: Bruce Woodburn (brucew_at_dccnet.com)
Date: 06/21/04 

Date: Sun, 20 Jun 2004 22:12:24 -0700

I don't know if this is going to help...

Imagine you start with a normal eye that is in focus at distance. To convert
it to a myopic eye, you put an imaginary plus sphere lens inside the eye at
the plane of the natural lens, about 5.5 millimeters behind the cornea. To
compensate, a minus spectacle lens is placed 14mm in front of the cornea.
You now have a reversed Galilean Telescope (paired plus and minus lenses)
which makes the image smaller. The opposite situation is true for hyperopes,
who see a magnified image (about 2.5% for each diopter)

When you look through a minus lens that is "too strong", your natural lens
accommodates by becoming more plus. This increases the strength of the
reversed Galilean Telescope and reduces the image size. It's not so much the
minus lens that is making the image smaller as the accommodation of your
lens. If accommodation is paralyzed with eye drops, the image would go
blurry, but remain closer to the same size.

I think the "2.5% per diopter" rule is pretty close for both minus and plus
lenses.

Bruce

PS: A Galilean Telescope is a 2 lens telescope (plus objective, minus
eyepiece) arrange so their focal points coincide. Magnification (or
minification) is the ratio of their strength in diopters (e.g.: a +1 diopter
objective and a -4 diopter eyepiece gives a 4 power scope.) This power
calculation doesn't apply to the armchair experiment described here because
of the complexity of the optical system.

"andrew Judd" <andrewedwardjudd@hotmail.com> wrote in message
news:64b5c3aa.0406180339.222c938c@posting.google.com...
> Hi
>
> Can somebody help out with a calculation to show the following?
>
> For your average near sighted person with a pair of glasses sitting
> with the surface of the lens nearest to the eye 14mm from the surface
> of the eye, what is the exact relationship between retinal image size
> and amount of minus sphere that they have in their glasses?
>
> Note that small changes in distance from the eye of a pair of glasses
> does not noticably change image size according to my perception of it.
> What does change image size noticably is the power of the sphere in
> front of the eye. Minus makes it smaller and plus bigger. I have
> different amounts of minus sphere by 2D in my glasses and easily see a
> size difference between eyes.
>
> Note that the near sighted eye is elongated by a tiny amount for each
> diopter of near sightedness present. This amount is about 0.33 mm per
> diopter of near sightedness. I think this consideration can be
> ignored for the purposes of the calculation?
>
> Internal changes in dioptric power of the eye do not cause very large
> amounts of image size difference according to the image size
> calculator shown below.
>
> http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
>
> I imagine that some kind of telescope power calculation is needed
> here?
>
> So having achieved this calculation, what is then the amount of minus
> sphere in the glasses required to reduce the retinal image by 100%
> while the near sighted eye is looking at infinity?
>
> This might help?
>
> eye schematic
>
> http://hyperphysics.phy-astr.gsu.edu/hbase/vision/eyescal.html
>
> Image formation concepts
>
> http://hyperphysics.phy-astr.gsu.edu/hbase/vision/imagformcon.html#c1
>
> Thanks
>
> Andrew

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