### Author Topic: Physics - Eye is 60 diopters - estimated.  (Read 3007 times)

#### OtisBrown

• Hero Member
•     • Posts: 1766 ##### Physics - Eye is 60 diopters - estimated.
« on: April 10, 2013, 04:46:43 PM »
Dear John,

John Link>  According to http://en.wikipedia.org/wiki/Human_eye#Dimensions the adult eye has a diameter of about 24 mm. If we assume neither myopia nor hyperopia then I would conclude that the total optical power of the relaxed eye is almost 42 diopters, since 1000/24 = 41.67. But according to http://en.wikipedia.org/wiki/Diopter#In_vision_correction the total optical power of the relaxed eye is approximately 60 dioptres.

John Link>  How do we reconcile the difference between 42 and 60 diopters? Is my calculation incorrect? Is the information in wikipedia incorrect?

Otis> You made the assumption for a lens in air.  The eye is not a lens "in air".  In my next post I will show how the correct calculation is done.

#### OtisBrown

• Hero Member
•     • Posts: 1766 ##### Re: Physics - Eye is 60 diopters - estimated.
« Reply #1 on: April 10, 2013, 04:49:03 PM »
Subject: How to calculate the power of an eye with a refractive state of zero.

This should explain the dimensions and total power of a normal eye with a zero refractive state.

http://www.i-see.org/otis_brown/chapter_03.html

Best,

Otis

#### johnlink

• Full Member
•   • Posts: 177 ##### Re: Physics - Eye is 60 diopters - estimated.
« Reply #2 on: April 11, 2013, 07:21:34 AM »

You made the assumption for a lens in air.  The eye is not a lens "in air".

Thank you, Otis, for posting that link to exactly what I’ve been looking for. I see that the calculation you present yields a focal power of 57 diopters, which I consider to be in agreement with the statement in  http://en.wikipedia.org/wiki/Diopter#In_vision_correction that the optical power of the relaxed eye is approximately 60 dioptres. You used the following equation:

Focal Power = (Refractive Index) / (Focal Length)

However, according to http://en.wikipedia.org/wiki/Focal_power, focal power is defined as the reciprocal of the focal length, i.e.,

Focal Power = 1 / (Focal Length)

So we now have a pair of conflicting definitions. How do we resolve that? And how is the lens being in air or not relevant?