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Wize University Physics Textbook (Master) > Geometric Optics

The Eye

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The Eye




We can apply physics knowledge about refraction, lenses, and images to the eye.

Light first enters at the cornea: this interface is where we see the most refraction. The approximate indices of refraction for various parts of the eye are:
  • Cornea: 1.38
  • Lens: 1.40
  • Aqueous humor and vitreous humor: 1.33


The lens is able to alter its focal length, depending on whether we are viewing something at a distance or nearby. The goal is to have an image form crisply on the fovea (which is part of the retina layer at the back of the eye).

  • The far point is at infinity
  • The near point is at 25 cm25 \ \text{cm} away from the eye

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Myopia

Myopia is the result of an eye lens that focuses short of the retina (near-sightedness). Their far point is too near.

A myopic person requires a diverging lens (negative focal length and power) to correctly focus light at the retina. Their glasses\contacts will create images of far away objects at the far point (the images of far away objects are brought closer).

 1do+1di=1f \boxed{ \ \dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f} \ }
  • do=∞d_o=\infin (the eye doctor will place an object far away)
  • did_i will be the negative of your actual far point (negative since the image formed by the glasses is virtual)
  • ff will be the focal length of the glasses needed (negative since it's a diverging lens)








Hyperopia

Hyperopia is the result of an eye lens that focuses behind the retina (far-sightedness). Their near point is too far.

A hyperopic person needs a converging lens (positive focal length and power) to correctly focus light at the retina. Their glasses\contacts will create images of nearby objects at the near point of the eye (the images of close objects are brought farther away).

 1do+1di=1f \boxed{ \ \dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f} \ }
  • do=0.25 md_o=0.25 \ \text{m} (the eye doctor will place an object at the normal near point)
  • did_i will be the negative of your actual near point (negative since the image formed by the glasses is virtual)
  • ff will be the focal length of the glasses needed (positive since it's a converging lens)




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Wize Concept
  • For myopia (near-sightedness), a diverging lens is needed to take a far away object (located at infinity, the normal far point), and create an image of it at your far point. Then your eyeball will see the image of the far away object at its own far point.
  • For hyperopia (far-sightedness), a converging lens is needed to take a nearby object (located at about 2525 cm, the normal near point), and create an image of it at your near point. Then your eyeball will see the image of the nearby object at its own near point.



Wize Concept
  • When using glasses (or contacts), the eye lens and the glasses effectively become a two-lens system. The image created by the glasses (or contacts) acts as the object for the eye lens.



Lens Power

Lens power is defined as:

 D=1f \boxed{ \ {D}=\dfrac{1}{f}} \


where ff is the focal length in meters. The units for lens power are called diopters (1 D=1 m−11\ D=1\ m^{-1}).


Exam Tip
Because focal length and lens power are inverses of each other, the shorter the focal length, the higher the power, and vice-versa. This is because a smaller radius implies a larger curvature.

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Example: Myopia


Your friend has myopia and their far point is 300300 cm.

a) Do they need a diverging or converging lens to correct their vision?
b) What lens power would correct this person's vision?
c) Repeat part b), but assume that the lens is going to sit 22 cm away from the eye. Comment on your result.



Part a)


Diverging lenses are used for myopic (near-sighted) people.

(By the way, a 300300 cm far point is actually pretty good; you might not even wear glasses for that, except for driving).


Part b)


The lens will be used to shift the image of the object placed at infinity (the normal far point) to the person's far point of 33 m.

This means that d0=∞d_0=\infin which makes 1do=0\dfrac{1}{d_o}=0. We also have di=−3d_i=-3 (negative because it's a virtual image).

Use the lens equation to find the power:

D=1f=1di+1doD=\dfrac{1}{f}=\dfrac{1}{d_i}+\dfrac{1}{d_o}

=1−3+0=\dfrac{1}{-3}+0

=−0.333=-0.333 (D)

Note: The weakest prescription you can get is −0.25-0.25 D


Part c)


The object needs to be seen 300300 cm from the eye. Since the lens is 22 cm away from the eye, the object placed at infinity has to have the image at 298298 cm from the lens.

So di=−2.98d_i= -2.98 and we get:

D=1f=1di+1doD=\dfrac{1}{f}=\dfrac{1}{d_i}+\dfrac{1}{d_o}

=1−2.98+0=\dfrac{1}{-2.98}+0

=−0.336=-0.336 (D)


Note: The glasses needed will be the same as before. For such a weak prescription we don't even need to account for the distance from the eyeball to the lens, which matters more for stronger prescriptions.


Practice: Contacts vs. Glasses


a) Consider a myopic person with a contact lens prescription of −4.00-4.00 D. What is her far point?

b) Another person wears glasses with the same prescription that are placed 1.751.75 cm from the eye. What is her far point?