Optics and the Human Eye - Lecture Notes

Optics and the Human Eye

Lenses are crafted from glass, ground and polished to achieve a smooth surface.
A spherical surface on one or both sides defines a lens.
Thin lenses have a thickness that is small compared to their radius of curvature.
They can be converging (thicker in the center) or diverging (thicker at the edge).

Ray tracing helps locate an image using three key rays:
- Ray 1: Enters parallel to the axis and exits through the focal point.
- Ray 2: Enters through the focal point and exits parallel to the axis.
- Ray 3: Passes through the center of the lens and is undeflected.

For a diverging lens, the same three rays (1-3) are used.
The image formed is upright and virtual.
A virtual image does not have light rays passing through it, like a mirror image.
The eye can perceive both real and virtual images, but a virtual image cannot be projected on a screen.

The thin-lens equation relates object distance (do), image distance (di), and focal length (f):
\frac{1}{do} + \frac{1}{di} = \frac{1}{f} = P
P represents the power of a lens, measured in diopters (1D = 1/m).
A strong lens (large P, small f) bends incident rays through large angles.

Magnification (m) is defined as: m = \frac{hi}{ho} = -\frac{di}{do}
- h_i is the image height.
- h_o is the object height.
The sign of m indicates whether the image is upright (positive) or inverted (negative).
f (and hence P) is positive for converging lenses and negative for diverging lenses.

The cornea and lens function together as a single thin lens.
The lens adjusts its power (f) to focus images on the retina (at a fixed distance d_i).
Near point: The closest distance at which an object is clearly in focus.
Far point: The farthest distance at which an object is clearly seen.

Visual acuity is the ability to distinguish two points with a small angular separation (
\thetao ). \thetao ≈ 5 × 10^{-4} \text{ radians} = 0.03°
This corresponds to the smallest feature that the unaided human eye can perceive and relates to the separation of cones in the retina.
To distinguish two small objects as separate, there must be at least one unexcited cone between two excited cones.

Given an average near point of 25 cm for a normal human eye, the separation of cones in the retina can be calculated as follows:
\theta ≈ \sin \theta ≈ \tan \theta = \frac{hi}{di}
hi = di \theta = 0.25 × 5 × 10^{-4} = 0.000125 \text{m} = 0.125 \text{mm}

When focusing on an object, the size of the image on the retina depends on the angle subtended by the object, \theta.
To enlarge the image and distinguish finer features, objects can be moved closer to the eyes, up to the near point limit.

An object placed at a distance less than or equal to f forms a virtual image.
The image is enlarged, making it suitable for simple magnifiers and eyepieces in optical instruments.
Magnification Formula:
M = \frac{\theta'}{\theta} = \frac{25 \text{cm}}{f}
where
\theta' ≈ \frac{h}{f}
and
\theta ≈ \frac{h}{25 \text{cm}}
When the eye is relaxed (focused at ∞) and the object is close to the focal point.

Problem: A converging lens with a focal length of 8 cm is used as a magnifying glass by a person with normal vision. What is the angular magnification when the eye is relaxed (image at infinity)?
Solution:
M = \frac{0.25}{f} = \frac{0.25}{0.08} = 3.1

As seen with the compound microscope: the image formed by the first lens becomes the object for the second lens.
The total magnification is the product of the individual magnifications
Consider a converging (f1) and diverging lens (f2) in contact. The sun’s rays focus to f{com}: \frac{1}{f{com}} = \frac{1}{f1} + \frac{1}{f2}
P{com} = P1 + P_2

Near point of normal human eye is taken to be X_n = 25cm
What is the power of the eye when focused on the near point?
Pn = \frac{1}{X} = \frac{1}{do} + \frac{1}{d_i} = \frac{1}{0.25} + \frac{1}{0.02} = 54D
The near point varies with age, e.g:
- 3 years: 7cm
- 20 years: 10cm
- 60 years: 100cm

Far point of normal human eye is taken to be X_f = ∞
What is the power of the eye when focused on the far point?
P{far} = \frac{1}{X} = \frac{1}{do} + \frac{1}{d_i} = \frac{1}{\infty} + \frac{1}{0.02} = 50D
The power of accommodation is the variation of the power of the eye from the near to the far point.
For a normal eye this is:
Pn − Pf = 54 − 50 = 4D
Young children have a greater power of accommodation; it decreases with age as the near point recedes. Most elderly people need corrective glasses for reading or close work.

Correcting nearsightedness:
- Use of diverging lens.
Correcting farsightedness:
- Use of converging lens.
  Spectacles typically have power of 1-3 D

Susan’s near point is at 32 cm. What power corrective lenses are required for reading? Assume a lens very close to the eye.
Unaided eye:
P{unaided} = \frac{1}{f{unaided}} = \frac{1}{0.32} + \frac{1}{0.02} = 3.1 + 50.0 = 53.1D
With spectacles: P{spectacles} = \frac{1}{f{spectacles}} = \frac{1}{f{unaided}} + \frac{1}{f{lens}} = 54.0D P{lens} = \frac{1}{f{lens}} = 54.0 − \frac{1}{f_{unaided}} = 54.0 − 53.1 = +0.9D
- i.e. a converging lens