Optics and the Human Eye - Lecture Notes

Optics and the Human Eye

Lenses

• 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).

Converging Lens

• A converging lens focuses parallel rays of light.
• It is thicker in the center than at the edge.
• Lenses form images, as seen in the human eye and optical devices.

Diverging Lens

• A diverging lens makes parallel light diverge.
• It is thicker at the edge than in the center.
• The focal point is where diverging rays converge when projected back.
• Diverging lenses are primarily used to correct eye defects.

Thin Lens - Ray Tracing

• 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.

Real Object – Virtual Image

• 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.

Thin Lens Equation

• The thin-lens equation relates object distance ($d<em>o$), image distance ($d</em>i$), and focal length ($f$):
  $\frac{1}{d<em>o} + \frac{1}{d</em>i} = \frac{1}{f} = P$
• $P$ 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

• Magnification (m) is defined as: $m = \frac{h<em>i}{h</em>o} = -\frac{d<em>i}{d</em>o}$
  • $h_i$ 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.

Physics of the Eye

• 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

• Visual acuity is the ability to distinguish two points with a small angular separation (
  $\theta<em>o$ ). $\theta</em>o ≈ 5 × 10^{-4} \text{ radians} = 0.03°$
• 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.

Example – Visual Acuity

• 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:
  $\theta ≈ \sin \theta ≈ \tan \theta = \frac{h<em>i}{d</em>i}$
  $h<em>i = d</em>i \theta = 0.25 × 5 × 10^{-4} = 0.000125 \text{m} = 0.125 \text{mm}$

Viewing Close Objects – Near Point

• 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.

Simple Magnifying Lens

• 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.

Example - Relaxed viewing

• 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:
  $M = \frac{0.25}{f} = \frac{0.25}{0.08} = 3.1$

Compound Microscope

• The compound microscope consists of an objective and an eyepiece lens.
• The objective lens forms an image.
• This objective image acts as the object for the eyepiece lens.
• The overall magnification is the product of the two lenses:
  $m = m<em>o m</em>e$

Combinations of lenses

• 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 ($f<em>1$) and diverging lens ($f</em>2$) in contact. The sun’s rays focus to $f<em>{com}$: $\frac{1}{f</em>{com}} = \frac{1}{f<em>1} + \frac{1}{f</em>2}$
  $P<em>{com} = P</em>1 + P_2$

Power of accommodation

• 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?
  $P<em>n = \frac{1}{X} = \frac{1}{d</em>o} + \frac{1}{d_i} = \frac{1}{0.25} + \frac{1}{0.02} = 54D$
• The near point varies with age, e.g:
  • 3 years: 7cm
  • 20 years: 10cm
  • 60 years: 100cm

Power of accommodation

• 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?
  $P<em>{far} = \frac{1}{X} = \frac{1}{d</em>o} + \frac{1}{d_i} = \frac{1}{\infty} + \frac{1}{0.02} = 50D$
• 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:
  $P<em>n − P</em>f = 54 − 50 = 4D$
• 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.

Vision Correction

• Myopia (Nearsightedness):

  • Cause: Eye too long or lens too strong.

• Hyperopia (Farsightedness):

  • Cause: Eye too short or lens too weak.

Vision Correction

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

Example – Vision Correction

• 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:
  $P<em>{unaided} = \frac{1}{f</em>{unaided}} = \frac{1}{0.32} + \frac{1}{0.02} = 3.1 + 50.0 = 53.1D$
• With spectacles: $P<em>{spectacles} = \frac{1}{f</em>{spectacles}} = \frac{1}{f<em>{unaided}} + \frac{1}{f</em>{lens}} = 54.0D$$P<em>{lens} = \frac{1}{f</em>{lens}} = 54.0 − \frac{1}{f_{unaided}} = 54.0 − 53.1 = +0.9D$
  • i.e. a converging lens

Summary

• Converging lens: focuses the parallel incoming rays at a focal point
• Diverging lens: incoming rays spread to appear to come from a point
• Thin lens equation:
  $\frac{1}{d<em>o} + \frac{1}{d</em>i} = \frac{1}{f} = P$
• Magnification:
  $m = \frac{h<em>i}{h</em>o} = -\frac{d<em>i}{d</em>o}$
• Real image: light passes through the image
• Virtual image: light does not pass through
• Human eye
  • Visual acuity: minimum angular separation of two points
  • Accommodation: ability of the lens to change its focal length
  • Near point of a normal human eye is taken to be $X_n$ =25 cm.
  • Far point of normal human eye is taken to be $X_f$ = ∞.
  • Nearsightedness/Farsightedness can be corrected by diverging/converging lenses respectively.