PHYSICS OF THE HUMAN EYE
PARTS OF THE EYE
HOW THE EYE WORKS
• Light enters the eye through the pupil and
onto the lens.
• The light then passes through the lens and
converges and focused unto the retina.
• The retina, which is located at the back of the
eye acts as screen or film on which the image
is produced.
REFRACTION OF LIGHT IN THE EYE
• When viewing distant objects the lens adjust
itself to become flatter.
• Less refraction is needed to focus an image
unto the retina when viewing far objects.
• When viewing near objects the lens adjust
itself to become fatter.
• More refraction is needed to focus an image
unto the retina when viewing near objects.
IMPORTANT FEATURES OF THE EYE
•
ACCOMMODATION – This is ability of the lens to adjust itself to provide better focus
of image unto the retina.
•
SHORT-SIGHTEDNESS (Myopia) – The ability to see objects clearly that are near. Far
objects are blurred.
•
LONG-SIGHTEDNESS (Hyperopia) – The ability to see objects clearly that are far
away. Near objects are blurred.
•
ASTIGMATISM – This is the unevenness or asymmetry in how the eye focus light
from objects in different planes. This is due mainly to unevenness in the shape of the
cornea but can also be due to irregularities with the lens or retina.
•
CATERACTS - A cataract is a clouding of the lens in the eye that affects vision. Most
cataracts are related to aging. Cataracts are corrected by glasses with converging
lens or by removing the lens through surgery.
FAR AND NEAR SIGHTEDNESS
CORRECTING NEAR-SIGHTEDNESS
CORRECTING NEAR-SIGHTEDNESS
• A diverging or concave lens is placed in front of
the eyes to enable persons to see far objects.
• In a near-sighted person the eye lens bends the
light too much preventing it from being focused
on the retina.
• The diverging lens spreads out the light entering
the eye thereby reducing the over convergence of
the eye lens.
CORRECTING FAR-SIGHTEDNESS
CORRECTING FAR-SIGHTEDNESS
• A converging or convex lens is placed in front of
the eyes to enable persons to see near objects.
• In a far-sighted person the eye lens does not
bends the light enough to focus the image on the
retina.
• A converging lens bends the light entering the
eyes thereby compensating for the under
convergence of the eye lens.
DEPTH OF FIELD & DEPTH OF FOCUS
DEPTH OF FIELD & DEPTH OF FOCUS
• Depth of Field – Is the distance in front of the
eye between the persons near point and far
point where an object can still be seen clearly.
• Depth of Focus - Is the distance towards the
back of the eye where the image is in focus.
IMAGE FORMATION IN SIMPLE
CAMERA
• Camera: Image is real, diminished & inverted.
IMAGE FORMATION IN MAGNIFYING
GLASS
• Magnifying glass
POWER OF A LENS
•
POWER RANGE OF THE EYE
• A person with normal (ideal) vision can see
objects clearly at distances ranging from 25 cm to
essentially infinity
• The eye is most relaxed when viewing distant
objects
• Vision of very distant objects is called totally
relaxed, while close vision is
termed accommodated, with the closest vision
being fully accommodated.
POWER RANGE OF THE EYE
• Calculate the power of the eye when viewing
objects at the greatest and smallest distances
possible with normal vision, assuming a
lens-to-retina distance of 2.00 cm (a typical
value).
• Strategy
• For clear vision, the image must be on the retina,
and so di = 2.00 cm here. For distant vision, do ≈
∞, and for close vision, do = 25.0 cm
POWER RANGE OF THE EYE
POWER RANGE OF THE EYE
POWER RANGE OF THE EYE
• For an eye with this typical 2.00 cm
lens-to-retina distance, the power of the eye
ranges from 50.0 D (for distant totally relaxed
vision) to 54.0 D (for close fully
accommodated vision), which is an 8%
increase.
Questions
• Correcting Nearsightedness
What power of spectacle lens is needed to correct the
vision of a nearsighted person whose far point is 30.0
cm? Assume the spectacle (corrective) lens is held 1.50
cm away from the eye by eyeglass frames.
• Correcting Farsightedness
What power of spectacle lens is needed to allow a
farsighted person, whose near point is 1.00 m, to see an
object clearly that is 25.0 cm away? Assume the
spectacle (corrective) lens is held 1.50 cm away from the
eye by eyeglass frames.
SOLUTION 1 (Myopia)
• Strategy
• You want this nearsighted person to be able to see
very distant objects clearly. That means the spectacle
lens must produce an image 30.0 cm from the eye for
an object very far away. An image 30.0 cm from the
eye will be 28.5 cm to the left of the spectacle lens
(see Figure 2). Therefore, we must get di = −28.5 cm
when do ≈ ∞. The image distance is negative, because
it is on the same side of the spectacle as the object.
• P = - 3.51 D
SOLUTION 1 (Hyperopia)
• Strategy
• When an object is held 25.0 cm from the person’s
eyes, the spectacle lens must produce an image 1.00
m away (the near point). An image 1.00 m from the
eye will be 98.5 cm to the left of the spectacle lens
because the spectacle lens is 1.50 cm from the eye
(see Figure 3). Therefore, di=−98.5 cm. The image
distance is negative, because it is on the same side of
the spectacle as the object. The object is 23.5 cm to
the left of the spectacle, so that do=23.5 cm.
• P = 3.24 D
MORE QUESTIONS
• What is the far point of a person whose eyes have a relaxed
power of 50.5 D?
• What is the near point of a person whose eyes have an
accommodated power of 53.5 D?
• The far point of a myopic administrator is 50.0 cm. (a) What
is the relaxed power of his eyes? (b) If he has the normal
8.00% ability to accommodate, what is the closest object he
can see clearly?
• A myopic person sees that her contact lens prescription is
−4.00 D. What is her far point?