The document discusses the physics of the human eye. It describes the eye's parts and their functions, including the cornea, lens, retina, and ciliary muscles. It explains visual acuity and the eye's ability to focus at different distances through accommodation. Accommodation occurs when the ciliary muscles contract or relax to change the lens's focal length and allow vision from long distances to short distances. The document also covers topics like myopia, hyperopia, presbyopia, and how lenses can correct different vision problems by changing the eye's effective focal length.

1. Physics of the Human EyePhysics of the Human Eye

2. Its PartsIts Parts

4. VISUAL ACUITYVISUAL ACUITY
• A person with a normal eye has visualA person with a normal eye has visual
acuity ofacuity of 20/20 (in feet) or 6/6 (in20/20 (in feet) or 6/6 (in
meters)meters)
• A person with a VA of 20/50 means theA person with a VA of 20/50 means the
person sees at 20 ft what a normal personperson sees at 20 ft what a normal person
can see at 50 ft.can see at 50 ft.

5. NORMAL NEAR POINT OF THENORMAL NEAR POINT OF THE
EYEEYE
• NORMAL FAR POINTNORMAL FAR POINT of the eye is atof the eye is at INFINITYINFINITY
Receding Near point of the Eye with AgeReceding Near point of the Eye with Age

6. FocusingFocusing
• The cornea and eye lens form a compoundThe cornea and eye lens form a compound
lens system, producing alens system, producing a realreal invertedinverted imageimage
on the retina.on the retina.
– From air to corneaFrom air to cornea (n=1.376): large bending, the(n=1.376): large bending, the
main focusing.main focusing.
– From cornea to eyelensFrom cornea to eyelens (n=1.406), less focusing(n=1.406), less focusing
power. (Eyelens can develop white cloudiness whenpower. (Eyelens can develop white cloudiness when
getting old: Cataracts.)getting old: Cataracts.)
• The eye has a limited depth of field. We cannotThe eye has a limited depth of field. We cannot
see things close and far at the same time.see things close and far at the same time.

7. AccommodationAccommodation
• The eye focusing is not done by changingThe eye focusing is not done by changing
the distance between the lens and retinathe distance between the lens and retina..
Rather, it is done byRather, it is done by changing the focalchanging the focal
length of the eye lens.length of the eye lens. Ciliary muscles helpCiliary muscles help
to change the shape of the lens:to change the shape of the lens:
accommodation.accommodation.
– Muscles relax,Muscles relax, long focal length, see objects farlong focal length, see objects far
way;way; Muscles tenseMuscles tense, short focal length see, short focal length see
objects close.objects close.
– Normal eyes can see 25cm to infinity, however,Normal eyes can see 25cm to infinity, however,
if the cornea bulges too much or too little. Theif the cornea bulges too much or too little. The
accommodation does not help. (myopia oraccommodation does not help. (myopia or
hyperopia)hyperopia)

8. • The DiopterThe Diopter
•
The power of a lens is measured byThe power of a lens is measured by
opticians in a unit known as a diopter.opticians in a unit known as a diopter.
A A diopterdiopter is the reciprocal of the focal is the reciprocal of the focal
length.length.
diopters = 1/(focal length indiopters = 1/(focal length in
meters)meters)

9. • A lens system with a focal length of 1.8 cm (0.018 m) is a 56-diopter lens.A lens system with a focal length of 1.8 cm (0.018 m) is a 56-diopter lens.
A lens system with a focal length of 1.68 cm is a 60-diopter lens. A healthyA lens system with a focal length of 1.68 cm is a 60-diopter lens. A healthy
eye is able to bring both distant objects and nearby objects into focuseye is able to bring both distant objects and nearby objects into focus
without the need for corrective lenses. That is, the healthy eye is able towithout the need for corrective lenses. That is, the healthy eye is able to
assume both a small and a large focal length; it would have the ability toassume both a small and a large focal length; it would have the ability to
view objects with a large variation in distance. The maximum variation inview objects with a large variation in distance. The maximum variation in
the power of the eye is called the the power of the eye is called the Power of AccommodationPower of Accommodation . If an eye. If an eye
has the ability to assume a focal length of 1.80 cm (56 diopters) to viewhas the ability to assume a focal length of 1.80 cm (56 diopters) to view
objects many miles away as well as the ability to assume a 1.68 cm focalobjects many miles away as well as the ability to assume a 1.68 cm focal
length to view an object 0.25 meters away (60 diopters), then its Power oflength to view an object 0.25 meters away (60 diopters), then its Power of
Accommodation would be measured as 4 diopters (60 diopters - 56Accommodation would be measured as 4 diopters (60 diopters - 56
diopters).diopters).
• The healthy eye of a young adult has a Power of Accommodation ofThe healthy eye of a young adult has a Power of Accommodation of
approximately 4 diopters. As a person grows older, the Power ofapproximately 4 diopters. As a person grows older, the Power of
Accommodation typically decreases as a person becomes less able toAccommodation typically decreases as a person becomes less able to
view nearby objects. This failure to view nearby objects leads to the needview nearby objects. This failure to view nearby objects leads to the need
for corrective lenses.for corrective lenses.

11. 1.381.38

12. 1.341.34

13. 1.411.41

14. • Quantitative Relationships Between dQuantitative Relationships Between doo, d, dii and fand f
• The use of the The use of the lens equation and magnification equationlens equation and magnification equation can provide an idea of the quantitative relationship can provide an idea of the quantitative relationship
between the object distance, image distance and focal length. For now we will assume that the cornea-lensbetween the object distance, image distance and focal length. For now we will assume that the cornea-lens
system has a focal length of 1.80 cm (0.0180 m). We will attempt to determine the image size and imagesystem has a focal length of 1.80 cm (0.0180 m). We will attempt to determine the image size and image
location of a 6-foot tall man (hlocation of a 6-foot tall man (hoo=1.83 m) who is standing a distance of approximately 10 feet away (d=1.83 m) who is standing a distance of approximately 10 feet away (doo= 3.05= 3.05
meters). (The lens equation is derived geometrically upon the assumption that the lens is a thin lens. Themeters). (The lens equation is derived geometrically upon the assumption that the lens is a thin lens. The
lens of the eye is anything but thin and as such the lens equation does not provide a truly accurate model oflens of the eye is anything but thin and as such the lens equation does not provide a truly accurate model of
the eye lens. Despite this fact, we will use the equation as a simplified approximation of the mathematics ofthe eye lens. Despite this fact, we will use the equation as a simplified approximation of the mathematics of
the eye.)the eye.)
• Like all problems in physics, begin by the identification of the known information.Like all problems in physics, begin by the identification of the known information.
• ddoo = 3.05 m = 3.05 m
• hhoo = 1.83 m = 1.83 m
• f = 0.0180 mf = 0.0180 m
• Next identify the unknown quantities that you wish to solve for.Next identify the unknown quantities that you wish to solve for.
• ddii = ??? = ???
• hhii = ??? = ???
• To determine the image distance, the lens equation can be used. The following lines represent the solutionTo determine the image distance, the lens equation can be used. The following lines represent the solution
to the image distance; substitutions and algebraic steps are shown.to the image distance; substitutions and algebraic steps are shown.
• 1/f = 1/do + 1/d1/f = 1/do + 1/dii
• 1/(0.0180 m) = 1/(3.05 m) + 1/d1/(0.0180 m) = 1/(3.05 m) + 1/dii
• 55.6 m55.6 m-1-1
= 0.328 m = 0.328 m-1 -1
+ 1/d+ 1/dii
• 55.2 m55.2 m-1-1
= 1/d = 1/dii
• ddii = 0.0181 m = 1.81 cm= 0.0181 m = 1.81 cm

15. • To determine the image height, the magnification equation isTo determine the image height, the magnification equation is
needed. Since three of the four quantities in the equationneeded. Since three of the four quantities in the equation
(disregarding the M) are known, the fourth quantity can be(disregarding the M) are known, the fourth quantity can be
calculated. The solution is shown below.calculated. The solution is shown below.
• hhii/h/hoo = - d = - dii/d/doo
• hhi i /(1.83 m) = - ( 0.0181 m)/(3.05 m)/(1.83 m) = - ( 0.0181 m)/(3.05 m)
• hhi i = - (1.83 m) • (0.0181m)/(3.05 m)= - (1.83 m) • (0.0181m)/(3.05 m)
• hhi i = -0.0109 m = -1.09 cm= -0.0109 m = -1.09 cm
• From the calculations in this problem it can be concluded thatFrom the calculations in this problem it can be concluded that
if a 1.83-m tall person is standing 3.05 m from your cornea-if a 1.83-m tall person is standing 3.05 m from your cornea-
lens system having a focal length of 1.8 cm, then the imagelens system having a focal length of 1.8 cm, then the image
will be inverted, 1.09-cm tall (the negative values for imagewill be inverted, 1.09-cm tall (the negative values for image
height indicate that the image is an inverted image) andheight indicate that the image is an inverted image) and
located 1.81 cm from the "lens".located 1.81 cm from the "lens".

16. • Now of course, if the person is standing furtherNow of course, if the person is standing further
away (or closer to) the eye, the image size andaway (or closer to) the eye, the image size and
image distance will be adjusted accordingly. This isimage distance will be adjusted accordingly. This is
illustrated in the following table for the same 6-footillustrated in the following table for the same 6-foot
tall (1.83 m) person.tall (1.83 m) person.
Object Distance Image Distance Image Height
1.00 m 1.83 cm 3.35 cm
3.05 m 1.81 cm 1.09 cm
100 m 1.80 cm 0.329 cm

18. EYE ILLNESSESEYE ILLNESSES
 MYOPIA/NEARSIGHTEDNESSMYOPIA/NEARSIGHTEDNESS
HYPEROPIA/FARSIGHTEDNESSHYPEROPIA/FARSIGHTEDNESS
PRESBYOPIAPRESBYOPIA
ASTIGMATISMASTIGMATISM

19. Normal VisionNormal Vision

20. Nearsighted EyesNearsighted Eyes

22. Correcting NearsightednessCorrecting Nearsightedness
• The far point of a certain myopic eye is 50 cm in front ofThe far point of a certain myopic eye is 50 cm in front of
the eye. Find the focal length and power of the eyeglassthe eye. Find the focal length and power of the eyeglass
lens that will permit the wearer to see clearly an object atlens that will permit the wearer to see clearly an object at
infinity. Assume that the lens is worn 2 cm in front of theinfinity. Assume that the lens is worn 2 cm in front of the
eye.eye.

23. Farsighted EyeFarsighted Eye

25. Correcting FarsightednessCorrecting Farsightedness
• The near point of a certain hyperopic eye is 100 cm inThe near point of a certain hyperopic eye is 100 cm in
front of the eye. Find the focal length and power of thefront of the eye. Find the focal length and power of the
contact lens that will permit the wearer to see clearly ancontact lens that will permit the wearer to see clearly an
object that is 25 cm in front of the eye.object that is 25 cm in front of the eye.

26. Presbyopic EyesPresbyopic Eyes

30. ThankThank
youyou
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