Download physics of eyes

The physics of the eye and vision :
The human eye :1. Cornea : The transparent part of the eye .
2. visual axis .
3. iris .
4. Aqueous humor .
5. The Lens .
6. vitreous humor .
7. vetina .
8. Foves centralis .
9. optic nerve .
10. blind spot .
11. ciliary muscle .
The following special features of the humans eye :1. The eye looking straight ahead has a wide angle of
vision .
2. The lens of the eye has variable focal length .
3. The eye can operate effectively over a range of light
intensity of about 1010 =1 daylight to very dark
night .
4. The eye has an automatic aperture adjustment (the
iris).
5. The eye has a self – regulating pressure system
about 20 mm Hg.
6. The image appears upside down on the light
sensitive retina at the back of the eyeball but the
brain automatically corrects for this .
7. The Muscles of the eye permit flexible movement
up and down sideways and diagonally .
The six muscles of the right eye permit a wide variety
of motion .
- U - D pair of muscle controls up and down .
- L – R pair of muscle controls Left and right
movement .
- R pair of muscle controls rotation .
Focusing elements of the eye
1. The Cornea ( C ): The Cornea focuses by bending
(refracting ) the light rays .
2. The Lens ( L ) : The Lens is variable in shape and
has the ability to focus objects at various distances
but the cornea is a fixed focus element .
The amount of Bending depend on :
a. The curvatunes of its surfaces :
Human eye is convex lens , crystalline , the focal
length of this lens is change by special muscle around the
lens ciliary muscle and the focal length depend on the
radius of the curvature ( R ) .
F
1
R
2
b. The speed of light in the lens compared with the
optical part of the eye .
Part of eye
C
A. humor
L. cover
L. center
V. humor
N( refrective index)
1.37
V

C
n 0  0 
1.33
V
 V
1.38
1.41
 o = wave length in
vacuum
1.33
 = wave length in
medium
Light detector of the eye :
Retina :
The light sensitive part of the eye converts the light
images into electrical nerve impulses that are sent to the
brain .
The physical aspects of the retina :
The absorption of visible light with energy > 1 cv
causes a photochemical reaction in the photorecepter and
initiates the action potential then the light image convert
into electrical signal sent to the brain through the optic
nerve .
- retina covers the back half of the eyeball .
- all the cisions takeplace in avery small area in the
yelow spot (0.3 mm dia) called the fovea centralis .
- The image on the retina , real , not erect and is very
small due to short image distance about (2 cm)  20
mm .
- Simple relation between the object and image size
and object and image distance comes from the ratio
of the lengths of sides of similar triangles .
Where P= object distance
Q=Image distance = 20 mm = 2 cm = 0.02 m
I = Image size , O = Object size .
O
P
Q
2

 I  Q   I   P O   P cm O


 


- Cones and rods (Type of photorecepter ) in the
retina .
- Cones { 6 x 106 in each eye } used for faylight
(photopic vision).
- Reds { 120 x 106 in each eye } cover most retina
used for night scotopic (vision).
Nods and Cones sensitivity :
- Cones has max at  = 550 nm ( yellow – green
region).
- The reds has max at = 510 nm (blue – green
region).
- The nods and Cones are equal sensitive at  = 650
to 700 nm (red region).
- The reds are much more sensitive than the cones .
Diffraction effects on the eye :
The iris produces a diffraction pattern on the retina
like the diffraction of light pass through a small opening
( slit ) . When mono chromatic light ( = 555 nm) from a
distant point source is brought to a focus at the Fovea
centralis in the retina .
From the relation d sin  = n
Where d = dia of the pupil of the iris .
 = angle of the diffraction of the light
n = order of the diffraction
 = Wave length of the light .
The diffraction pattern on the retina produced by a
pupil ( 3.0 mm dia.) consists of a central bright spot
( 8MM ) diameter surrounded by a ring of light of
reduced intensity (I).
From the equation :-  = 1.22λ / D
 The angular spread 2  of the central bright spot of the
retina :
2  = 2 x ( 1.22λ / d ) ; if d = 3 mm ,  = 555 nm
2  = 2 x 1.22 x 555 x 10-9 / 3 x 10-3
= 4.5 x 10-4 radians
 The diameter of the central bright spot at the retina = 2
 x retna distance .
=2xQ
= 2  x 20 mm
= 4.5 x 10-4 redians x 20 mm
 8 mm , 1m = 106 mm
The resolution of the eye :
Visial acuity :
The minimum angular seperation of the two
equidistance points of light which can just be resolved by
the eye .
Visual acuity = tan  = d / p   = tan-1 d / p
If d or the two points are closes than this minimum
value the eye will merely see the light as coming from
asingle point source .
The resolution by using objective Lens of microscope
of diameter (d) :
From the Fig. Below the angle subtended by the
images at the center of the lens is  .
 =
1.22
, If the two point object
d
distance ( d ) examined in a microscope will be resolved
and seen as separate point images :
If d =

2n sin i
n = refractive index
,  = wavelength of the light ,
of medium between object and objectivelens , I = The
semi- angle of the cone formed by joining the object to
the perimeter of the objectivelens 2n sin I = numberical
aprerture of microscope .
1.22
for eye a = diameter of the pupil .
a
Example : What is the limit of resolution and the useful
magnification of a microscope viewing object in air ,  =
600 nm , I = 80o .
=
- The limit of resolution d 

6 10 7 M

 3 10 7 M
2n sin i 2 1 sin 80
- Useful magnification M
0.1mm
10 4 M
M

 300
d (resolution ) 3 10 7 M
0.1 mm = limit of resolution of eye at the least distance
of distinct vision .
The Test of Vision :
The Property of our eyes to determine whether we
need corrective lenses test by using eye charts .
The eye chars :
Snellen chart to test vision is usually viewed from
20 ft [6 meter].
The kind of test :
1)
20 6

20 6
It means that you can read detail from 20 fi that a person
with good vision can read from 20 ft ,
2)
20 6

40 12
, you can read from 20ft that a person with
good vision can read from 40 ft .
3)
20 6

60 18
, you can read from 40 ft that a person
with good vision from 60 ft .
The strength ( power) of optic system :
The strength ( power ) [ refractive power]:
Is the reciprocal of the focal length of this system in
meter .
 the power ( P) 
1
100
or
F (meter)
f (cm)
f = the focallength in (M)
P
1
D
F ( m)
, where D is unit of the strength of this
system when F in Meter .
 D  diopter 
1
F (M )
1) The power of thin lens
A- Spherical Lens :
a. The strength ( power) of converging Lens
Converging Lens with positive focal length ( +F)
 Powr  p 
1
  Ddioptricst rength
 F (M )
if
F  0.1M 
1
 10 D
F
if
F  100cm 
1
1

 1D
F 1M
the lens of 1 diopter can bend or focused the ray at focus
of 1 meter .
 thepower
1
F
b. The strength of diverging lens ( concave lens ) :
diverging lens with negative focal length ( -f )
`
1
 Powr  p 
 F (M )
F  0.5Mcm 
 D
1
1

 2 D
F
0.5M
B. The cylindrical lens :
- + ve Lens
cylinder + convex = + ve cylinder
P
- + ve Lens
1
 D
F
- Ve cylinder = cylinder + concve
P
1
 D
F
2. The power of human eye :
a. The power of human eye lens at far point :
for point :- The point at which distant objects are
focused when the focusing muscle is relaxed
p = for point
 the power of eye at far point P
P 

1
1 1 1 1 1
    
 F ( )
F P Q  Q
1
1

F
Q
 P 
1
1

 50 D
Q
0.02 M
b. The power of human eye lens at near point :
near point :The closest point at which objects can be focused
when the lens is its thicknest .
PN = poher at near point = 1 / FN

1 1
1
1
 

P Q 0.25M Q
p=25cm near point
PN = 4 + 50 = 54 D .
c. The power of accommodation of the H . eye lens .
The ability of the lens by using ciliery muscle to change
the focal length between the far point and near point .
Pacc = PN - Pf
= 54 D – 50 D = 4 D
3. The power of the combination Lens :The focal length of two lens with f1 asnd f2
1
1
1


F
f1
f2
Pcomb = sum of diopters of the different lens .
D = D1 + D2
Ex:
- Pcomb = Peye + Plens
= eye + )(
= +D –D = 0
if P of eye ( +1 D) = P of lens ( -1 D)
Ex:
-
Pcomb = Peye + Plens = Oeye + Olens = +D +D using
for correct the vision of H. eye .
The total refractive power of the eye :Total refractive power of the eye = 59 D
P of the amterior surface of the cornea = 48 D
The posterior surface of cornea = - 4 D because it is
concave surface .
the total power refractive of cornea = 48 – 4 = 44 D
 2 / 3 of total power of eye .
The total power of the crystalline lens of the eye
when it surrounded by fluid on each side is equal to 15 D
.
the total power of the eye = PC + PL = 44 + 15 = 59 D
the normal power of eye = 59 D.
the curvature of
accommodation .
lens
is
very
important
for
Color visions and chromatic aberraction :
* Color vision :
- Three types of cones ( Red , Green , Blue ) .
- Red cones + Green cones  yellow sensation .
- Red + Green + Blue  White
if one of color sets is gone , the person is color blind .
Color blind : 890 male , 0.5 – 1% female .
* Chromatic aberration :
It is caused by change (n) refractive index with the wave
length ( ) :
n = C / V ……….(1)
f = C ………..…..(2)
n  
0
f
f

0
wave.length.in.air


wave.length.in.lens
nB = 1.523
nR = 1.517
Ocular chromatic aberration :-
- different colors are focused differently by the eye .
- the distance between fB and fR is 0.7 mm , which is
more than twice the thickness of the retina .
- about ( 2.5 D) difference in focusing power between
Blue – Red .
The defective vision :The emmetropic eye : the normal eye with good
vision , N . P = 0.25 m , F . P =  and Pacc = 4 D .
The ametropia : defective eye sight due to focusing
( refractive ) problems ) .
The types of ametropia :
1. Myopia ( near – sightedness) short sight :
The causes :
1. Long an eyeball d > 0.02 m .
or 2. The large curvature of the cornea .
The eye point :
( f . P) far point of this eye < f . P of normal (  )
N . P  0.25 m or little value than 0.25 m .
The correction of this defect by using diverging lens
with focal length in magnitude is equal to the f . p of
myopic eye . The prescribed lens must produced a virtual
image at the f . P of the eye of an object at (  ) infinty
so that the relaxed eye can focus on the retina the rays
coming from the lens and strinking it .
In the myopic eye , the parallel rays are focused by
the relaxed eye to a position in front of the retina only
near object see clearly .
2. Hypeopia ( far – sightedness ) , hypermetropia ,
long sight .
The causes :1. short an eyeball Q < 0.02 m .
or 2. small curvatare of the cornea .
The eye point :N . P near point > N . P of normal eye ( 0.25m) .
F . P 
The correction :
By using converging lens . the parallel rays are
focused to a point behind the retina , using the corrective
lens which will bring the eye's near point back to the
normal value by produce virtual image of an object
placed at new near point (25m) which will appear to the
eye to lie at the old near point .
Ex : The near point of eye is 100 cm in front of the eye .
What lens should be used to see clearly an object 25m in
front of the eye .
Defect : hyperopia
Correction: using converging lens
Object at near point (25m) , U= 25 cm  virtual image
at the old new point v= - 1m .
(lens ) 
1
1
1
1
1




f
u
v
25 1
f= 33 cm . That's a converging lens of focal length 33 cm
.
Ex: A man has a near point at 50 cm and far point at  :
1. What is his useful accomdating power ?
2. What spectacle lens used ?
The solution:
1. When eye focused at 
P 
1
1
1


f
 D
And when eye is accommodating to it max .
PN 
1
1
1


FN
0.50 M
D
 useful accommodating power :PN - P = 2 D
2. The man has long sight ( hyperiopia )
+ Ve lens ( converging ) is used to bring his near point
back to normal value ( 25cm) , This spectacle produce a
virtual image of an object placed at the new near point (
25cm) which will appear to the man to lie at old near
point .
 the eye has fully accommodated .
Ex: A myopic male has near and far point of 20 and
250cm . What spectacle lens is used and where is his
near point .
The solution :
The defect is myopia . The correction by using -Ve
lens diverging concave lens ).
The correction Lens :
This lens produce a virtual image at the far point of
the eye of an object at  . The relaxed eye can focus on
to the retina , the rays coming from the lens and striking
it .
The lens has focal length F , U =  from the law of
the
lens
,
V= -250 .
1
1
1
1
1




f
u
v

250Cm
F= - 250 cm  Power = - 0.4 D
 the correction lens has strength of – 0.4 D .
The kind of the lens is diverging .
The near point when wearing the lens will be at distance
(d) from the eyes .
The virtual image produced by the lens must be at
the near point of the eye in order that the fully
accommodating  V = - 20 cm and U= d
1
1
1


F
u
v
1
1
1



250Cm
d
20
d= 21.7 cm .
Q1: A lens system consists of two lenses of focal lengths
10 cm and 12 cm a distance of 15 cm apart . If an
object is placed 15 cm from the stronger lens ,
where is the final image located ?
Q2: An object is placed 12 cm from – Ve lens of focal
length (-8cm) find the position , the properties and
the magnification of the image ?
Q3: If the distance between an object of its images in
+Ve lens is 50 cm at the images is four times the
size of the object and inverted , what is the focal
length of the lens ?
Q4: Two thin converging lens of power 20 D and 7 D are
placed 30 cm a part . An object is placed 10 cm
from the stronger lens on the opposite side from the
seconds lens . Where is the final image located after
the passage of light through both lenses , and what is
the total magnification produced ?
Q5: A compound microscope consists of objective and
eye lenses of focal length 0.40 cm and 1.0 cm
respectively . An object is placed 0.41 cm from the
objective lens and the final image is formed at
infinity () . Calculate the magnifying power of the
instrument and 15 tube length .
Q6: What is the power of the lens system of an human
eye when the object viewed is : a. at infinity ()
b. 500 cm from the eye , c. 25 cm
Q7: A man has a near point of 60 cm what lenses would
be prescribed for him ?
Q8: A man is prescribed spectacle lenses of focal length
of +75 cm . Where is the near point of theis eye
located ?
Q9: A man while wearing spectacles of focal length
( -2×102 cm ) see clearly all objects lying between
25 cm and  . Where are the near and far points of
this unaided eye located ?
Q10: If the cones in the foveal region (Fovea cenralis)
are seperated by distance of 2 x 10-6 m and Q = 0.02
M , estimate the minimum possible value of the
visual acuity of the eye , where  of green light =
5×10-7 m and iris diameter (pupil) of 5 mm ?
Q11: Find the velocity of blue light in the center of the
human lens ?
Q12 : Find the raius of curvatcere of the cornea by using
Keratometer ?
Q13: The accomodation power of left eye is 2 D when
the object at 50 cm from the eye , What is properties
of this eye and find the radius of curvature of the
right left :
a. the eye at rest .
b. at mid–distance between far point and near point .
c. at  .
d. at high dioptric power .
Q14: By using emmetropic eye to focus object in the
plane mirror 3 meter away find :
1. dioptric power of this eye .
2. visual acuity of eye for this object of length 1
meter .
Q15: Discuss and compare with the normal value the
following below :
1. visual acuity .
2. resolution of eye , the diffraction pattern in the
eye when the diameter of pupil is 4 mm ,  =
585 nm .
3. The accommodation power at eye ≥ 45 year .