Transcript Optics I

Hong Kong Polytechnic University
Thick Lens
Thick Lens:
Thick lens is a lens whose thickness along its optical axis cannot be ignored without
leading to serious errors in analysis.
Six Cardinal points: applicable to any optical system
• first and second system focal points (F1 and F2);
• first and second principal points (H1 and H2);
• first and second nodal points (N1 and N2).
Planes normal to the axis at these points are called cardinal planes.
Optics 1----by Dr.H.Huang, Department of Applied Physics
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Thick Lens
Basic Equations:
Focal length:
1 nL  n nL  n nL  n nL  n t



f1
nR2
nR1
nnL
R1 R2
f2  
n
f1
n
Position of principal planes:
n L  n
r
f1t
nL R2
s
nL  n
f 2t
nL R1
Position of nodal points:
 n n  n 
v  1   L
t  f1
 n nL R2 
Thick lens formula:
Properties:

n n n 
w  1   L
t  f 2
 n n L R1 
f1 f 2
 1
so si
H1H 2  N1 N 2
F1H1  N 2 F2
m
H1 N1  H 2 N 2
nsi
nso
Distances are directed,
positive or negative by a
sign convention that makes
distances directed to the
left negative and distances
to the right, positive.
If n=n, then r=v and s=w
F2 H 2  N1F1
Optics 1----by Dr.H.Huang, Department of Applied Physics
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Thick Lens
Example: Determine the focal lengths and the principal points for a 4-cm thick, biconvex
lens with refractive index of 1.52 and radii of curvature of 25 cm, when the lens caps the
end of a long cylinder filled with water (n=1.33).
Solution: nL=1.52, n=1, n=1.33, t=4 cm,
The first refraction surface is a convex plane, so R1=25cm.
The second one is a concave plane, so R2=25cm. According to equation
1 nL  n nL  n nL  n nL  n t



f1
nR2
nR1
nnL
R1 R2
we can get f1=35.74cm to the left of the first principal plane. From
f2  
n
f1
n
we have f2=47.53cm to the right of the second principal plane. Equations
r
n L  n
f1t
nL R2
s
nL  n
f 2t
nL R1
give us r=0.715cm and s=2.60cm.
Optics 1----by Dr.H.Huang, Department of Applied Physics
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Thick Lens
Example: (a) Find the cardinal points and sketch a ray diagram for the hemispherical glass
lens shown in the figure. The radii of curvature are R1=3 cm and R2=, and the lens in air has
a refractive index of 1.50. (b) Sketch a ray diagram to locate the image of a 1-cm tall object 3
cm to the left of the first principal point for this lens.
Solution: (a) nL=1.50, n=n=1, t=3 cm
1 nL  n nL  n nL  n nL  n t
1




f1
nR2
nR1
nnL
R1R2
6
r
f1  6 cm
f 2  6 cm
nL  n
f1t  0
nL R2
s
nL  n
f 2t  2 cm
nL R1
v= 0 cm and w=2 cm
(b) so=3 cm,
f1 f 2
 1
so si
si=6 cm
m
nsi
2
nso
Optics 1----by Dr.H.Huang, Department of Applied Physics
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Hong Kong Polytechnic University
Thick Lens
Example: Find the cardinal points and sketch a ray diagram for the thick lens shown in the
figure. The radii of curvature are R1=10 cm and R2=25 cm and its thickness on axis is 5 cm.
The lens is situated in air and has a refractive index of 1.60.
Solution: nL=1.60, n=n=1, t=5 cm
1 nL  n nL  n nL  nnL  n t
8.85




f1
nR2
nR1
nnL
R1R2 100
f1  11.3 cm
f 2  11.3 cm
nL  n
1.6  1
r
f1t 
11.3  5  0.847 cm
v  r  0.847 cm
nL R2
1.6  25
n n
1.6  1
 11.3 5  2.12 cm w  s  2.12 cm
s L
f 2t  
nL R1
1.6 10
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Thick Lens
Ray Tracing:
Each ray is traced independently, using only the laws of reflection and refraction
together with geometry, to determine the image forming quality of an optical system.
Ray tracing procedures are often limited to meridional rays, that is, rays that pass
through the optical axis of the system.
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Thick Lens
Homework:
1. A biconvex lens of 5 cm thickness and index 1.60 has surfaces of radius 40 cm. If this
lens is used for objects in water, with air on its opposite side, determine its effective focal
length and sketch its focal and principal points.
2. A double concave lens of glass with n=1.53 has surfaces of 5D (diopters) and 8D,
respectively. The lens is used in air and has an axial thickness of 3 cm. (a) Determine the
position of its focal and principal planes. (b) Also find the position of the image, relative
to the lens center, corresponding to an object at 30 cm in front of the first lens vertex. (c)
calculate the paraxial image distance assuming the thin-lens approximation. What is the
percent error involved?
3. A biconcave lens has radii of curvature of 20 cm and 10 cm. Its refractive index is 1.50
and its central thickness is 5 cm. Describe the image of a 1-in. tall object, situated 8 cm
from the first vertex.
4. An equiconvex lens having spherical surfaces of radius 10 cm, a central thickness of 2 cm,
and a refractive index of 1.61 is situated between air and water (n=1.33). An object 5 cm
high is placed 60 cm in front of the lens surface. Find the cardinal points for the lens and
the position and size of the image formed.
5. A hollow, glass sphere of radius 10 cm is filled with water. Refraction due to the thin
glass walls is negligible for paraxial rays. (a) Determine its cardinal points and make a
sketch to scale. (b) Calculate the position and magnification of a small object 20 cm from
the sphere.
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Vergence and Vision
Muscular action controls the curvature of the lens surfaces, and hence the lens
power. In this way focusing for different object distances is achieved, a process
known as accommodation. In fact, most of the curvature change takes place at the
front surface of the lens, so that the distance between the lens and the retina is
almost constant during accommodation. There is a standard least distance of
distinct vision (25cm) below which a clear retinal image cannot be formed.
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Vergence and Vision
Left to right: unaccommodated emmetropia; myopia; and hyperopia
Far point in (left) an unaccommodated myopic eye and (right) an unaccommodated
hypermetropic eye.
Rays from a distant object can be made to focus on the center of the macula (M)
of a myopic eye (left) or a hypermetropic eye (right) by arranging a diverging or a
converging lens, respectively.
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Vergence and Vision
Example: Suppose a simplest possible model of the eye, consisting of a single convex thin
lens and a screen. The power of the single lens is +60.0 D. The relaxed eye without
accommodation is focused on a distant object. (a) What is the lens-retinal separation for the
unaccommodated, emmetropic eye? (b) Where does the image of a distant object occur if the
lens has, instead, a power of +50.0 D? Of +70.0 D? (c) How much dioptric power does each
require in the form of an additional contact lens to correct the problem?
Solution: (a) S=S+P=0+60=60 D, s=1/60=0.0167 m
(b) S=50 D, s=1/50=0.02 m;
S=70 D, s=1/70=0.0143 m
(c) Hyperopic eye requires +10 D (convex lens). Myopic eye requires 10 D (concave lens).
Example: In the simple thin lens-retina model of the unaided eye, let us suppose the
distance between lens and retina is 18 mm for a female patient. What dioptric power is
required for an emmetrope to focus clearly on a distant object?
Solution: For a distant object S=0 and S=S+P=P
Since s=18 mm, S=1/0.018=55.6 D, so P=55.6 D
Example: If, instead, the unaided eye has a more strong convergent power P=+62.0 D, the
light focuses before it reaches the retina, and the 18-mm eye is myopic. What then is its far
point (farthest point at which an object is well-focused on the retina)?
Solution: s=0.018 mm, so S=1/0.018=55.56 D. Since S=S+P, we have S=6.44 D
s=1/S=0.155 m.
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Hong Kong Polytechnic University
Vergence and Vision
Example: What correction is needed for the myope described above?
Solution: Suppose the correction power is PC so that distant objects can be focused on
retina without accommodation. For a distant object S=0, S=S+PC+P=55.56D, so PC=6.44D
Example: The near point (nearest point at which an object is well-focused on the retina) of
a certain hyperopic eye is 100 cm in front of the eye. To see clearly an object that is 25 cm in
front of the eye, what contact lens is required?
Solution: The lens should form an image 100 cm in front of the eye.
S=1/0.25=4 D, S=1 D, so P=SS=+3 D.
Example: The far point of a certain myopic eye is 50 cm in front of the eye. To see clearly
an object at infinity, what eyeglass lens is required? Assume that the lens is worn 2 cm in
front of the eye.
Solution: The eyeglass should form an image 502=48 cm in front of the eye.
S=0, S=1/0.48=2.08 D, so P=SS=2.08 D.
Example: A nearsighted eye has near and far points of 12cm and 17cm, respectively. (a)
What lens power is needed for this person to see distant objects clearly, and (b) what then
will be the near point? Assume that the lens is 2.0cm from the eye (typical for eyeglass).
Solution: (a) s=(172)=15 cm, S=6.67D, S=0, so P=SS=6.67 D.
(b) s=(122)=10 cm, S=10 D, so S=SP=3.33 D; s=30 cm in front of the glass.
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Vergence and Vision
Effective Power:
Vx 
n
n f
P
nP
nP




f  x 1  x f 1  x f n  xn f n  xP
in air:
Vx 
P
1  xP
This is the effective power or effectivity of the thin lens of power P at the distance x
beyond the lens.
Lens Correction: Suppose the contact lens correction PC and the spectacle lens
correction PS. They are equivalent when,
PC 
PS
1  xPS
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Vergence and Vision
Example: A negative thin lens in air, of focal length 10.0 cm, is subject to incident plane
waves. What is the curvature or vergence of the wave front 25 cm beyond the lens?
 10
Solution: V  P 
 2.86 D
1  xP
x
1  0.25 10
Example: A positive lens in air, of focal length 10 cm, is subject to incident plane waves.
Find the curvature or vergence of the wave front both at 5 cm and at 20 beyond the lens.
P
10
Solution: For x=0.05m, Vx 

 20 D
1  xP 1  0.05 10
P
10
For x=0.2 m, Vx 

 10.0 D
1  xP 1  0.2 10
Example: Suppose the prescription power of a contact lens for hyperopia is +10 D. If the
person is to be fitted instead with spectacle lenses, worn at a distance of 14 mm from the
cornea, what is the required prescription?
PS
PS
Solution:
PC 
1  xPS

1  0.014PS
 10
 PS  8.77 D
Example: Suppose that a myope uses spectacle lenses of 10 D and wishes to change to
contact lenses. His cornea is at a distance of 13 mm from the spectacle lens. What should be
the contact lens power to provide equivalent correction?
 10
Solution: P  PS 
 8.85 D
C
1  xPS
1  0.013 10
Optics 1----by Dr.H.Huang, Department of Applied Physics
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Vergence and Vision
Accommodation:
the difference between the power Pa of the accommodated eye and the power Pu of
the unaccommodated eye,
A0  Pa  Pu
The accommodation is always greater than zero.
Example: An emmetrope with +60.0 D of unaccommodated power focuses on an object at
his near point of 25.0 cm. How much accommodation is needed to see the object at the near
point clearly?
Solution: For unaccommodated emmetrope, S=0+Pu=+60 D,
To see the near point, S=1/0.25=4.0 D, Pa=SS=+64.0 D; So, A0=+4.0 D
Example: A young female myope with a far point at 50 cm views an object at 25 cm from
the eye. What accommodation does she require?
Solution: To see the far point, S=1/(0.5)+Pu=Pu2, so Pu=S+2
To see the object at 25 cm from the eye, S=1/(0.25)+Pa=Pa4, so Pa=S+4.
Therefore, A0=42=+2.0 D
If the vergence from the far point is Vfp, and the vergence from the new object
position is Vx, then, to see the object clearly, accommodation should be,
A0  V fp  Vx
Optics 1----by Dr.H.Huang, Department of Applied Physics
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Vergence and Vision
Range of Vision: The range between the far point and the near point.
Example: Determine the near points and far points for the following eyes, given that each
can provide a maximum accommodation of 8.0 D. (a) An emmetrope. (b) A myope with a
“refractive error” (need for a refractive correction of 5.0 D. (c) A hyperope with a refractive
error of 5.0 D. (d) A hyperope with a refractive error of 12.0 D.
Vnp  8 D
Solution: (a) Far point at infinity. A0  V fp  Vnp  0 Vnp  8
Near point at vnp=12.5 cm
(b) V =5.0 D,
fp
far point at vfp=20 cm.
Vnp=VfpA0=13D,
near point at vnp=7.69 cm
(c) Vfp=5.0 D,
far point at vfp=20 cm.
Vnp=VfpA0=3 D,
near point at vnp=33.3 cm.
(d) Vfp=12.0 D,
far point at vfp=8.33 cm.
Vnp=VfpA0=4 D,
near point at vnp=25.0 cm.
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Vergence and Vision
Homework:
1. An emmetrope with an unaccommodated eye is represented by a +64.0 D thin lens and
retina. (a) What is the lens-retinal separation for this unaccommodated eye? (b) Where
does the image of a distant object occur if the lens has instead a power of +60.0 D? Of
+68.0 D? (c) How much dioptric power does each require in the form of an additional
contact lens to correct his vision problem?
2. In the simple, thin lens-retina model of the eye, suppose the distance between lens and
retina is 16.5 mm for a particular eye. (a) What dioptric power is required for an
emmetrope fo focus on a distant object? (b) If, instead, the lens has a more strongly
convergent power of +65.0 D, what is its far point? (c) What correction is needed in an
external lens?
3. A person suffers from myopia requiring a correction of 4.50 D. The maximum
accommodation possible is +6.0 D. Without correction, can this myope read the fine print
in a book held 40 cm from the eye?
4. The prescription power of a spectacle lens for a myope is 12.0 D. If the myope switches
to contact lenses from the spectacle lenses at 14 mm from the cornea, what is the new
prescription?
5. An emmetrope with an unaccommodated eye is represented by a +62.0 D thin lens and
retinal screen. Determine the retinal size of the image of a distant object that subtends an
angle of 20 at the lens.
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