Transcript Optics I - Department of Applied Physics

Hong Kong Polytechnic University
Optics I (AP219)
Dr. Haitao Huang (黄海涛)
Department of Applied Physics, Hong Kong PolyU
Tel: 27665694; Office: CD602
Lecture Notes can be downloaded from http://ap.polyu.edu.hk/apahthua
Text Books:
Optics, A.H.Tunnacliffe and J.G.Hirst, (Association of British Dispensing
Opticians 1996)
Optics & Vision, L.S.Pedrotti and F.L.Pedrotti, (Prentice-Hall 1998)
Optics 1----by Dr.H.Huang, Department of Applied Physics
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Reference Books:
Introduction to Optics, 2nd Edition, F.L.Pedrotti and L.S.Pedrotti (Prentice Hall
1993)
Optics, 4th Edition, E.Hecht (Addison Wesley )
Optics, 3rd Edition, A.Ghatah (McGraw Hill 2005)
Fundamentals of Physics, 7th Edition, D.Halliday, R.Resnick, and J.Walker (John
Wiley & Sons 2005)
University Physics, 11th Edition, H.D.Young and R.A.Freedman (Pearson 2004)
Physics for Scientists and Engineers, R.D.Knight (Pearson 2004)
Physics for Scientists and Engineers, 3rd Edition, D.C.Giancoli, (Prentice Hall
2000)
College Physics, A.Giambattista, B.Mc.Richardson, and R.C.Richardson,
(McGraw Hill 2004)
Advanced Studies:
The Feynman Lectures on Physics, Definitive Edition, R.P.Feynman,
R.B.Leighton, and M.Sands (Addison-Wesley 2006)
Principles of Optics, 7th Edition Expanded, M.Born et al.
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Assessment Weighting:
Final Exam: 60% (A minimum grade D is required)
The final exam will be in the form of open-book test. You are allowed to take at most
one book plus the lecture notes.
Mid-term test: 20% (1.5 hour on 20/10/2006)
Course work: 40%
Lab reports: 10%
Homework: 10%
Features:
• Many physics concepts will be learned.
• Plenty of exercises and examples are provided in each chapter.
• Course related experiments are provided to help students get a better
understanding on course materials.
• The PowerPoint of each chapter can be downloaded from my website.
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Introduction
Duality in Nature:
Rays of light can be regarded as streams of (very small) particles emitted from a
source of light and traveling in straight lines.
Geometrical Optics: when wavelength can be neglected as compared with
dimension of the relevant components in the optical system.
Light also takes a wave motion, spreading out from a light source in all directions
and propagating through an all-pervasive medium. Light can be viewed as the
electromagnetic radiation in a particular region of spectrum.
Physical Optics: when wave character can be ignored.
Index of Refraction:
n
c
v
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Introduction
Law of Reflection
When a ray of light is reflected at an
interface dividing two uniform media, the
reflected ray remains within the plane of
incidence includes the incident ray and
the normal to the point of incidence.
ir
Law of Refraction (Snell’s Law)
When a ray of light is refracted at an
interface dividing two uniform media, the
transmitted ray remains within the plane
of incidence and the sine of the angle of
refraction is directly proportional to the
sine of the angle of incidence.
n sin i  n sin i
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Introduction
Huygens’ Principle
Wavefront: the locus of the points which are in the same phase.
Example: a small stone dropped in a calm pool of water. Circular ripples spread out
from the point of impact, each point on the circumference of the circle (whose center
is at the point of impact) oscillates with the same amplitude and same phase and
thus we have a circular wavefront.
Each point of a wavefront is a source of secondary disturbance and the
wavelets emanating from these points spread out in all directions with the
speed of the wave. The envelope of these wavelets gives the shape of the new
wavefront.
The ray is always perpendicular
to the wavefront.
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Introduction
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Introduction
Fermat’s Principle
The Fermat’s Principle is also called the “principle of least time”. The field of
geometrical optics can be studied by using Fermat’s principle which determines the
path of the rays.
The ray will correspond to that path for which the time taken is an extreme in
comparison to nearby paths, i.e., it is either a minimum or a maximum or
stationary.
The mathematical form of the Fermat’s principle is,   nds  0
where n is the refractive index and the integration is done along the path.
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Introduction
Principle of Reversibility:
Any actual ray of light in an optical system, if reversed in direction, will retrace the
same path backward. This principle is directly derived from the Fermat’s principle.
Reflection in Plane Mirrors
Specular reflection: the reflection from a perfectly smooth surface which obeys the
law of reflection.
Specular reflection from the xy-plane: the
reflected ray PQ remains within the plane of
incidence, making equal angles with the
normal at P. If the direction of the incident ray
is described by its unit vector, r1=(x, y, z), then
the reflected ray is r2=(x, y, z)
Corner reflector: Three successive
reflections from orthogonal planes.
r1= (x, y, z)  r2= (-x, -y, -z)
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Introduction
A virtual image of a point object S is
formed at point S since the rays of
light appear to come from that point.
SA=SA
The image is the same size as the
object. The image is reversed,
laterally inverted and upright.
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Introduction
Refraction through Plane Surfaces:
n1 sin 1  n2 sin 2
For rays making a small angle with the normal to the
surface, a reasonably good image can be located. In
this approximation, we only allow paraxial rays to form
the image. Since n1>n2, the virtual image S of the point
object appears to be nearer than it actually is.
n 
s   2  s
 n1 
A critical angle of incidence C is reached when the
angle of refraction reaches 90°. For angles of
incidence larger than this critical angle, the incident ray
experiences total internal reflection, as shown.
 n2 
 C  sin  
 n1 
1
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Introduction
Example: A swimmer has dropped her goggles in the shallow
end of a pool, marked as 1.0 m deep. But the goggles don’t
look that deep. How deep do the goggles appear to be when
you look straight down into the water?
Solution: n1=1.00 for air and n2=1.33 for water
n1 sin 1  n2 sin 2
n1 tan1  n2 tan2
For small angle sin=tan,
x
x
n
n1  n2
d   1 d  0.75 m
d
d
n2
Example: Light strikes a flat piece of uniform thick glass at an
incident angle of 60°. If the index of refraction of the glass is 1.50,
(a) what is the angle of refraction A in the glass; (b) what is the angle
B at which the ray emerge from the glass?
Solution: n1=1.00 and n2=1.50
n1 sin 60  n2 sin  A
A=35.2°
n2 sin  A  n1 sin  B
B=60°
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Geometrical Optics
Imaging by an Optical System
An optical system includes any number of reflecting and/or refracting surfaces, of
any curvature, that may alter the direction of rays leaving an object.
The region may include any number of intervening media, but we assume each
individual medium is homogenous and isotropic, and so characterized by its own
refractive index. The rays from the object point O spread out radially in all directions
in real object space. In real object space the rays are diverging and spherical
wavefronts are expanding. Suppose that the optical system redirects these rays in
to the real image space, the wavefronts are contracting and the rays are converging
to a common image point I. From Fermat’s principle, the rays are said to be
isochronous. Further, by the principle of reversibility, if I is the object point, the image
point will be O. The points O and I are said to be conjugate points for the optical
system. Both points O and I can be imaginary.
Reflecting or refracting surfaces that form perfect images are called Cartesian
surfaces.
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Geometrical Optics
Spherical Mirror:
Spherical mirrors may be either concave or convex relative to an object point O,
depending on weather the center of curvature C is on the same or opposite side of
the surface.
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Geometrical Optics
Reflection at a Spherical Mirror:
One ray originating at O is normal to the
spherical surface at its vertex V and the other is
incident arbitrary at P. The intersection of the
two rays (extended backward) determines the
image point I conjugate to O. The image is
virtual, located behind the mirror surface.
   
2     
     2
sin   tan  
h h
h
  2
s s
R
1 1
2
 
s s
R
Similarly, for a concave mirror, we have
1 1 2
 
s s R
1 1 2
 
s s R
1. The object and image distances
are negative to the left of the
vertex and positive to the right.
2. The radius of curvature R is
positive when C is to the right of V
(convex mirror) and negative when
C is to the left of V (concave
mirror).
3. Vertical dimensions are positive
above the horizontal axis and
negative below.
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Geometrical Optics
Focal Length:
For an object at infinity, the image distance is the focal
length,
f 
R
2
The focal length is positive for a convex mirror and
negative for a concave mirror. The mirror equation is,
1 1 1
 
s s f
m
Lateral magnification:
hi
h0
We assign a (+) magnification to the case where the
image has the same orientation as the object and a ()
magnification when the image is inverted relative to the
object.
s
m
i
s0
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Geometrical Optics
Graphic diagram for spherical mirrors:
(a) Rays parallel to the principal axis are reflected though F;
(b) Rays passing through F are reflected parallel to the principal axis;
(c) Rays passing through C is reflected back along the same path.
Real object
Image position
Nature of image
Image size
At 
At F
Real, inverted
Very small
Between  and C
Between F and C
Real, inverted
Smaller
At C
At C
Real, inverted
Same
Between C and F
Between C and 
Real, inverted
Larger
At F
At 
—
—
Between F and A
Behind mirror
Virtual, upright
Larger
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Geometrical Optics
Example: An object 3 cm high is placed 20 cm from (a) a convex and (b) a concave
spherical mirror, each of 10 cm focal length. Determine the position and nature of the image
in each case.
Solution: (a) Convex mirror: f=10 cm and s=20 cm.
From mirror equation,
m
s 1

s 3
1 1 1
we get s=6.67 cm
 
s s f
The image is virtual, 6.67 cm to the right of the mirror vertex, and is erect and 1 cm high.
(b): Concave mirror: f=10 cm and s=20 cm.
From mirror equation we obtain s=20 cm and, therefore, m=1
The image is real, 20 cm to the left of the mirror vertex, and is inverted and the same
size as the object.
Example: A convex rearview car mirror has a radius of curvature of 16 m. Determine the
location of the image and its magnification for an object 10.0 m from the mirror.
Solution: The focal length f=R/2=8.0 m; the object distance s=10.0 m.
1 1 1
 
s s f
s=4.4 m;
m
s
 0.44
s
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Geometrical Optics
Example: A 1.50-cm-high diamond ring is placed 20.0 cm from a concave mirror whose
radius of curvature is 30.0 cm. Determine (a) the position of the image, and (b) its size.
Solution: The focal length f=R/2=15.0 cm; the object distance s=20.0 cm.
1 1 1
s
 
s=60.0 cm; m    3.00
;
hi=mho=4.5 cm
s s f
s
Example: A 1.00-cm-high object is placed 10.0 cm from a concave mirror whose radius of
curvature is 30.0 cm. Determine the position of the image and the magnification.
Solution: The focal length f=R/2=15.0 cm; the object distance s=10.0 cm.
1 1 1
s
 
s=30.0cm;
m    3.00
s s f
s
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Geometrical Optics
Homework:
1. Show that if two plane mirrors meet at an angle , a single ray reflected successively from
both mirrors is deflected through an angle 2 independent of the incident angle. Assume
<90° and that only two reflections, one from each mirror, take place.
2. Show with ray diagrams that the magnification of a concave mirror is less than 1 if the
object is beyond the center of curvature C, and is greater than 1 if it is within this point.
3. A beam of light in air strikes a slab of crown glass (n=1.52) and is partially reflected and
partially refracted. Find the angle of incidence if the angle of reflection is twice the angle
of refraction.
4. A light beam strikes a piece of glass at a 60.00° incident angle. The beam contains two
wavelengths, 450.0nm and 700.0nm, for which the index of refraction of the glass is
1.4820 and 1.4742, respectively. What is the angle between the two refracted beams?
5. Fermat’s principle states that “light travels between two points along that path which
requires the least time, as compared to other nearby paths.” From Fermat’s principle
derive (a) the law of reflection and (b) the law of refraction (Snell’s law)
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Geometrical Optics
Curvatures and Vergences:
The curvature is usually expressed in
reciprocal of radius. Its unit is diopter
(m-1).
Sign Convention: The curvature is
negative if the wave front is diverging
and it is positive if the wave front is
converging.
Specifically we define vergence
S
1
s
S  
1
s
Mirror power: P  
mirror equation: S  S  
2
1

R
f
2
R
we have S  P  S 
The initial vergence S of the incident wave front at the mirror is modified by the
mirror power P to produce the final vergence S of emergent wave front at the mirror.
Magnification:
m
s S

s S
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Geometrical Optics
Example: An object 3 cm high is placed 20 cm from (a) a convex and (b) a concave
spherical mirror, each of 10 cm focal length. Determine the position and nature of the image
in each case.
Solution: (a) Convex mirror: P=1/f=10 D and S=1/s=5 D.
Then S=S+P=15 D.
So we get s=1/S=6.67 cm
m=S/S=1/3
The image is virtual, 6.67 cm to the right of the mirror vertex, and is erect and 1 cm high.
(b): Concave mirror: P=1/f=10 D and S=1/s=5 D.
Then S=S+P=5 D, therefore, s=1/S=20 cm
m=S/S=1
The image is real, 20 cm to the left of the mirror vertex, and is inverted and the same
size as the object.
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Geometrical Optics
Refraction at a Spherical Surface:
n1 sin 1  n2 sin 2
n11  n2 2
n1      n2     
h h
h h
n1     n2   
 s R
 s R 
n1 n2 n1  n2
 
s s
R
Employing the same sign convention for mirrors,
n1 n2 n1  n2
 
s s
R
Lateral magnification:
h 
h 
n1  0   n2  i 
 s 
 s 
hi n1s
m

hO n2 s
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Geometrical Optics
Example: A real object is positioned in air, 30 cm from a convex spherical surface of
radius 5 cm. To the right of the interface, the refractive index is 1.33. What is the image
distance and lateral magnification of the image?
Solution: s=30 cm and R=5 cm.

1 1.33 1  1.33



30
s
5
This gives s=40 cm which means the image is real.
m
140  1
n1s

1.3330
n2 s
indicating an inverted image.
As shown in the figure, suppose now the second medium is only 10 cm thick, forming a thick
lens, with a second, concave spherical surface, also of radius 5 cm. The refraction by the first
surface will be unaffected by this change. Inside the lens, before a real image is formed, rays
are intercepted and again refracted by the second surface to produce a different image.
We have s2=30 cm and R2=5 cm.
1.33 1 1.33  1
 
30 s2
5
So, s2=9 cm. It is a real image.
m
1.339  2
130 5
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Geometrical Optics
Vergence in Refraction:
n1
s
n
Vergence of emerging wave front leaving the spherical surface: S   2
s
n  n1
Power of spherical surface: P  2
R
S
S  P  S  and
Then we have
m
S
Vergence of incident wave front arriving at the spherical surface: S 
Example: Repeat the example on previous slide.
For the first spherical surface, we have s=30 cm, R=5 cm, n1=1 and n2=1.33.
S=n1/s=3.33 D; P=(n2n1)/R=6.67 D; S=S+P=3.33 D; s=n2/S=0.4 m
m=S/S=1
For the second spherical surface, we have s2=30 cm, R=5 cm, n1=1.33 and n2=1.
S2=n1/s2=4.44 D; P=(n2n1)/R=6.67 D; S2=S2+P=11 D; s2=n2/S2=0.09 m
m=S2/S2=0.4
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Thin Lenses:
1st refracting surface
n1 n2 n1  n2
 
s1 s1
R1
2nd refracting surface
n2 n1 n2  n1
 
s2 s2
R2
For thin lens, s1=s2, then
Thin lens equation:
Focal length:
Geometrical Optics
 1 1
n1 n1
  n2  n1   
s1 s2
 R2 R1 
1 1 n2  n1  1 1 
  
 
s s
n1  R1 R2 
1 n2  n1  1 1 
  

f
n1  R1 R2 
This is the lensmaker’s equation.
simplified lens equation:
1 1 1
 
s s f
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Geometrical Optics
Ray Tracing:
can be used to locate the image of an object.
Real image: the rays after refraction actually
intersect.
Virtual image: the rays do not intersect after
refraction.
The focal length is positive for converging lenses
and negative for diverging lenses.
hi s
Lateral Magnification: m  
ho s
Example: What is (a) the position and (b) the size of a large 7.6-cm-high flower placed
1.00 m from a +50.0-mm-focal length camera lens?
Solution: f=5.00 cm; s=100 cm; ho=7.6 cm
1 1 1
 
s s f
s=5.26 cm;
hi  mho 
s
ho  0.40 cm
s
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Geometrical Optics
Example: Find and describe the intermediate and final images produced by a two-lens
system as show in the figure. Both are 15-cm focal length lenses and their separation is 60
cm. Let the object RO1 be 25 cm from the first lens.
Solution: 1st lens
f1=15 cm, s1=25 cm
1 1 1
 
s1=37.5 cm
s1 s1 f1
s
m1  1  1.5
s1
The first image is real, 37.5 cm to the right of the first lens, inverted and 1.5 times the size
of the object.
For 2nd lens, f2=15 cm, s2=(60s1)=22.5 cm
1 1
1
 
s2 s2 f 2
s2=9 cm,
m2 
s2
 0.4
s2
The final image is virtual, 9 cm to the left of the second lens and inverted. The overall
magnification is m=m1m2=0.6.
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Geometrical Optics
Vergence of the Thin Lens:
The power of the lens as a whole is just the sum of the powers of the two surfaces.
 1 1  n1
n2  n1 n1  n2
P  P1  P2 

 n2  n1    
R1
R2
 R1 R2  f
s S
The lateral magnification is m  
s S
Example: Repeat the previous one.
S  P1  P2  S 
Solution: 1st lens: f1=15 cm, s1=25 cm, S1=n1/s1=4 D, P1=n1/f1=6.67 D
S1=S1+P1=2.67 D; s1=n1/S1=0.375 m; m1=S1/S1=1.5
2nd lens: f2=15 cm, s2=22.5 cm, S2=n1/s2=4.44 D, P2=n1/f2=6.67 D
S2=S2+P2=11.11 D; s2=n2/S2=0.09 m; m2=S2/S2=0.4
Example: A convex meniscus lens is made from glass with
n=1.50. The radii of the two surfaces are 22.4 cm and 46.2 cm,
respectively. (a) What is the focal length? (b) Where will it focus
an object 2.00 m away?
Solution: R1=22.4 cm and R2=46.2 cm. (a) Using the
lensmaker’s equation, we can easily get f=87 cm.
(b) s=2 m, therefore, s=1.54 m.
Optics 1----by Dr.H.Huang, Department of Applied Physics
29
Hong Kong Polytechnic University
Geometrical Optics
Thin Lenses in Contact:
The vergence approach is of particular importance and interest to optometrists.
1 1 1
1 1
1




Consider two thin lenses in contact
s1 s1 f1
s2 s2 f 2
1 1 1 1
1
Since s1=s2, so
   
P  P1  P2
s2 s1 f1 f 2 f eq
several thin lenses in contact
P  P1  P2  P3  ...
1
1 1 1
    ...
f eq f1 f 2 f 3
Newtonian Equation for the Thin Lens:
hi
x

hO
f
hi
f

hO x
xx  f 2
f
x
m 
x
f
Optics 1----by Dr.H.Huang, Department of Applied Physics
30
Hong Kong Polytechnic University
Geometrical Optics
Example: An object is placed 10 cm from a 15-cm-focal-length converging lens. Determine
the image position and size.
Solution: f=15 cm and s=10 cm;
1 1 1
 
s s f
s=30cm
s
m   3 .0
s
Example: Where must a small insect be placed if a 25-cm-focal-length diverging lens is to
form a virtual image 20 cm in front of the lens?
Solution: f=25 cm and s=20 cm
1 1 1
 
s=100 cm
s s f
Example: To measure the focal length of a diverging lens, a converging lens is placed in
contact with it as shown. The Sun’s rays are focused by this combination at a point 28.5 cm
behind the lenses. If the converging lens has a focal length
fC of 16.0 cm, what is the focal length f of the diverging
lens? Assume both lenses are thin and the space between
them is negligible.
Solution:
1
1
1


fC f d
fT
fd=36.5 cm
Optics 1----by Dr.H.Huang, Department of Applied Physics
31
Hong Kong Polytechnic University
Geometrical Optics
Homework:
1. It is desired to magnify reading materials by a factor of 2.5 when a book is placed 8.0
cm behind a lens. (a) What type of lens is needed for this? (b) What is the power of the
lens in diopters?
2. Two 27.0-cm-focal-length converging lenses are placed 16.5 cm apart. An object is placed
35.0 cm in front of one. Where will the final image formed by the second lens be located?
What is the total magnifications?
3. A prescription for a corrective lens calls for +2.50 D. The lensmaker grinds the lens from
a “blank” with n=1.56 and a performed convex front surface of radius of curvature of 20.0
cm. What should be the radius of curvature of the other surface?
4. Two identical, thin, plano-convex lenses with radii of curvature of 15 cm are situated with
their curved surfaces in contact at their centers. The intervening space is filled with oil of
refractive index 1.65. The index of the glass is 1.50. Determine the focal length of the
combination. (Hint: Think of the oil layer as an intermediate thin lens.)
5. A sphere 5 cm in diameter has a small scratch on its surface. When the scratch is viewed
through the glass from a position directly opposite, where does the scratch appear and
what is its magnification? Assume n=1.50 for the glass. Solve by both image location and
vergence models.
Optics 1----by Dr.H.Huang, Department of Applied Physics
32