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Lecture 14 Images Chapter 34
•Geometrical Optics
•Fermats Principle
-Law of reflection
-Law of Refraction
•Plane Mirrors and Spherical Mirrors
•Spherical refracting Surfaces
•Thin Lenses
•Optical Instruments
-Magnifying Glass, Microscope, Refracting telescope
• Polling Questions
Left over from Lecture 13 and Chapter 33
1) Show electric and magnetic fields are out of phase in demo.
2) Reflection and Refraction
1) Red or green laser through smoky block
2) Show maximum bending angle
3) Total internal reflection
1) Show it with block
1) Dispersion through prism
2) Brewsters Law
1) Show using polaroid paper
Reflection and Refraction of Light
Dispersion: Different wavelengths have different velocities and
therefore different indices of refraction. This leads to different
refractive angles for different wavelengths. Thus the light is dispersed.
The frequency does not change when n changes.
v  f
 changes when medium changes
f does not change when medium changes
Snells Law
n1 sin 1  n2 sin 2
Red
1.00sin 1
1.509
1.00sin 90
sin  2 
1.509
 2  41.505deg
sin  2 
n1=1.000

n2=1.509
θ1
θ2

v1 sin 1  v2 sin 2
Snell’s Law at Work
Fiber Cable
Why is light totally reflected inside a fiber optics
cable? Internal reflection
n1 sin 1  n2 sin 2
(1.509)sin 1  (1.00)sin 90  1.00

1  sin
1
1
1.509
 41.505 deg
Equilateral prism
dispersing sunlight
late afternoon.
Sin θrefr= 0.866nλ
n increases with frequency
Blue light bends more than red
9
How much light is polarized when reflected from a surface?
Polarization by Reflection: Brewsters Law
Fresnel Equations
The parallel and perpendicular components of the electric vector of the reflected and transmitted
light wave is plotted below when light strikes a piece of glass. Now the intensity of polarized
light is proportional to the square of the amplitude of the oscillations of the electric field. So, we
can express the intensity of the incoming light as I0 = (constant) E 2.
E
What causes a Mirage
eye
sky
1.09
1.09
1.08
Index of refraction
1.08
1.07
1.07
1.06
Hot road causes gradient in the index of refraction that increases
as you increase the distance from the road
Inverse Mirage Bend
Geometrical Optics:Study of reflection and refraction
of light from surfaces
The ray approximation states that light travels in straight lines
until it is reflected or refracted and then travels in straight lines again.
The wavelength of light must be small compared to the size of
the objects or else diffractive effects occur.
Law of Reflection
I  R
Mirror
B
r
i
A
1
Fermat’s Principle
Using Fermat’s Principle you can prove the
Reflection law. It states that the path taken
by light when traveling from one point to
another is the path that takes the shortest
time compared to nearby paths.
JAVA APPLET
Show Fermat’s principle simulator
http://www.phys.hawaii.edu/~teb/java/ntnujava/index.html
Two light rays 1 and 2 taking different paths
between points A and B and reflecting off a
vertical mirror
B
Plane Mirror
2
A
1
Use calculus - method
of minimization
t  C1 ( h12  y 2  h22  (w  y)2 )
dt

dy
2y
h y
2
1
2
y
h12  y 2
Write down time as a function of y
and set the derivative to 0.
sin  I
2(w  y)

h  (w  y) )
2
2


I  R
2
0
(w  y)
h22  (w  y)2 )
sin  R
t(
1
1
h12  y 2 
h22  (w  y)2 )
v1
v2
dt
0
dy
1
sin  I 
v1
1
sin  R
v2
n1 sin  I  n 2 sin  R
Mirrors and Lenses
Plane Mirrors Where is the image formed
Plane mirrors
Angle of
Real side incidence
Normal
Angle of reflection
Virtual side
Virtual image
i=-p
eye
Object distance = - image distance
Image size = Object size
Problem: Two plane mirrors make an angle of 90o. How
many images are there for an object placed between
them?
mirror
eye
object
2
mirror
1
3
Using the Law of Reflection to
make a bank shot
Assuming no spin
Assuming an elastic collision
No cushion deformation
d
d
pocket
i=-p
magnification = 1
What happens if we bend the mirror?
Concave mirror.
Image gets magnified.
Field of view is diminished
Convex mirror.
Image is reduced.
Field of view increased.
Rules for drawing images for mirrors
• Initial parallel ray reflects through focal point.
•Ray that passes in initially through focal point reflects parallel
from mirror
•Ray reflects from C the radius of curvature of mirror reflects along
itself.
• Ray that reflects from mirror at little point c is reflected
symmetrically
1 1 1
r
f

 
2
p i f
i
Lateral magnification = ratio of image height/object height
m
p 
One simple way to measure focal length of a
concave mirror?
Spherical refracting surfaces
Using Snell’s Law and assuming small
Angles between the rays with the central
axis, we get the following formula:
n1 n2 n2  n1


p
i
r
Chapter 34 Problem 32
In Figure 34-35, A beam of parallel light rays from a laser is incident on a
solid transparent sphere of index of refraction n.
n1 n2 n2  n1


p
i
r
Fig. 34-35
(a) If a point image is produced at the back of the sphere, what is the index
of refraction of the sphere?
(b) What index of refraction, if any, will produce a point image at the
center of the sphere? Enter 'none' if necessary.
n1 n2 n2  n1


p
i
r
Apply this equation to Thin Lenses where the thickness is
small compared to object distance, image distance, and
radius of curvature. Neglect thickness.
Converging lens
Diverging lens
Simple Lens Model
1 1 1
 
f p i
Thin Lens Equation
Lensmaker Equation

1
1 1
 (n 1)(  )
f
r1 r2
Lateral Magnification for a Lens

What is the sign convention?
i
m
p
image height
m
object height
Sign Convention
Light
Virtual side - V
Real side - R
r1 r2
p
i
Real object - distance p is pos on V side (Incident rays are diverging)
Radius of curvature is pos on R side.
Real image - distance is pos on R side.
Virtual object - distance is neg on R side. Incident rays are converging)
Radius of curvature is neg on the V side.
Virtual image- distance is neg on the V side.
Rules for drawing rays to locate images from
a lens
1) A ray initially parallel to the central axis will pass through the focal point.
2) A ray that initially passes through the focal point will emerge from the lens
parallel to the central axis.
3) A ray that is directed towards the center of the lens will go straight
through the lens undeflected.
Real image ray diagram for a
converging lens
Virtual image ray diagram for
converging lens
Virtual image ray diagram for a
diverging lens
Example
Given a lens with a focal length f = 5 cm and object distance p = +10 cm,
find the following: i and m. Is the image real or virtual? Upright or inverted?
Draw the 3 rays.
Virtual side
.
F1
p
Real side
.
F2
Example
Given a lens with a focal length f = 5 cm and object distance p = +10 cm,
find the following: i and m. Is the image real or virtual? Upright or inverted?
Draw the 3 rays.
Virtual side
Real side
.
.
F1
F2
p
1 1 1
 
i f p
m
1 1 1
1
  
i 5 10
10
m
i  10 cm



y 
i

y
p
10
 1
10
Image is real,
inverted.
Example
Given a lens with the properties (lengths in cm) r1 = +30, r2 = -30, p =
+10, and n = 1.5, find the following: f, i and m. Is the image real or
virtual? Upright or inverted? Draw 3 rays.
Real side
Virtual side
.
F1
r1
r2
p
.
F2
Example
Given a lens with the properties (lengths in cm) r1 = +30, r2 = -30, p =
+10, and n = 1.5, find the following: f, i and m. Is the image real or
virtual? Upright or inverted? Draw 3 rays.
Real side
Virtual side
.
F1
r1
r2
p
.
F2
y 
i

y
p
1 1
1
 n  1  
f
 r1 r2 
1 1 1
 
i f p
1
1  1
 1
 1.5  1 

f
30

30

 30
1 1 1
1

 
i 30 10
15
m
f  30cm
i  15cm
Image is virtual,
upright.
m

15
 1.5
10
Example
A converging lens with a focal length of +20 cm is located 10 cm to the left of a
diverging lens having a focal length of -15 cm. If an object is located 40 cm to
the left of the converging lens, locate and describe completely the final image
formed by the diverging lens. Treat each lens
Separately.
Lens 1 Lens 2
+20
-15
f1
f2
f1 f2
40
10
Lens 1 Lens 2
+20
-15
f1
f2
f 1 f2
40
40
10
30
Ignoring the diverging lens (lens 2), the image formed by the
converging lens (lens 1) is located at a distance
1 1 1
1
1
  

. i1  40cm
i1 f1 p1 20cm 40cm
Since m = -i1/p1= - 40/40= - 1 , the image is inverted
This image now serves as a virtual object for lens 2, with p2 = - (40 cm - 10 cm) = - 30 cm.
Lens 1 Lens 2
+20
-15
f1
f2
f 1 f2
40
40
10
1 1 1
1
1
  

i2 f 2 p2 15cm 30cm
30
i2  30cm.
Thus, the image formed by lens 2 is located 30 cm to the left of lens 2. It is
virtual (since i2 < 0).
The magnification is m = (-i1/p1) x (-i2/p2) = (-40/40)x(30/-30) =+1, so the image
has the same size orientation as the object.
Optical Instruments
Magnifying lens
Compound microscope
Refracting telescope
Galileo - converging + diverging lens
Keplerian - converging + converging lens
Reflecting Telescope
Chapter 34 Problem 91
Figure 34-43a shows the basic structure of a human eye. Light
refracts into the eye through the cornea and is then further redirected
by a lens whose shape (and thus ability to focus the light) is
controlled by muscles. We can treat the cornea and eye lens as a
single effective thin lens (Figure 34-43b). A "normal" eye can focus
parallel light rays from a distant object O to a point on the retina at
the back of the eye, where processing of the visual information
begins. As an object is brought close to the eye, however, the muscles
must change the shape of the lens so that rays form an inverted real
image on the retina (Figure 34-43c).
(a) Suppose that for the parallel rays of Figure 34-43a and Figure
34-43b, the focal length f of the effective thin lens of the eye is
2.52 cm. For an object at distance p = 48.0 cm, what focal length
f' of the effective lens is required for the object to be seen clearly?
(b) Must the eye muscles increase or decrease the radii of
curvature of the eye lens to give focal length f'?