PHYS632_L14_ch_34_Im..
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Chapter 33 Continued
Properties of Light
•
•
•
•
Law of Reflection
Law of Refraction or Snell’s Law
Chromatic Dispersion
Brewsters Angle
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
Red
Snells Law
blue
Fiber Cable
Total Internal Reflection Simulator
Halliday, Resnick, Walker: Fundamentals of Physics,
7th Edition - Student Companion Site
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
Total Internal Reflection
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
Chromatic Dispersion
How does a Rainbow get made?
Primary rainbow
Secondary rainbow
Supernumeraries
Polarization by Reflection: Brewsters Law
Snells Law Example
47. In the figure, a 2.00-m-long vertical pole extends from the bottom
of a swimming pool to a point 50.0 cm above the water. What is the
length of the shadow of the pole on the level bottom of the pool?
Consider a ray that grazes the top of the
pole, as shown in the diagram below. Here
1 = 35o, l1 = 0.50 m, and l2 = 1.50 m.
The length of the shadow is x + L.
x is given by
1
x = l1tan1 = (0.50m)tan35o = 0.35 m.
l1
air
L is given by
water
2
L=l2tan
Use Snells Law to find
l2
shadow
L
x
Calculation of L
According to the law of refraction, n2sin2 =
n1sin1. We take n1 = 1 and n2 = 1.33
o
sin
sin
35
1
1
1
sin
25.55o
2 sin
1.33
n2
L is given by
L l2 tan 2 (1.50m) tan 25.55o 0.72m.
1
l1
air
water
2
The length of the shadow is L+x.
l2
L+x = 0.35m + 0.72 m = 1.07 m.
shadow
L
x
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
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
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
p i f
i
m
p
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
Thin Lenses
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
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
Object Distance=20 cm
and lens focal length is +10 cm
Find:
Image distance
Magnification
Height of image
Is image erect or inverted
Is it real or virtual
Draw the 3 rays
Object Distance =5 cm and
focal length is + 10 cm
Find:
Image distance
Magnification
Height of image
Is image erect or inverted
Is it real or virtual
Draw the 3 rays
p=Object Distance =10 cm
Virtual image ray diagram for
converging lens
Virtual image ray diagram for a
diverging lens
Example
24(b). 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.
24(e). 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
27. 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
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.
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.
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'?