Transcript PowerPoint Presentation - Chapter 15

Final Exam (chapters since Exam#2)
. time: Friday 05/03 3:30 pm- 5:30 pm
. Location: room 114 of physics building.
If you can not make it, please let me know by Friday 04/26 so that I can arrange a
make-up exam. If you have special needs, e.g. exam time extension, and has not contact
me before, please bring me the letter from the Office of the Dean of Students before
Friday 04/26. No requested will be accepted after that.
AOB
•30-40 problems.
•Prepare your own scratch paper, pencils, erasers, calculators etc.
•Use only pencil for the answer sheet
•No cell phones, no text messaging which is considered cheating.
•No crib sheet of any kind is allowed. Equation sheet will be provided and will also be
posted on the web.
1

Different wavelengths of visible light are associated
with different colors.
Violet is about 3.8 x 10-7 m.
 Wavelengths shorter than the violet comprise ultraviolet
light.
 Red is about 7.5 x 10-7 m.
 Wavelengths longer than the red comprise infrared light.
 In between, the colors are red, orange, yellow, green,
blue, indigo, and violet.

2
Color Mixing
 The process of mixing two different
wavelengths of light, such as red and green, to
produce a response interpreted as another
color, such as yellow, is additive color mixing.
Combining the three primary colors
blue, green, and red in different
amounts can produce responses in
our brains corresponding to all the
colors we are used to identifying.
Red and green make yellow, blue
and green make cyan, and blue and
red make magenta.
Combining all three colors
produces white.
3
Color Mixing
 The pigments used in paints or dyes work by
selective color mixing.

They absorb some wavelengths of light more than
others.
When light strikes an object, some
of the light undergoes specular
reflection: all the light is reflected
as if by a mirror.
The rest of the light undergoes
diffuse reflection: it is reflected in
all directions.
Some of the light may be
selectively absorbed, affecting the
color we see.
If red light is absorbed, we see
blue-green.
4
Why is the sky blue?
The white light coming from the sun
is actually a mixture of light of
different wavelengths (colors).
The shorter wavelengths of blue
light are scattered by gas molecules
in the atmosphere more than longer
wavelengths such as red light.
The blue light enters our eyes after
being scattered multiple times, so
appears to come from all parts of the
sky.
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Why is the sunset red?
The shorter wavelengths of blue light
are scattered by gas molecules in the
atmosphere more than longer
wavelengths such as red light.
When the sun is low on the horizon, the
light must pass through more
atmosphere than when the sun is
directly above.
By the time the sun’s light reaches our
eyes, the shorter wavelengths such as
blue and yellow have been removed by
scattering, leaving only orange and red
light coming straight from the sun.
6
Mirage and Rainbow
water droplet
 red is outside.
 intensity
max at 42
rainbow
7
Thin film interference
Thin film interference occurs when light
is reflected from the top surface and the
underneath surface.
This provides the two beams of
coherent light that interfere.
Since we normally observe this with
white light we see colors because the
path difference varies depending on the
angle of observation
So different wavelengths (colors) have
constructive and destructive interference
at different places on the film.
8
Coatings for lenses
As light passes from one transparent medium to another a few
percent of the light will be reflected.
This is a particular problem in optical systems like lenses where
there may be many glass elements.
For example if 96% of the light is transmitted at a surface after 8
surfaces only 72% of the light remains and the other 28% will be
scattered everywhere.
Thin coatings are put on glass surfaces so that for particular
wavelengths the light reflected from the top surface is exactly
cancelled by the light from the bottom surface.
This is only true for a single wavelength and to reduce the
reflections for a range of wavelengths requires multiple thin film
layers very often just λ/4 thick.
9
Polarization of Electromagnetic Waves
Polarization is a measure of the degree to which the electric field (or the
magnetic field) of an electromagnetic wave oscillates preferentially along
a particular direction.
Linear combination
of many linearly
partially
polarized rays of
polarized
random orientations
unpolarized
linearly
polarized
Looking at E
head-on
components
equal y- and zamplitudes
unequal y- and zamplitudes
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Polarizer: polarization by absorption
An electric field component parallel to the
transmission axis is passed by a polarizer; a
component perpendicular to it is absorbed.
transmission
axis
dichroism (tourmaline, polaroid,…)
So if linearly polarized beam with E is
incident on a polarizer as shown,
E y  E cos 
I  I 0 cos 2 
Zero if =/2, I0 if =0
If unpolarized beam is incident instead,
The intensity will reduce by a factor of two.
I  I 0 cos 2 
 I0 / 2
The light will become polarized along the transmission axis
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Quiz (Bonus: 20 points)
 A beam of un-polarized lights with
intensity I is sent through two polarizers
with transmission axis perpendicular to
each other. What’s the outgoing light
intensity?
a) ½ I
b) 2 I
c) 0
d) 1.5 I
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Example: two polarizers
This set of two linear
polarizers produces LP
(linearly polarized) light.
What is the final
intensity?


P1 transmits 1/2 of
the unpolarized
light:
I1 = 1/2 I0
P2 projects out the
E-field component
parallel to x’ axis:
E2  E1 cos 
I  E2
1
I 2  I1 cos   I 0 cos 2 
2
2
= 0 if  = /2
(i.e., crossed)
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7B-22 Polarizer Effects
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Quiz (bonus: 20 points): Unpolarized light of intensity I0 is
sent through 3 polarizers, each of the last two rotated 45 from
the previous polarizer so that the last polarizer is perpendicular
to the first. What is the intensity transmitted by this system?
(hint: sin2(45)=0.5)
a) 0.71 I0
b) 0.50 I0
c) 0.25 I0
d) 0.125 I0
e) 0
1
𝐼 = 𝐼0 × × 𝑐𝑜𝑠 2 (45) × 𝑐𝑜𝑠 2 (45)
2
= 0.125𝐼0
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Image by Reflection
from
a Plane Mirror
point object
r
i
 Only small fraction of reflected
rays received.
 Virtual image at same distance
from but on the other side of the
mirror as the object
•it is called a virtual image since
no rays actually go through the
image,
extended object
• An extended object can be
broken into infinite number of
point objects.
• Image has the same height and
orientation as the object.
16
7A-05 Candle Illusion
17
“Full Length” Mirror
1
AC
2
1
GH  CD  CE
2
FG  BC 
1
1
FH   AC  CE   AE
2
2
Only half the object (and image) size is needed.
18
Quiz (bonus: 20 points)
A person is standing still 2 meters in front of a
mirror, then the mirror is moved 1 meter towards
him and then stopped. What’s the distance between
the person and his image before and after the
mirror is moved?
a) 2m and 1m
b) 3m and 2m
c) 4m and 2m
d) 5m and 4m
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