20/08/2010
1
i r
1 1 2 2sin sinn n
Waves
Ensure Sound is on:
Light Doesn’t Just Bounce
It Also Refracts!
Reflected: Bounces (Mirrors!)
Refracted: Bends (Lenses!)
0
1 1 1
id d f
i r
1
2
n2
n1
i = r
n1 sin(1) = n2 sin(2)
cv
n
Speed of light in
mediumIndex of refraction
Speed of
light in
vacuum
1n v c so
Index of Refraction
300,000,000 m/second: it’s not
just a good idea, it’s the law!
always!
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2
n1
n2
When light travels from one medium to another the speed
changes v=c/n, but the frequency is constant. So the
light bends:
n1 sin(1)= n2 sin(2)
1
2
1) n1 > n2
2) n1 = n2
3) n1 < n2
Preflight
Compare n1 to n2.
Snell’s Law
1 < 2
sin1 < sin2
n1 > n2
n1
n2
Snell’s Law Practice
n1sin 1 n2 sin 2
no
rmal
2
A ray of light traveling through the air (n=1) is incident on water
(n=1.33). Part of the beam is reflected at an angle r = 60. The
other part of the beam is refracted. What is 2?
1 r
Usually, there is both reflection and refraction!
sin(60) = 1.33 sin(2)
2 = 40.6 degrees
1 =r =60
Total Internal Reflection
normal
2
1
n2
n1
Recall Snell’s Law: n1 sin(1)= n2 sin(2)
(n1 > n2 2 > 1 )
1 = sin-1(n2/n1) then 2 = 90
c
Light incident at a larger angle will
only have reflection (i = r)
ir
―critical angle‖
The light ray MUST
travel in the direction
from larger n to
smaller n for a
critical angle to be
formed.
Review Equations (1)
cn
v
1 1 2 2sin sinn n
c f
11 1 2
2 2 2 1
sin
sin
v n
v n
n is the refractive index
and v is the speed in the
medium
Snell’s Law
Wave equation
Alternative
versions of
Snell’s Law
21 2
1
sin ,c
nn n
n
Critical angle for
total internal
reflection
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Light travels from air into water. If it enters water at 600 to the surface, find the
angle at which it is refracted and the speed at which it moves in the water
(nwater=1.33)
Understanding
60n=1.00
n=1.33 1 1 2 2sin sinn n
30
12 1
2
1 12 1
2
sin sin
sin sin
n
n
n
n
1
2
1.00sin sin 30
1.33
22
cn
v
cv
n
8
8
3.0 10
1.33
2.3 10
m
sv
m
s
22
Determine the critical angle for an air-water boundary (nwater=1.33, nair=1.00)
Understanding
21 2
1
1 2
1
sin ,
sin
c
c
nn n
n
n
n
1 1.00sin
1.33
49
c
A wave moves from medium 1 to medium 2. given that θ2=300, 2=0.50 cm,
v1=3.0 cm/s, and the index of refraction for n2 = 1.3 and for n1 = 1.0, find v2, 1,
θ1, and f
Understanding
2 1
1 2
n
n
21 2
1
1.30.50
1.0
0.65
n
n
cm
cm
2 1
1 2
n v
n v
2 1
1 2
12 1
2
1.03.0
1.3
2.3
n v
n v
nv v
n
cm
s
cm
s
2
2
vf
2.3
0.50
4.6
cm
sfcm
Hz
12
1 2
sin
sin
n
n
22 1
1
1 21 2
1
1
sin sin
sin sin
1.3sin sin 30
1.0
41
n
n
n
n
n2
n1
d
d
d d
n2
n1
Apparent depth:
Apparent Depth
actual fish
apparent fish
d d
n2
n1
Apparent Depth
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Can the person standing on the edge of the pool be
prevented from seeing the light by total internal reflection ?
1) Yes 2) No
Understanding
“There are millions of light ’rays’ coming from the light. Some of the rays will be totally reflected back into the water,but most of them will not.”
As we pour more water into bucket, what
will happen to the number of people who
can see the ball?
1) Increase 2) Same 3) Decrease
Understanding Refraction
As we pour more water into bucket, what
will happen to the number of people who
can see the ball?
1) Increase 2) Same 3) Decrease
Understanding Refraction Fiber Optics
Telecommunications
Arthroscopy
Laser surgery
Total Internal Reflection only
works if noutside < ninside
At each contact w/ the glass air interface, if the light hits at
greater than the critical angle, it undergoes total internal
reflection and stays in the fiber.
ninside
noutside
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Fiber Optics
At each contact w/ the glass air interface, if the light hits at
greater than the critical angle, it undergoes total internal
reflection and stays in the fiber.
We can be certain that
ncladding < ninside
ninside
Add ―cladding‖ so
outside material doesn’t
matter!
ncladding
noutside
Polarization
Transverse waves have a polarization
(Direction of oscillation of E field for light)
Types of Polarization
Linear (Direction of E is constant)
Circular (Direction of E rotates with time)
Unpolarized (Direction of E changes randomly)
x
y
Linear PolarizersLinear Polarizers absorb all electric fields perpendicular to their
transmission axis.
Dochroic (Polaroid H)
materials have the property
of selectively absorbing
one of two orthogonal
components of ordinary
light. The crystal actual
align opposite to what you
would intuitively expect
waves to be blocked. I.e.
vertical alignment block
vertical electric fields, by
absorbing their energy and
oscillating the electrons in a
vertical manner. Horizontal
waves will pass through.
Reflected Light is Polarized
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Unpolarized Light on
Linear Polarizer
Most light comes from electrons accelerating in
random directions and is unpolarized.
Averaging over all directions: Stransmitted= ½ Sincident
Always true for unpolarized light!
2S E
Intensity is
proportional
to the cross
product of E
and B
Linearly Polarized Light on Linear
Polarizer (Law of Malus)
Etranmitted = Eincident cos()
Stransmitted = Sincident cos2()
TA
is the angle between the
incoming light’s polarization, and
the transmission axis
Transmission axisIncident E
ETransmitted =Eincidentcos()
Recall S is intensity
and2S E
Intensity
2
2
0
2 2
0
2
0
2
0
0
cos
cos
1
2
1
2
1
2
trans
S E
S E
E
E
E
S
The average of 2cos
2 22
0 0
2
0
1 cos 21 1cos
2 0 2 2
1 1 1sin 2
2 2 4
1
2
1
2
d d
2 2
2 2
2
2
cos sin 1
cos sin cos 2
2cos 1 cos 2
1 cos 2cos
2
Question
Unpolarized light (like the light from the sun) passes through a polarizing
sunglass (a linear polarizer). The intensity of the light when it emerges is
1. zero
2. ½ what it was before
3. ¼ what it was before
4. ⅓ what it was before
5. need more information
Now, horizontally polarized light passes through the same glasses (which are
vertically polarized). The intensity of the light when it emerges is
1. zero
2. ½ what it was before
3. ¼ what it was before
4. ⅓ what it was before
5. Need more information
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Answer for 1
Unpolarized light (like the light from the sun)
passes through a polarizing sunglass (a linear
polarizer). The intensity of the light when it
emerges is
1. zero
2. 1/2 what it was before
3. 1/4 what it was before
4. 1/3 what it was before
5. need more information 0
1
2transS S
Answer for 2
Now, horizontally polarized light passes through the
same glasses (which are vertically polarized). The
intensity of the light when it emerges is
• zero
• 1/2 what it was before
• 1/4 what it was before
• 1/3 what it was before
• need more information
Since all light hitting the
sunglasses was
horizontally polarized
and only vertical rays
can get thru the
glasses, the glasses
block everything
Law of Malus – 2 Polarizers
Cool Link
unpolarized
light
E1
45
I = I0
TA
TA
0
TA
E0
I3
B1
unpolarized
light
E1
45
I = I0
TA
TA
0
TA
E0
I3
B1
1) Intensity of unpolarized light incident on linear polarizer is reduced by ½ . S1 = ½ S0
S = S0
S1
S2
2) Light transmitted through first polarizer is vertically polarized. Angle between it and
second polarizer is =90º. S2 = S1 cos2(90º) = 0
unpolarized
light
E1
45
I = I0
TA
TA
0
TA
E0
I3
B1
unpolarized
light
E1
45
I = I0
TA
TA
0
TA
E0
I3
B1
Law of Malus – 3 Polarizers
2) Light transmitted through first polarizer is vertically polarized. Angle between it and
second polarizer is θ=45º
3) Light transmitted through second polarizer is polarized 45º from vertical. Angle
between it and third polarizer is θ=45º
I2= I1cos2(45)
I3 = I2 cos2 (45º)
= ½ I0 cos4 (45º)
I1= ½ I0
. I2 = I1 cos2 (45º) = ½ I0 cos2 (45º)
Angle is 45 degrees with
respect to last TA
I1=½I01) Light will be vertically polarized with:
Remember you can use I
or S for intensity.
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0
TA
TA
S1
S2
S0
60
TATA
S1
S2
S0
60
ACT: Law of Malus
A B
1) S2A > S2
B 2) S2A = S2
B 3) S2A < S2
B
S1= S0cos2(60)
S2= S1cos2(30) = S0 cos2(60) cos2(30)
S1= S0cos2(60)
S2= S1cos2(60) = S0 cos4(60)
E0E0
Brewster’s Angle
Light
The incident light can have the electric field of the
light waves lying in the same plane as the that is is
traveling. Light with this polarization is said to be
p-polarized, because it is parallel to the plane.
Light with the perpendicular polarization is said to
be s-polarized, from the German senkrecht—
perpendicular
Brewster’s angle
…when angle between reflected
beam and refracted beam is
exactly 90 degrees, reflected
beam is 100% horizontally
polarized !
Reflected light is partially polarized (more horizontal than
vertical). But…
2
1
tan B
n
n
n1 sin B = n2 sin (90-B)
n1 sin B = n2 cos (B)
horiz. and vert.
polarized
B B
90º-B
90º
horiz.
polarized
only! n1
n2
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Polarized so that the oscillation is in
same direction as to be flat with
reflecting surface. (polarized in the
plane of reflecting surface)
Brewster’s Angle
Since then
1 1 2 2
1 2
2 1
1 2
1 1
21
1
sin sin
sin
sin
sin
cos
tan
n n
n
n
n
n
n
n
1 2 90
2 1sin cos
Polarizing sunglasses are often considered to be better
than tinted glasses because they…
Preflight
block more light
block more glare
are safer for your eyes
are cheaperWhen glare is around
B, it’s mostly horiz.
polarized!
Polarizing sunglasses (when worn by someone standing
up) work by absorbing light polarized in which direction?
horizontal
vertical
Brewster’s Angle from air off glass
tanglass
air
n
n
1.5
tan1.0
56
56
Air n=1.0
Glass n=1.5
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How Sunglasses Work
Any light ray that comes from a reflection will
be slightly horizontally polarized, the glasses
will only allow vertically (or vertical
component of the unpolarized light) to pass
through to the eye. Thus reducing the glare.
ACT: Brewster’s Angle
When a polarizer is placed between the light source and
the surface with transmission axis aligned as shown, the
intensity of the reflected light:
(1) Increases (2) Unchanged (3) Decreases
T.A.
Dispersion
prism
White light
Blue light gets
deflected more
nblue > nred
The index of refraction n depends on color!
In glass: nblue = 1.53 nred = 1.52
Skier sees blue coming up
from the bottom (1), and
red coming down from the
top (2) of the rainbow.
Understanding
Which is red?
Which is blue?
Rainbow Power point
(1)
(1)
(2)
(2)
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LIKE SO! In second rainbow
pattern is reversedInterference Requirements
Need two (or more) waves
Must have same frequency
Must be coherent (i.e. waves must have definite
phase relation)
Interference for Sound …
For example, a pair of speakers, driven in phase,
producing a tone of a single f and :
l1 l2
But this won’t work for light rays--can’t get coherent pair of
sources
hmmm… I’m just
far enough away that l2-l1=/2, and I
hear no sound at
all!
Interference for Light …
Can’t produce coherent light from
separate sources. (f 1014 Hz)
Single source
Two different paths
Interference
possible here
• Need two waves from single source
taking two different paths
– Two slits
– Reflection (thin films)
– Diffraction
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ACT: Young’s Double Slit
Screen a distance
L from slits
Single source of
monochromatic
light
d
2 slits-
separated
by d
1) Constructive
2) Destructive
3) Depends on L
L
Light waves from a single source travel through 2 slits
before meeting on a screen. The interference will be:
The rays start in
phase, and travel the
same distance, so
they will arrive in
phase.
Understanding
Screen a distance
L from slits
Single source of
monochromatic
light
d
2 slits-
separated
by d
1) Constructive
2) Destructive
3) Depends on L
The rays start out of
phase, and travel the
same distance, so
they will arrive out of
phase.
L
½ shift
The experiment is modified so that one of the waves has its
phase shifted by ½ . Now, the interference will be:
Young’s Double Slit Concept
Screen a distance
L from slits
Single source of
monochromatic
light
d
2 slits-
separated
by d
L
At points where the
difference in path
length is 0, ,2, …,
the screen is bright.
(constructive)
At points where the
difference in path
length is
the screen is dark.
(destructive)
3 5, ,
2 2 2
Young’s Double Slit Key IdeaL
Two rays travel almost exactly the same distance.(screen must be very far away: L >> d)
Bottom ray travels a little further.
Key for interference is this small extra distance.
d
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d
Path length difference =
Young’s Double Slit Quantitative
d sin(
d
Constructive interference
Destructive interference(Where m = 0, 1, 2, …)
sin(θ) tan(θ) = y/L
sind m
1
sin2
d m
Need < d
y
L
m Ly
d
1
2m L
yd
(Where m = 0, 1, 2, …)
L is much larger than d so yellow lines are parallel
If You only know the Distances between
Sources and Point (or Large Distances)
Pm
1 2 , 0,1,2m mP S P S m m
1 2
1, 0,1,2
2m mP S P S m m
Constructive Interference
Destructive interference
m=0
Central
Maximum m=1 (first order maximum) m=0 (first order minimum)
Equations
1 2 , 0,1,2m mP S P S m m Constructive interference when angle not given
1 2
1, 0,1,2
2m mP S P S m m
Destructive interference when angle not given
sin , 0,1,2,d m m Constructive interference, angle given
1
sin , 0,1,2,2
d m m
Destructive interference, angle given
, 0,1,2,y
d m mL
Constructive interference, distance from
screen given
1, 0,1,2,
2
yd m m
L
Destructive interference, distance from
screen given
d
L
Understanding
y
When this Young’s double slit experiment is placed under
water. The separation y between minima and maxima
1) increases 2) same 3) decreases
…wavelength is shorter under water.
1 1
2 2m L L
m Ly
d d d
1 2
2 1
n
n
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UnderstandingIn the Young double slit experiment, is it possible to see interference
maxima when the distance between slits is smaller than the wavelength
of light?
1) Yes 2) No
Need: d sin = m => sin = m / d
If d then / d > 1
so sin > 1
Not possible!
Understanding
If one source to the screen is 1.2x10-5 m farther than the other source
and red light with wavelength 600 nm is used, determine the order
number of the bright spot.
Path length difference =5
1 2 1.2 10m mP S P S m
1 2
1 2
5
7
1.2 10
6.00 10
20
m m
m m
P S P S m
P S P Sm
m
m
We don’t know the
angle, so we use
Understanding
Light of wavelength 450 nm shines through openings 3.0 um apart. At what
angle will the first-order maximum occur? What is the angle of the first order
minimum?
sind m
1
7
1
6
sin
1 4.50 10sin
3.0 10
8.6
m
d
m
m
Since we are asked for the angle
1
sin2
d m
1
7
1
6
1
2sin
10 4.50 10
2sin
3.0 10
4.3
m
d
m
m
First order max starts at m=1,
first order min starts at m=0
Understanding
A point P is on the second-order maximum, 65 mm from one source and 45 mm
from another source. The sources are 40 mm apart. Find the wavelength of the
wave and the angle the point makes in the double slit.
65 45 2
20
2
10
mm mm
mm
mm
1 2m mP S P S m sind m
1
1
sin
2 10sin
40
30
m
d
mm
mm
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Understanding
For light of wavelength 650 nm, what is the spacing of the bright bands on a screen
1.5 m away if the distance between slits is 6.6x10-6 m?
Since we want the spacing on the
bands and are given the distance
the screen is from the slits
y m
L d
7
6
1 6.50 101.5
6.6 10
0.15
my L
d
mm
m
m
Therefore the distance between the principle bright band and the first order
band is 0.15m (this is actually the distance between any two bright bands
d
Path length difference 1-2
Multiple Slits(Diffraction Grating – N slits with spacing d)
L
= d sin
sind m
Constructive interference for all paths when
d
Path length difference 1-3 = 2d sind
1
2
3
Path length difference 1-4 = 3d sin
2
3
4
d
UnderstandingL
d
1
2
3
All 3 rays are interfering constructively at the point shown. If the intensity
from ray 1 is I0 , what is the combined intensity of all 3 rays?
1) I0 2) 3 I0 3) 9 I0
When rays 1 and 2 are interfering destructively, is the intensity
from the three rays a minimum?
1) Yes 2) No
Each slit contributes amplitude
Eo at screen. Etot = 3 Eo. But I a
E2. Itot = (3E0)2 = 9 E0
2 = 9 I0this one is still there!
these add to zero
3
2
3
2
3
2
3
2 dsin
Three slit interference
I0
9I0
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For many slits, maxima are still at sin md
Region between maxima gets suppressed more and
more as no. of slits increases – bright fringes become
narrower and brighter.
10 slits (N=10)
dsin
inte
nsity
20
2 slits (N=2)
dsin
inte
nsity
20
Multiple Slit Interference (Diffraction Grating)
Peak location
depends on
wavelength!
N slits with spacing d
Constructive Interference Maxima are at:
sin md
* screen
VERY far
away
Diffraction Grating
Same as for Young’s
Double Slit !
UnderstandingCompare the spacing produced by a diffraction grating with 5500 lines
in 1.1 cm for red light (620 nm) and green light (510 nm). The distance
to the screen is 1.1 m
First we need to
determine d the distance
between the grates
2
6
1.1 10
5500
2.0 10
md
lines
m
Now for the spacing, y
Ly
d
7
6
1.1 6.20 10
2.0 10
0.34
red
m my
m
m
7
6
1.1 5.10 10
2.0 10
0.28
green
m my
m
m
Note: the larger
spacing for red than
for green
UnderstandingA diffraction grating has 1000 slits/cm. When light of wavelength 480
nm is shone through the grating, what is the separation of the bright
bands 4.0 m away? How many orders of this light are seen?
First we need to
determine d the distance
between the grates
5
distance (m)
0.01
1000
1.0 10
dslits
m
m
Now for the spacing, y
Ly
d
7
5
4.0 4.8 10
1.0 10
0.192
red
m my
m
m
For max we will
set the angle θ
to 900
5
7
sin
sin 90
1.0 10
4.8 10
20.8
20
d m
dm
d
m
m
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Single Slit Interference?!Wall
Screen with opening
shadow
bright
This is not what is actually
seen!
Diffraction Rays
Huygens’ Wave Model
In this theory, every point on a wave front is considered a point source for tiny
secondary wavelets, spreading out in front of the wave as the same speed of
the wave itself.
•
•
Diffraction/ HuygensEvery point on a wave front acts as a source of tiny
wavelets that move forward.
We will see maxima and
minima on the wall.
Light waves originating at
different points within
opening travel different
distances to wall, and can
interfere!
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Huygens’ Principle
The Dutch scientist Christian
Huygens, a contemporary of
Newton, proposed Huygens’
Principle, a geometrical way
of understanding the behavior
of light waves.
Huygens Principle: Consider a wave front of light:
1. Each point on the wave front is a new source of
a spherical wavelet that spreads out spherically
at wave speed.
2. At some later time, the new wave front is the
surface that is tangent to all of the wavelets.
Christian Huygens
(1629 - 1695)
1st minima
Central
maximum
Single Slit Diffraction
As we have seen in the
demonstrations with light
and water waves, when light
goes through a narrow slit, it
spreads out to form a
diffraction pattern.
Now, we want to understand
this behavior in more detail.
W
sin2
w
2
W
Rays 2 and 2 also start W/2 apart and have the same path
length difference.
1st minimum at:
When rays 1 and 1
interfere destructively.
sin2 2
w
Under this condition, every ray originating in top half of slit interferes destructively
with the corresponding ray originating in bottom half.
Single Slit Diffraction
sin2 2
w sinW sin
w
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19
w
Rays 2 and 2 also start w/4 apart and have the same path
length difference.
2nd minimum at
Under this condition, every ray originating in top quarter of slit interferes destructively
with the corresponding ray originating in second quarter.
Single Slit Diffraction
w
4
wsin( )
4
When rays 1 and 1
will interfere destructively.
wsin( )
4 2
wsin( )
4 2
sin( ) 2w
2sin( )
w
Analyzing Single Slit Diffraction
For an open slit of width w, subdivide the opening into segments and imagine a
Hyugen wavelet originating from the center of each segment. The wavelets
going forward (=0) all travel the same distance to the screen and interfere
constructively to produce the central maximum.
Now consider the wavelets going at an angle such that w sin @ w . The
wavelet pair (1, 2) has a path length difference Dr12 = /2, and therefore will
cancel. The same is true of wavelet pairs (3,4), (5,6), etc. Moreover, if the
aperture is divided into m sub-parts, this procedure can be applied to each sub-
part. This procedure locates all of the dark fringes.
thsin ; 1, 2, 3, (angle of the m dark fringe)m mm mw
@
w w
w
2
Conditions for Diffraction Minima
th
sin ; 1, 2, 3,
(angle of the m dark fringe)
m mm mw
@
w w
w w w w w w
w 2w w
Condition for quarters of slit to destructively interfere
sin mw
(m=1, 2, 3, …)
Single Slit Diffraction Summary
Condition for halves of slit to destructively interfere sin( )w
sin( ) 2w
Condition for sixths of slit to destructively interfere sin( ) 3w
THIS FORMULA LOCATES MINIMA!!
Narrower slit => broader pattern
All together…
Note: interference only occurs when the (w)idth >
mym
L w
If you do not
know the angle
20/08/2010
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UnderstandingA slit is 2.0x10-5 m wide and is illuminated by light of wavelength 550 nm.
Determine the width of the central maximum, in degrees and in centimetres when
L=0.30m?
We will find the position of the first order minimum, which is
one-half the total width of the central maximum.
sin mm w
1
7
1
5
sin
sin
sin
1 5.50 10sin
2.0 10
1.6
m
m
m
m w
m
w
m
w
m
m
The full width is twice this, 3.20
mwym
L
7
5
1 5.50 10 0.30
2.0 10
0.00825
m
m Ly
w
m m
m
m
Therefore the width of Central
Maximum is 2(.825cm)=1.65 cm
UnderstandingLight is beamed through a single slit 1.0 mm wide. A diffraction pattern is
observed on a screen 2.0 m away. If the central band has a width of 2.5 mm,
determine the wavelength of the light.
To use the equations,
we need half of the
width
33
1
2.5 101.25 10
2
my m
mwym
L
1
3 3
7
1.0 10 1.25 10
2.0 1
6.25 10
wy
Lm
m m
m
m
Example: Diffraction of a laser
through a slit
Light from a helium-neon laser ( = 633 nm) passes through a narrow slit and
is seen on a screen 2.0 m behind the slit. The first minimum of the diffraction
pattern is observed to be located 1.2 cm from the central maximum.
How wide is the slit?
11
(0.012 m)0.0060 rad
(2.00 m)
y
L
74
3
1 1
(6.33 10 m)1.06 10 m 0.106 mm
sin (6.00 10 rad)w
@
1.2 cm
m
Lmw
y
7
4
2
2.0 1 6.33 101.06 10
1.2 10
m mm
m
or
Width of a Single-Slit
Diffraction Pattern
; 1,2,3, (positions of dark fringes)m
m Ly m
w
2(width of diffraction peak from min to min)
Lwidth
w
width
-y1 y1 y2 y30
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Understanding
Two single slit diffraction patterns are shown. The distance from
the slit to the screen is the same in both cases.
Which of the following could be true?
1
2
(a) The slit width w is the same for both; 12.
(b) The slit width w is the same for both; 12.
(c) The wavelength is the same for both; width w1<w2.
(d) The slit width and wavelength is the same for both; m1<m2.
(e) The slit width and wavelength is the same for both; m1>m2.
sin mw
(m=1, 2, 3, …) my
mL w
Understanding
What is the wavelength of a ray of light whose first side diffraction
maximum is at 15o, given that the width of the single is 2511 nm?
sin mw
sinw
m
We want the first maximum, therefore m is approximately 1.5
2511 sin 15
1.5
430
nm
nm
(degrees)
(degrees)
(degrees)
Combined Diffraction
and Interference (FYI only)
So far, we have treated
diffraction and interference
independently. However, in a
two-slit system both
phenomena should be present
together.
2
2
2slit 1slit
sin4 cos ;
sin ;
sin .
I I
a ay
L
d dy
L
a
a
a
d
a a
Notice that when d/a is an integer, diffraction minima will fall on top of ―missing‖
interference maxima.
Maxima and minima will be a series of bright and dark
rings on screen
Central maximum
sin 1.22D
First diffraction minimum is at
Hole with diameter D
light
Diffraction from Circular Aperture
1st diffraction minimum
20/08/2010
22
UnderstandingA laser is shone at a screen through a very small hole. If you
make the hole even smaller, the spot on the screen will get:
(1) Larger (2) Smaller
Which drawing correctly depicts the pattern of light on the
screen?
(1) (2) (3) (4)
sin 1.22D
sin
1.22D
I
Intensity from Circular Aperture
First diffraction minima
These objects are just resolved
Two objects are just resolved when the maximum of one is
at the minimum of the other.
Resolving Power
To see two objects distinctly, need objects » min
min
objects
Improve resolution by increasing objects or decreasing min
objects is angle between objects and aperture:
tan objects d/y
sin min min = 1.22 /D
min is minimum angular separation that aperture
can resolve:
D
dy
20/08/2010
23
Understanding
Astronaut Joe is standing on a distant planet with binary
suns. He wants to see them but knows it’s dangerous to look
straight at them. So he decides to build a pinhole camera by
poking a hole in a card. Light from both suns shines through
the hole onto a second card.
But when the camera is built, Astronaut Joe can only see one
spot on the second card! To see the two suns clearly, should
he make the pinhole larger or smaller?
Decrease min = 1.22 / D
Want objects > min
Increase D !
min minsin 1.22D
ACT: Resolving Power
How does the maximum resolving power of your eye
change when the brightness of the room is decreased.
1) Increases 2) Constant 3) Decreases
When the light is low, your pupil dilates (D can increase
by factor of 10!) But actual limitation is due to density of
rods and cones, so you don’t notice an much of an
effect!
Summary
Interference: Coherent waves
Full wavelength difference = Constructive
½ wavelength difference = Destructive
Multiple Slits
Constructive d sin() = m m=1,2,3…)
Destructive d sin() = (m + 1/2) 2 slit only
More slits = brighter max, darker mins
Huygens’ Principle: Each point on wave front acts as
coherent source and can interfere.
Single Slit:
Destructive: w sin() = m m=1,2,3…)
Thin Film Interference
n1 (thin film)
n2
n0=1.0 (air)
t
1 2
Get two waves by reflection off of two different interfaces.
Ray 2 travels approximately 2t further than ray 1.
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Reflection + Phase Shifts
n1
n2
Upon reflection from a boundary between two transparent
materials, the phase of the reflected light may change.
• If n1 > n2 - no phase change upon reflection.
• If n1 < n2 - phase change of 180º upon reflection.
(equivalent to the wave shifting by /2.)
Incident wave Reflected wave
Thin Film Summary (1)
n1 (thin film)
n2
n = 1.0 (air)
t
1 2
Ray 1: 1 = 0 or ½
Determine , number of extra wavelengths for each ray.
If |(2 – 1)| = ½ , 1 ½, 2 ½ …. (m + ½) destructive
If |(2 – 1)| = 0, 1, 2, 3 …. (m) constructive
Note: this is
wavelength in film!
(film= o/n1)+ 2 t/ film
Reflection Distance
Ray 2: 2 = 0 or ½
This is important!
Thin Film Summary (2)
n1 (thin film)
n2
tvacuum
filmn
1 2n n
1 2n n
2 filmt m
12
2filmt m
Phase shift after reflection
No phase shift after reflection
No Phase shift
Phase shift
Thin Film Practice
nglass = 1.5
nwater= 1.3
n = 1.0 (air)
t
1 2
1 = ½
2 = 0 + 2t / glass = 2t nglass/ 0= 1
Blue light (o = 500 nm) incident on a glass (nglass = 1.5) cover slip (t = 167 nm)
floating on top of water (nwater = 1.3).
Is the interference constructive or destructive or neither?
Phase shift = 2 – 1 = ½ wavelength
Reflection at air-film interface only
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ACT: Thin Film
nglass =1.5
nplastic=1.8
n=1 (air)
t
1 2
1 = ½
2 = ½ + 2t / glass = ½ + 2t nglass/ 0= ½ + 1
Blue light = 500 nm is
incident on a very thin film
(t = 167 nm) of glass on
top of plastic. The
interference is:
(1) constructive
(2) destructive
(3) neither
Phase shift = 2 – 1 = 1 wavelength
Reflection at both interfaces!
Understanding
The gas looks:
• bright
• dark
A thin film of gasoline (ngas=1.20) and a thin film of oil (noil=1.45) are floating on water (nwater=1.33). When the thickness of the two films is exactly one wavelength…
t =
nwater=1.3
ngas=1.20
nair=1.0
noil=1.45
1,gas = ½
The oil looks:
• bright
• dark
2,gas = ½ + 2 1,oil = ½ 2,oil = 2
| 2,gas – 1,gas | = 2 | 2,oil – 1,oil | = 3/2
constructive destructive
UnderstandingLight travels from air to a film with a refractive index of 1.40 to water (n=1.33). If
the film is 1020 nm thick and the wavelength of light is 476 nm in air, will
constructive or destructive interference occur?
nfilm=1.4
nwater=1.33
n=1.0 (air)
t
1 2
Ray 1: Since nair<nfilm, we have a phase shift of /2 when it reflects
of the film.
Ray 2: Since nfilm>nwater, we have
no phase shift. So we need only
determine the number of extra
wavelengths travelled by the ray.
74.76 10
1.40
340
airfilm
filmn
m
nm
2
20
2 10200
340
6
film
t
nm
nm
1 = ½
Phase shift = |2 – 1|= 5½ wavelength Destructive interference
UnderstandingA camera lens (n=1.50) is coated with a film of magnesium fluoride (n=1.25).
What is the minimum thickness of the film required to minimize reflected light of
wavelength 550 nm?
nfilm=1.25
nlens=1.50
n=1.0 (air)
t
1 2
Ray 1: Since nair<nfilm, we have a phase shift of /2 when it reflects
of the film.
Ray 2: Since nfilm<nlens, we have a phase shift of /2 when it
reflects of the lens.
1 = ½
2 = ½ + 2t / film
0
0
2 1
2
4
550
4 1.25
110
film
film
tn
tn
nm
nm
We need 1 and 2 to be
out by a minimum ½ if
we require destructive
interference
vacuumfilm
n
Therefore minimum thickness is 110 nm
0
2 1
2
filmtn
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Light
Color
Color Addition & Subtraction
Spectra
What do we see? Our eyes can’t detect intrinsic light from objects
(mostly infrared), unless they get ―red hot‖
The light we see is from the sun or from artificial light (bulbs, etc.)
When we see objects, we see reflected light
immediate bouncing of incident light (zero delay)
Very occasionally we see light that has been absorbed, then re-emitted at a different wavelength
called fluorescence, phosphorescence, luminescence
Colours Light is characterized by frequency, or more
commonly, by wavelength
Visible light spans from 400 nm to 700 nm
or 0.4 m to 0.7 m; 0.0004 mm to 0.0007 mm, etc.
White light White light is the combination of all wavelengths, with
equal representation
―red hot‖ poker has much more red than blue light
experiment: red, green, and blue light bulbs make
white
RGB monitor combines these colors to display white
blue light green light red lightwavelength
combined, white light
called additive color
combination—works
with light sources
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Introduction to Spectra We can make a spectrum out of light,
dissecting its constituent colors
A prism is one way to do this
A diffraction grating also does the job
The spectrum represents the wavelength-by-wavelength content of light
can represent this in a color graphic like that above
or can plot intensity vs. wavelength
Why do things look the way
they do?
Why are metals shiny? Recall that electromagnetic waves are generated from
accelerating charges (i.e., electrons)
Electrons are free to roam in conductors (metals)
An EM wave incident on metal readily vibrates electrons on
the surface, which subsequently generates EM radiation of
exactly the same frequency (wavelength)
This indiscriminate vibration leads to near perfect reflection,
and exact cancellation of the EM field in the interior of the
metal—only surface electrons participate
What about glass? Why is glass clear?
The clear piece of glass is transparent to
visible light because the available electrons in
the material which could absorb the visible
photons have no available energy levels
above them in the range of the quantum
energies of visible photons. The glass atoms
do have vibrational energy modes which can
absorb infrared photons, so the glass is not
transparent in the infrared. This leads to the
greenhouse effect.