Transcript PHYS632_L15_ch_35_In..

Lens 1 Lens 2
Actual ray diagram purple.
Dashed lines are virtual rays
f1 f2
-15
+20
f1
f2
40
40
10
30
1 1 1
1
1
1 1
  

=+
i2  30cm.
i2 f2 p2 15cm 30cm
15 30
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.
Lecture 15 Interference Chp. 35
• Topics
– Interference is due to the wave nature of light
– Huygen’s principle, Coherence
– Young’s Interference Experiment and demo
– Intensity in double slit experiment
– Change in wavelength and phase change in a medium soap bubble
– Interference from thin films
• Demos
• Polling
Huygen’s Principle, Wavefronts and Coherence
2
E  Em sin( x  2 f t)
E  Em sin( kx  t)

E  Em sin(
2

(x  c t))
Examples of coherence
are:
k
Laser light
Small spot on tungsten filament
Wavefront
Most light is incoherent:
Two separate light bulbs
Two headlight beams on a car
Sun is basically incoherent
In order to form an interference pattern, the
incident light must satisfy two conditions:
(i) The light sources must be coherent. This means that the plane
waves from the sources must maintain a constant phase relation.
For example, if two waves are completely out of phase with φ = π
, this phase difference must not change with time.
(ii) The light must be monochromatic. This means that the light
consists of just one wavelength λ = 2π/k .
Light emitted from an incandescent
lightbulb is incoherent because the
light consists of waves of different
wavelengths and they do not
maintain a constant phase
relationship. Thus, no interference
pattern is observed.
An important question to consider is once
light is in phase what are the ways it can get
out of phase
1. The selected rays travel different distances.
2. Rays go through different material with different
index of refraction
3. Reflection from a medium with greater index of refraction
First what do we mean by out of phase or in phase
In Phase
Constructive interference
Out of Phase by
180 degrees or 
radians or /2
Destructive interference
In between
1) Lets consider the first case. How does light get out of
phase when selected rays travel different distances.
This leads to the famous Young’s double slit experiment
Young’s Double Slit
Interference Experiment
m=2
q
y
m=1
m=0
m=1
m=2
D
If you now send the light from the two openings onto a screen, an
interference pattern appears, due to differing path lengths from each
source
• we have constructive interference if paths differ by any number of
full wavelengths
• destructive interference if difference is half a wavelength longer or
shorter
Constructive interference
Constructive interference
Destructive interference
We want to find the pathlength difference
  path length difference
 =d sinq
Constructive interference
Destructive interference
How do we locate the vertical position of the fringes on the screen?
We want to relate y to m, L and the angle
1) L >> d
2) d >> λ
These tell us that θ is small
Therefore,
y
tan q 
L
y  L tan q
y is the distance to the m’th fringe
ym  L tan q
tanq ; sin q
ym  L sin q
dsinq  m
m
sinq =
ym 
ym
Maxima
d
m L
d
Maxima
Minima
m +/- ym
m +/- ym
0
1
2
3
0
1
2
3
0
L/d
2L/d
3L/d
L/2d
3L/2d
5L/2d
7L/2d
1
d sin q  (m  ) m=0,1,2,3..
2
ym 
(m  1 / 2) L
d
Young’s Double Slit
Interference Experiment
m=2
q
y
m=1
m=0
m=1
m=2
D
Problem 13E.
Suppose that Young’s experiment is performed with blue-green light
of 500 nm. The slits are 1.2mm apart, and the viewing screen is 5.4
m from the slits. How far apart the bright fringes?
From the table on the previous slide we see that the separation
between bright fringes is S  L 
d

500  10-9 m
S=L =(5.4 m)
d
0.0012 m
S=0.00225 m=2.25 mm
How far off the axis is the m=3 bright fringe?
y =3L/d = 3S=6.75 mm
What is the Intensity Distribution along the screen
The electric field at P is sum of E1
and E2. The intensity is
proportional to the square of the
net amplitude.
I
represents the correlation between the two waves. For
incoherent light, as there is no definite phase relation
between E1 and E2 and cross term vanishes
and incoherent sum is
For coherent sources, the cross term is non-zero. In fact, for
constructive interference E1 = E2 and I = 4I1
For destructive interference E1 = -E2 and and correlation term = - I1,
the total intensity becomes I = I1 -2I1 + I1 = 0
Suppose that the waves emerged from the slits are coherent sinusoidal
plane waves. Let the electric field components of the wave from slits 1
and 2 at P be given by
We have dropped the kx term by assuming that P is at the origin (x =0)
and we have acknowledged that wave E2 has traveled farther by giving
it a phase shift (ϕ) relative to E1
For constructive interference, with path difference of δ = λ would
correspond to a phase shift of ϕ=2π. This then implies
E0
E

E2
E0
t
E1
For constructive interference, with path difference of δ = λ would
correspond to a phase shift of ϕ=2π. This then implies
The intensity I is proportional to the
square of the amplitude of the total
electric field
Let I 0  E02
Then..
I  E2

E 2  4E02 cos 2 ( )
2
I  4I 0 cos2 12 

What about the intensity of light along the screen?
I  4I0 cos 2 12 

2d

sin q
2) How does a wave get out of phase when passing through different
media with different indices of refraction? You must consider the
thickness of the two materials in numbers of wavelength N1 and N2.
Note that the velocity and wavelength changes but not the frequency.
Vacuum
N2 
Vacuum
N1 
c=f
n=c/v
n2
L
n1

Ln2

Ln1
N2  N1 
vn= f n f /n
Path length difference =
L
Ln2  n1 



Ln2  n1 

Concept of path length difference, phase and index of
refraction
Vacuum
Vacuum
c=f
n=c/v
vn= f n f /n
Rays are in phase if
Ln2  n1 
Rays are out of phase if
2, 3

 m
Ln2  n1 

1
  m  
2

where m1,2,3..
where m=1,
3) The wave changes phase by 180 degrees when reflection from a medium
with greater index of refraction. See cartoon below.
Demonstrate
using rope
E stands for the
Electric field of the
wave
n1
Explain using wave
machine
air
E
n2
water
What are the conditions for Constructive Interference in
reflection from a soap bubble?
First consider phase change upon reflection
180 deg phase change for
Ray 1when reflecting
from the soap film surface
but not ray 2
eye
Reflection
1
2
air 1.0
n 
soap 1.30
air 1.00

L
n
Now consider the path length differences
Suppose the soap thickness L is such that
1 
2L   n
2 

Answer


1 
2L  m  n 2

2 
where m  0, 1, 2, ...
We must add a half wavelength to account for the 180 deg phase change for
constructive interference.
Path difference must = integral
 number of wavelengths plus 1/2 a wavelength.
What are the conditions for Constructive
Interference in transmission from a soap bubble?
For constructive interference path difference
must = integral number of wavelengths
Transmission
air 1.0
2L  m
L
soap 1.30
air 1.00

n
m = 1,2,3,4….
2
1

2nL
m
n =1.30

eye
No phase changes upon reflection
Demonstrate with soap bubble
Why does it look black on top?
What are the conditions for destructive interference?
39. A disabled tanker leaks kerosene (n = 1.20) into the Persian Gulf,
creating a large slick on top of the water (n = 1.30).
(a) If you are looking straight down from an airplane while the Sun is
overhead at a region of the slick where its thickness is L=460 nm, for
which wavelength(s) of visible light is the reflection brightest
because of constructive interference?
1
2
air 1.0
Kerosene 1.20
Water 1.30
180 deg phase
change
L
(b) If you are scuba diving directly under this same region of the
slick, for which wavelength(s) of visible light is the transmitted
intensity strongest? (Hint: use figure (a) with appropriate indices of
refraction.)
air 1.0
L
Kerosene 1.20
Water 1.30
2
1
Scuba diver

2L  m 


1  

2 n 2
Solving for 
4n 2 L
2m  1

4n2 L
4(1.2)(460)

 2208 nm
2m  1
1
m=0
4n2 L
4(1.2)(460)


 736 nm m =1
2m  1
3
4n2 L
4(1.2)(460)


 441.6 nm m  2
2m  1
5
Visible spectrum is 430 nm - 690 nm
We note that only the 441.6 nm
wavelength (blue) is in the visible range,
27. S1 and S2 in Fig. 36-29 are point
sources of electromagnetic waves of
wavelength 1.00 m. They are in phase
and separated by d = 4.00 m, and they
emit at the same power.
x
(a) If a detector is moved to the right
along the x-axis from source S1, at
what distances from S1 are the first
three interference maxima detected?
Consider what the path difference is between the detector and S1
and the detector and S2
detector
detector
x
d 2  x2 .
For constructive interference we have
path difference = d 2  x 2  x  m
The solution for x of this equation is
m  1,2,3..
Solve for x
d 2  x 2  x  m
d 2  x 2 = m  x
Now square both sides
d  x = ( m  x)
2
2
2
d 2  x 2  m 2 2  2mx  x 2 Now cancel x 2
d 2  m 2 2  2mx
solve for x
d 2  m 2 2
x
for m = 1, 2, 3,..
2m
m=3
m=2
m=1
16  3
For m  3 x 
 1.17m.
23 2
16  2
For m  2 x 
 3.0m.
22
d 2  m 22
x
.
2m
16  m 2
x
2m
2

16  1
For m  1 x 
 7.5m.
2
1
  
2


What about m = 4 ? This corresponds to x=0. Path difference =4 meters.
Where do the minima occur?
path difference = d 2  x 2  x  (m  12 )
m=3
m=2
m=1
m=1
m=0
m=0
m=1
m=2
m=3
x=15.75 m
x =4.55 m
 m
x=1.95
x= 0.55 m
m  0,1,2,3
d 2  (m  12 ) 2 2
x
.
1
2(m  2 ) 
16  (m  12 ) 2
x
.
1
2(m  2 )
Although the amplitudes are the same at the sources, the waves travel
 of minimum intensity and each
different distances to get to the points
amplitude decreases in inverse proportion to the square of the distance
traveled. The intensity is not zero at the minima positions.
P0
I1 
4 x 2
P0
I2 
4  (d 2  x 2 )
I1
x2
(0.55) 2
1
 2

~
I2 d  x 2 4 2  (0.55) 2 64
Demo with speakers using sound waves
Set oscillator frequency to 1372 Hz,
Then wavelength of sound is 343/1372 =0.25 m
Set speakers apart by 1m. Then maxima occur at
1  m 2 (1 / 16)
x
2m(1 / 4)
16  m 2
x
8m
m4
m3
m2
m 1
d 2  m 22
x
.
2m
x0
16-9 7
x=

m  0.33m
24
24
16-4 12
 m  0.75m
16
16
16-1 15
x=
 m  2.0m
7
7
x=