Transcript Week 3

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Optics News: Kepler Mission
successfully launched Saturday!
• Solar orbit
• Watching 105
stars for
planetary transits.
• Expects to find
50 Earth like
planets in a few
years.
• 95Megapixel CCD
• Why cant we
directly image
distant planets?
• Find out this
week!
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Planet Mercury passing in
front of the sun.
• Brief
dimming
of the
sun by
about 1
part in
105
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Kepler
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Week 3
• Interference and Diffraction of
Light
• Young’s Double Slit Experiment
• Thin Film Interference
• Michelson Interferometer
• Single Slit Diffraction
• Rayleigh’s Criterion
• Diffraction Gratings
Assignment 2
• Chap 33 # 14, 22, 27, 34,50, 59
• Chap 34 #6, 88, 92
My Error?
–How is the energy distributed for waves in a
string?
–Standing waves: I was right: energy
alternates between kinetic and potential
energy.
–Travelling wave: text book argues that max
KE coincides with Max PE because position of
max velocity coincides with position of max
stretching. Figure 16-12
–Localisation of energy is often difficult in
physics
From week 1
Interference of Waves
• Waves sum to
– a maximum (constructive)
– or minimum (destructive)
y1(x,t)  ym sin( kx  t)
y2 (x,t)  ym sin( kx  t  )
– Use
– Combined wave is
sin A  sin B  2 sin
1
1
( A  B) cos ( A  B)
2
2
1
1
y ( x, t )  [2 ym cos  ] sin( kx  t   )
2
2
n
– New phase, new amplitude (zero for 
Phasor Method for Interference
From week 1, phasor: a vector, length equals
wave amplitude, direction: relative phase angle
(compared with some standard wave)
For multiple waves, resultant is
vector sum of their phasors
This case:
(2/3) radians
behind
Vector sum of
the two phasors
Near destructive
interference: =0.95
Interference from Thin Films
• Light reflects off
both faces of film.
• Light reflecting at b
has to travel a bit
further.
• There is a phase
difference: outgoing
waves may experience
constructive or
destructive
interference.
Fig. 35-15
DEMO bubbles
Interference Thought Experiment
• Light is quantised: photons : energy E=hf.
• Since c=f E =hc/. h= Planck’s constant = 6 x 10-34
• Green LED light (=500nm) power 1mW. Just visible from
1km distance.
• Energy per photon = 6 x 10-34 x 3 x 108/500 x 10-9
= 3.6 x 10-19 J
• Photons per second from LED = power/energy per photon
= 10-3/3.6 x 10-19 ~ 3 x 1015 per second
• Photons per second entering your eye = number per
second emitted x area of pupil/area of 1km sphere ~ 1000
photons/second. Or 1 photon/millisecond.
• But photons travel 300km in 1 millisecond.
• So in 1km path no photons at all most of the time.
• Similarly in a 10-6 meter thckness soap bubble film
photons only present a tiny fraction of the time.
How can there be
interference when only 1
photon is present at a time?
• Seems like interference of waves
• Not interference of photons
• Interference of possibilities
• We will continue to treat light as waves but
remember that the reality is the strange and
absurd concept of the interference of something
intangible: we call it a wavefunction but we don’t
know what it is.
Reflection Phase Shifts
n1
n1
n1 > n2
n1 < n2
n2
Important to understand phase
change at reflection
n2
Reflection
Reflection Phase Shift
Off lower index
0
Off higher index
0.5 wavelength 
Fig. 35-16
Rope: high refractive index, low velocity
Wall or thin string: low refractive index,
high velocity
No phase change on reflection
Fig. 35-15
(35-15)
Film Thickness Much Less Than 
r1
r2
If L is much less than l, for example L < 0.1, then phase
difference due to the path difference 2L can be
neglected.
Phase difference between r1 and r2 will always be ½
wavelength  destructive interference  film will appear
dark when viewed from illuminated side.
ie: zero reflection as film thickness goes to zero
Note choice of reference phase is
arbitrary
Length of resultant phasor is
independent of choice of reference
phasor direction
(35-17)
Frequency, Wavelength in
different media
• Frequency of waves remains constant
(except from Doppler effect)
• Refractive index n = c/v
• When waves travel slow wavelength must
reduce.
n v
v

  n    n 
 c
c
n
• Hence phase changes c/v times greater
for a given distance.
Equations for Thin-Film Interference
Three effects can contribute to the phase
difference between r1 and r2.
1. Differences in reflection conditions.

2
2. Difference in path length traveled.
0
3. Differences in the media in which the waves
travel. One must use the wavelength in each
medium (/ n) to calculate the phase.
Fig. 35-17
½ wavelength phase difference to difference in reflection of r1 and r2
odd number
odd number
2L 
 wavelength =
 n 2 (in-phase waves)
2
2
2 L  integer  wavelength = integer  n 2 (out-of-phase waves)
n 2 

n2
2L  m 
2L  m

n2
1
2


n2
for m  0,1, 2,
for m  0,1, 2,
(maxima-- bright film in air)
(minima-- dark film in air)
(35-16)
Last week
Why Does Light Refract
• Life Saver on the Beach
• Which trajectory?
Lifesaver
sand
drowning
Sea
Law of Refraction
1
2
1 v1
t 


v1 v2
2 v2
1
sin 1 
(for triangle hce)
hc
2
sin  2 
(for triangle hcg )
hc
c
sin 1 1 v1
Index of Refraction: n 


v
sin  2 2 v2
c
c
n1 
and n2 
v1
v2
sin 1 c n1 n2


sin  2 c n2 n1
Fig. 35-3
Law of Refraction:
n1 sin 1  n2 sin2
(35-3)
Why does angle of incidence
equal angle of reflection?
1
n…
m…
N ∞
Path
length
Trajectory number
– Consider trajectories where angle
of incidence was not equal to angle
of reflection.
– Now remember phasors
– Consider the sum of all trajctories
– At edges phase angle increases
steadily: phasors create circles
that add to zero
– Near centre phase shifts approach
zero:phasors add to give a finite
resultant
Paint Black Stripes on the Mirror
1
n…
m…
N ∞
Path
length
Trajectory number
1
n…
m…
– Consider the same sum of all trajectories.
– Now black out mirror for bits that make
up  phase shift of the phasors
– Now the phasor circles are only half
circles.
– At edges where the phase angle increases
steadily we have cut out the phasors
components for half the circles so there
is now a large resultant.
– Now trajectories for which angle of
incidence is not equal to angle of
reflection are allowed.
– This is a diffraction grating.
N…
∞
trajectory number
Wavelength and Index of Refraction
n v
v

  n    n 
 c
c
n
v
cn c
fn 

  f The frequency of light in a medium is the same
as it is in vacuum.
n  n 
Since wavelengths in n1 and n2 are different, the
two beams may no longer be in phase.
L
L
Ln1
Number of wavelengths in n1: N1 


Fig. 35-4
n1  n1

L
L
Ln2
Number of wavelengths in n2 : N 2 


n 2  n2

Ln2 Ln2 L
Assuming n2  n1: N 2  N1 

  n2  n1 



(35-4)
Diffraction and the Wave Theory of Light
Diffraction pattern from a single narrow slit.
Side or secondary
maxima
Light
Central
maximum
These patterns cannot be
explained using geometrical
optics (Ch. 34)!
Fresnel Bright Spot.
Light
Bright
spot
(36-2)
DEMO ripple tank
Diffraction
For plane waves entering a single slit, the waves emerging from the slit
start spreading out, diffracting.
Fig. 35-7
(35-6)
Young’s Double Slit Experiment
Fig. 35-8
(35-7)
Calculating Fringes
The phase difference between two waves can change if the waves travel paths of
different lengths.
What appears at each point on the screen is determined by the path length difference
DL of the rays reaching that point.
Path Length Difference: DL  d sin
Fig. 35-10
(35-8)
Diffraction by a Single Slit: Locating the Minima
When the path length difference between rays r1
and r2 is /2, the two rays will be out of phase when
they reach P1 on the screen, resulting in destructive
interference at P1. The path length difference is the
distance from the starting point of r2 at the center of
the slit to point b.
For D>>a, the path length difference between rays
r1 and r2 is (a/2) sin .
Fig. 36-4
(36-3)
Diffraction by a Single Slit: Locating the Minima,
cont'd
Repeat previous analysis for pairs of rays, each separated by a
vertical distance of a/2 at the slit.
Setting path length difference to /2 for each pair of rays, we
obtain the first dark fringes at:
a

sin    a sin   
2
2
(first minimum)
For second minimum, divide slit into 4 zones of equal widths
a/4 (separation between pairs of rays). Destructive interference
occurs when the path length difference for each pair is /2.
a

sin    a sin   2 (second minimum)
4
2
Dividing the slit into increasingly larger even numbers of zones,
we can find higher order minima:
a sin   m , for m  1, 2,3
Fig. 36-5
(minima-dark fringes)
(36-4)
Intensity in Single-Slit Diffraction,
Qualitatively
To obtain the locations of the minima, the slit was equally divided into N zones,
each with width Dx. Each zone acts as a source of Huygens wavelets. Now
these zones can be superimposed at the screen to obtain the intensity as a
function of , the angle to the central axis.
To find the net electric field E (intensity a E2) at point P on the screen, we
need the phase relationships among the wavelets arriving from different zones:
 phase   2


 difference   
  path length 
  difference 


 2 
D  
  Dx sin  
  
N=18
1st side
max.
= 0
1st min.
 small
Fig. 36-6
(36-5)
Intensity in Single-Slit Diffraction, Quantitatively
Here we will show that the intensity at the screen due to
a single slit is:
 sin a 
I    I m 
(36-5)

 a 
1
a
where a   
sin  (36-6)
2

2
In Eq. 36-5, minima occur when:
a  m ,
for m  1, 2,3
If we put this into Eq. 36-6 we find:
a
m 
sin  , for m  1, 2,3

or a sin   m , for m  1, 2,3
Fig. 36-7
(minima-dark fringes)
(36-6)
Proof of Eqs. 36-5 and 36-6
If we divide the slit into infinitesimally wide zones Dx, the arc of the phasors
approaches the arc of a circle. The length of the arc is Em.  is the difference in
phase between the infinitesimal vectors at the left and right ends of the arc.  is
also the angle between the 2 radii marked R.
E
.
The dashed line bisecting f forms two triangles, where: sin  
2R
Em
.
In radian measure:  
R
Em
1
sin
Solving the previous 2 equations for E one obtains: E 
2 .
1
1
2
2

The intensity at the screen is therefore:
I   E2
 sin a 
 2  I    I m 

Im
Em
a


2
 is related to the path length difference across the
entire slit:
Fig. 36-8
 2 
     a sin  
  
(36-7)
Diffraction by a Circular Aperture
Distant point
source, e,g., star
d
lens
sin   1.22

d

(1st min.- circ. aperture)
Image is not a point, as
expected from geometrical
optics! Diffraction is
responsible for this image
pattern.
a
Light
Light
sin   1.22


a
(1st min.- single slit)
a

(36-8)
Rayleigh’s Criterion
Rayleigh’s Criterion: Two point sources are barely resolvable if their angular
separation R results in the central maximum of the diffraction pattern of one
source’s image centered on the first minimum of the diffraction pattern of the
other source’s image.
Fig. 36-10
 R small




1
 R  sin 1.22   1.22
d
d

(Rayleigh's criterion)
(36-9)
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– Southern Cross
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– Brightest is
nearest star a-Cen
– Increase aperture
d and it resolves
into a binary
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