Download The refractive index is constant

What is the resolution of a grating in the first order
of 4000 groves/mm if 1 cm of the grating is
illuminated?
Are 489 and 489.2 nm resolved?

 2 
ave
2
R

2  1

1
R  nF
Constructive/Destructive interference
1.
Laser
2.
FT instrument
3.
Can be used to select wavelengths
4. Can be used to obtain information
about distances
5. Holographic Interference filter.
Change in path length results
In phase lag


E 0  x, y   E o 0  x, y  cos 2ft  0  x, y 
The photo plate contains all the information
Necessary to give the depth perception when
decoded
Interference Filter
Holographic Notch Filter
Can create a filter using
The holographic principle
To create a series of
Groves on the surface
Of the filter. The grooves
Are very nearly perfect
In spacing
Constructive/Destructive interference
1.
Laser
2.
FT instrument
3.
Can be used to obtain information
about distances
4.
Interference filter.
5.
Can be used to select wavelengths
End
Section on Using Constructive
and Destructive
Interference patterns based on
phase lags
Constructive/Destructive interference
1.
Laser
2.
FT instrument
3. Can be used to obtain information
about distances
4.
Interference filter.
5.
Can be used to select wavelengths
Begin Section
Interaction with Matter
In the examples above have assumed that there is no interaction with
Matter – all light that impinges on an object is re-radiated with it’s
Original intensity
Move electrons around (polarize)
Re-radiate
“virtual state”
Lasts ~10-14s
Move electrons around (polarize)
Re-radiate
This phenomena causes:
1. scattering
2. change in the velocity of light
3. absorption
First consider propagation of light in a vacuum
c
1
 0 0
c is the velocity of the electromagnetic wave in free space
 0 Is the permittivity of free space which describes the
Flux of the electric portion of the wave in vacuum and
Has the value
 0  8.8552 x10 12
C2
N  m2
force
capacitance
kg  m
N
s
It can be measured directly from capacitor measurements
0
Is the permeablity of free space and relates the current
In free space in response to a magnetic field and is defined as
 0  4 x10  7
N  s2
C2
c
1
 0 0
c
 0  4 x10  7
N  s2
C2
 0  8.8552 x10 12
1


C 
12
7 N  s
 8.8552 x10
2   4  x10
2 
N m 
C 

2
c
1
2
111
. x10 17
s
m2
2
1

3.33485x10
m
c  2.9986 x10
s
8
9
s
m

C2
N  m2
sqrt dielectric
1
 0 0
c
1.000034
1.000131
1.000294
1.00E+00
1.8
1.0005
1.6
1.0004
1.4
1
velocity 

1
1.0002
0.8
1.0001
0.6
1
r 
c
vvelocity



 Ke
0 0
0
0.4
0.9999
0.2
0
0.9998
1.51
4.63E+00
5.04
5.08
8.96E+00
sqrt dielectric
Dielectric constant
Typically
r 
 ~  0 so
c
velocity
This works pretty well for gases (blue line)
 Ke
Says: refractive index is proportional to the dielectric
Maxwell’s relation constant
Index of refraction
Index of refraction
1.0003
1.2
Our image is of electrons perturbed by an electromagnetic field which causes
The change in permittivity and permeability – that is there is a “virtual”
Absorption event and re-radiation causing the change
It follows that the re-radiation event should be be related to the ability to
Polarize the electron cloud
10-14 s to polarize the electron cloud and re-release electromagnetic
Radiation at same frequency
 vol molecule2 
Is  Io 

4



SCATTERING
particle
 8 4 2

2
I s  I o  4 2 1  cos  
 r

 Polarizability of electrons
a) Number of electrons
b) Bond length
c) Volume of the molecule,
which depends upon
the radius, r
= vacuum 
Io = incident intensity
Angle between incident and scattered
light
1.2
1
Relative Intensity
Light in
Most important parameter is the
relationship to wavelength
0.8
0.6
0.4
0.2
0
0
45
90
135
180
225
Angle of Scattered Light
270
315
360
At sunset the shorter wavelength is
Scattered more efficiently, leaving the
Longer (red) light to be observed
Better sunsets in polluted regions
Blue is scattered
Red is observed
Long path allow more of the blue
light (short wavelength) to be
scattered
 vol molecule2 
Is  Io 

4



What is the relative intensity of scattered
light for 480 vs 240 nm?
What is the relative intensity of scattered
light as one goes from Cl2 to Br2? (Guess)
Our image is of electrons perturbed by an electromagnetic field which causes
The change in permittivity and permeability – and therefore, the speed of the
Propagating electromagnetic wave.
It follows that the index of refraction should be related to the ability to
Polarize the electron cloud
r 
c
Refractive index = relative speed of radiation
velocity
Refractive index is related to the relative permittivity
(dielectric constant) at that Frequency
 2  1  Pm

2
M
 1
NA 
2 
Pm 
 

3o 
3kT 
0
Where  is the mass density of the sample, M is the molar
mass of the molecules and Pm is the molar polarization
Is the permittivity of free space which describes the
Flux of the electric portion of the wave in vacuum and
Has the value
Where  is the electric dipole moment operator
 is the mean polarizabiltiy
  1 N A 
   N A

 
 
2
3kT  3 0 M
  1 3 0 M 
2
2
 2  1  N A

2
  1 3 0 M
Point – refractive index
Is related to polarizability
Clausius-Mossotti equation
2e 2 R 2

3 E
  1  N A

2
  1 3 0 M
2
Where e is the charge on an electron, R is the radius of
the molecule and ∆E is the mean energy to excite an
electron between the HOMO-LUMO
 2  1   N A   2e 2 R 2 



2
  1  3 0 M   3 E 
The change in the velocity of the electromagnetic radiation is a function of
1.mass density (total number of possible interactions)
2. the charge on the electron
3. The radius (essentially how far away the electron is from the nucleus)
4. The Molar Mass (essentially how many electrons there are)
5. The difference in energy between HOMO and LUMO
2  1

2
 1
  N A   2e 2 R 2 



3

M
3

E

 0 
An alternative expression for a single atom is
2
Ne
  1
 o me
2
Molecules per
Unit volume
Each with
J oscillators

j
Transition probability that
Interaction will occur
fj
2
0j
  2j  i j 
A damping force term that account for
Absorbance (related to delta E in prior
Expression)
Natural
Frequency of
The oscillating electrons
In the single atom j
Frequency of incoming electromagnetic
wave
If you include the interactions between atoms and ignore absorbance you get
2  1

2
 1
  N A   2e 2 R 2 



3

M
3

E

 0 
2  1
Ne 2

2
  2 3 o me
when
when
c
r  
v
Ke

j
fj
2
0j
  2j
 20 j   2j
 2j   20 j
The refractive index is constant
The refractive index depends on omega
And the difference
 20 j   2j
Gets smaller so the
Refractive index rises
REFRACTIVE INDEX VS 
Anomalous dispersion near absorption bands
which occur at natural harmonic frequency of
material
Normal dispersion is required for lensing materials
What is the wavelength of a beam of light that is
480 nm in a vacuum if it travels in a solid with a
refractive index of 2?
r 
c
velocity
 frequency 


frequency
frequency
r 
c
vacuum
   c
   v
c
v elocity
vacuum
media
elocity ,media
 frequency vacuum vacuum


 frequency media
media
t  21  '
n
t  '
2
t
Wavelength
In media
Filters can be constructed
By judicious combination of the
Principle of constructive and
Destructive interference and
Material of an appropriate refractive
index
'
t  '
vacuum

'
t
t  23  '
t
 n   vacuum 
t   

 2   
2t
 vacuum
n
What is (are) the wavelength(s) selected from
an interference filter which has a base width
of 1.694 m and a refractive index of 1.34?
2t
 vacuum
n
Holographic filters are better
INTERFERENCE WEDGES
2t1
n
2t2
2 
n
2t 3
3 
n
2t4
4 
n
1 
AVAILABLE WEDGES
Vis
Near IR
IR
400-700 nm
1000-2000 nm
2.5 -14.5 m
Using constructive/destructive interference to select for polarized light
The electromagnetic wave can be described in two components, xy, and
Xy - or as two polarizations of light.
Refraction, Reflection, and Transmittance Defined
Relationship to polarization
The amplitude of the spherically oscillating electromagnetic
Wave can be described mathematically by two components
The perpendicular and parallel to a plane that described the advance of
The waveform. These two components reflect the polarization of the wave
When this incident, i, wave plane strikes a denser surface with polarizable electrons
at an angle, i, described by a perpendicular to
The plane
It can be reflected
Air, n=1
Or transmitted
The two polarization components are
reflected and transmitted with
Different amplitudes depending
Upon the angle of reflection, r,
And the angle of transmittence, t
Glass
n=1.5
T
Let’s start by examing
The Angle of transmittence
i 
c
velocity 1
Snell’s Law
sin  i  t

sin  t  i
2
1
Less dense 1
Lower refractive index
Faster speed of light
More dense 1
Higher refractive index
Slower speed of light
sin 1 sin i velocityi t



sin 2 sin t velocityt i
What is the angle of refraction, 2, for a beam of
light that impinges on a surface at 45o, from air,
refractive index of 1, to a solid with a refractive
index of 2?
sin  i  t

sin  t  i
PRISM
1.535
1.53
refractive index of crown glass
Crown Glass
(nm)

400nm 1.532
450 nm 1.528
550 nm 1.519
590 nm 1.517
620 nm 1.514
650 nm 1.513
1.525
1.52
1.515
1.51
0
100
200
300
400
500
600
700
wavelength nm
Uneven spacing = nonlinear
POINT, non-linear dispersive device
Reciprocal dispersion will vary with wavelength, since refractive index varies with
wavelength
The intensity of light (including it’s component polarization) reflected as compared
to transmitted (refracted) can be described by the Fresnel Equations
Angle of transmittence
Is controlled by
The density of
Polarizable electrons
In the media as
Described by Snell’s Law
T
R  r 
R/ /  r/ / 
2
2
 i cosi  t cost 

 
 i cosi  t cost 
  t cos  i   i cos  t 

 
  i cos  i   t cos  t 
2
T  t  
2
2
T/ /  t / / 
2


2i cosi

 
 i cosi  t cost 


2i cosi

 
 i cost  t cosi 
The amount of light reflected depends upon the Refractive indices and
the angle of incidence.
We can get Rid of the angle of transmittence using Snell’s Law
sin  i  t

sin  t  i
Since the total amount of light needs to remain constant we also know that
R//  T//  1
R  T  1
2
Therefore, given the two refractive
Indices and the angle of incidence can
Calculate everything
2
Consider and air/glass interface
i
0.8
0.7
Perpendicular
Transmittance
0.6
0.5
0.4
0.3
Here the transmitted parallel light is
Zero! – this is how we can select
For polarized light!
0.2
Parallel
0.1
0
0
10
20
30
Angle of incidence
This is referred to as the polarization
angle
40
50
Total Internal Reflection
Glass
n=1.5
Here consider
Light propagating
In the DENSER
Medium and
Hitting a
Boundary with
The lighter
medium
Air, n=1
T
Same calculation but made the indicident medium denser so that wave is
Propagating inside glass and is reflected at the air interface
Discontinuity at 42o signals
Something unusual is
happening
1.2
Reflectance and Transmittence
Parallel
1
Perpendicular
0.8
0.6
0.4
0.2
0
0
20
40
60
Angle of incidence
80
100
t
All of the light is reflected
internally
ic
RT  r 
2
  i cos  i   t cos  t 

 
  i cos  i   t cos  t 
Set R to 1 &  to 90
The equation can be solved for the critical angle of incidence
sin c 
transmitted ,less dense
For glass/air
incident ,dense
1
sin c 
 0.666
15
.
c  a sin(0.666)  0.7297rads
 0.7297rads180

 418
. o
2
The angle at which the discontinuity occurs:
1. 0% Transmittance=100% Reflectance
2. Total Internal Reflectance
3. Angle = Critical Angle – depends on refractive index
0.8
1.69/1
0.7
% Transmittence
0.6
1.3/1
1.5/1
ni/nt
0.5
0.4
Critical Angles
1.697
36.27
1.5
41.8
1.3
50.28
0.3
0.2
0.1
0
0
10
20
30
37
42
40
50
Angle of Incidence
60
70
51
80
90
Numerical Aperture
NA  sin c 

2
incident ,dense
2
 transmitted
,less dense

The critical angle here is defined differently because we have to LAUNCH the
beam
sin  i  t

sin  t  i
Shining light directly through our sample
i=0
Using Snell’s Law the angle of transmittance is
 t 
0    sin  t
 i 
sin 1 0   t  0
cos0  1
RT  r 
R/ /  r/ / 
2
2
  i cos  i   t cos  t 

 
  i cos  i   t cos  t 
  t cos  i   i cos  t 

 
  i cos  i   t cos  t 
2
2
T
cos0  1
RT  r 
R/ /  r/ / 
2
2
 i  t 


 i  t 
 t  i 


 i  t 
 t  i 

R/ /  
 i  t 
2
2
same
2
The amount of light reflected depends
Upon the refractive indices of the medium
For a typical Absorption Experiment,
How much light will we lose from the cuvette?
Or another way to put it is how much light will get transmitted?
R/ / 
I reflected
I initial
I reflected
 t  i 


 i  t 
2
 t  i 
 I initial 

 i  t 
I transmitted  I initial  I reflected
I transmitted  I initial
I transmitted
   i 
 I initial  t

 i  t 
   2
i
 I initial 1   t
 





i
t  

2
2
Io
It=I’o
Water,
refractive index
1.33
It’ = I’’o
It’’ =
I’’’o
It’’’
Air, refractive index 1
Air
Glass, refractive index 1.5
Final exiting light
2
2
  15
  15
.  1 
.  1 
''
I t '''  I ''' o 1  
   I t 1  
 



 
15
.

1
15
.

1



2
2
  15
  15
.  133
.  
.  133
.  
I t ''  I o ' ' 1  
   I t ' 1  
 



 
15
.

133
.
15
.

133
.



2
2
  133
  133
.  15
.  
.  15
.  
I t '  I ' o 1  
   I t 1  
 



 
133
.

15
.
133
.

15
.



2
  15
.  1 
I t  I o 1  
 

 
15
.

1

2
2
  15
.  1    133
.  15
.  
I t '  I o 1  
  1  
 



 
15
.

1
133
.

15
.

 
2
2
  15
.  1    133
.  15
.  
I t ' '  I o 1  
  1  
 



 
15
.

1
133
.

15
.

 
2
  15
.  1 
I t '''  I o 1  
 

15
.  1 

2
2
2
  133
.  15
.  
 
1  

133
.  15
.  

2
2
  15
.  1 
I t '''  I o 1  
 

 
15
.

1

I glass / air 2
2
  0.5 2 
 I initial 1    
  2.5 
2
  133
.  15
.  
 
1  

 
133
.

15
.

2
2
  017
.  
 
1  

 
2
.
83

I glass / air 2  I initial 0.96 0.99
2
2
I glass/ air 2  I initial 0.915
We lose nearly 10% of the light
2
2
Key Concepts
Interaction with Matter
Light Scattering
 vol 2 
I s  Io  4 
  
Refractive Index
Is wavelength dependent
Used to separate light by prisms
r 
c
velocity
2  1
Ne 2

2
  2 3 o me

j
fj
2
0j
  2j
 2  1   N A   2e 2 R 2 



2
  1  3 0 M   3 E 
Refractive index based
Interference filters
2t

n
Key Concepts
Interaction with Matter
Snell’s Law
sin  i  t

sin  t  i
Describes how light is bent based differing refractive indices
Fresnell’s Equations describe how polarized light is transmitted and/or reflected
at an interface
R  r 
R/ /  r/ / 
2
2
 i cosi  t cost 

 
 i cosi  t cost 
  t cos  i   i cos  t 

 
  i cos  i   t cos  t 
2
T  t  
2
2
T/ /  t / / 
2


2i cosi

 
 i cosi  t cost 
2


2i cosi

 
 i cost  t cosi 
Used to create surfaces which select for polarized light
2
Key Concepts
Interaction with Matter
Fresnell’s Laws collapse to
sin c 
transmitted ,less dense
incident ,dense
Which describes when you will get total internal reflection (fiber optics)
And
R/ / 
I reflected
I initial
 t  i 


 i  t 
2
Which describes how much light is reflected at an interface
PHOTONS AS PARTICLES
The photoelectric effect:
The experiment:
1. Current, I, flows when Ekinetic > Erepulsive
2. E repulsive is proportional to the applied voltage, V
3. Therefore the photocurrent, I, is proportional to the applied voltage
4. Define Vo as the voltage at which the photocurrent goes to zero = measure of
the maximum kinetic energy of the electrons
5. Vary the frequency of the photons, measure Vo, = Ekinetic,max
KE m  h  
Energy of
Ejected
electron
Work function=minimum energy binding an
Electron in the metal
Frequency of impinging photon
(related to photon energy)
KE m  h  
To convert photons to electrons that we can measure with an electrical circuit use
A metal foil with a low work function (binding energy of electrons)
DETECTORS
Ideal Properties
1. High sensitivity
2. Large S/N
3. Constant parameters with wavelength
Selectrical signal  kPradiant power  kd
Where k is some large constant
kd is the dark current
Classes of Detectors
Name
comment
Photoemissive
single photon events
Photoconductive “ (UV, Vis, near IR)
Heat
average photon flux
Want low dark current
Very sensitive detector
Rock to
Get different
wavelengths
1. Capture all simultaneously
= multiplex advantage
2. Generally less sensitive
Sensitivity of photoemissive
Surface is variable
Ga/As is a good one
As it is more or less consistent
Over the full spectral range
Diode array detectors
-Great in getting
-A spectra all at once!
Background current
(Noise) comes from?
One major problem
-Not very sensitive
-So must be used
-With methods in
-Which there is a large
-signal
Photomultiplier tube
The AA experiment
Photodiodes
The fluorescence
experiment
Charge-Coupled Device (CCD detectors)
1. Are miniature – therefore do not need to “slide” the image across
a single detector (can be used in arrays to get a
Fellget advantage)
2. Are nearly as sensitive as a photomultiplier tube
1. Set device to accumulate
charge for some period of
time. (increase sensitivity)
2. Charge accumulated near
electrode
+V
3. Apply greater voltage
4. Move charge to “gate”
And Count,
5. move next “bin” of
charge and keep on counting
6. Difference is charge in
One “bin”
Requires special cooling, Why?
END
6. Really Basic Optics
Since polarizability of the electrons in the material also controls the dielectric
Constant you can find a form of the C-M equation with allows you to compute
The dielectric constant from the polarizability of electrons in any atom/bond
N  r  1

3 o  r  2
N = density of dipoles
= polarizability (microscopic (chemical) property)
r = relative dielectric constant
Frequency dependent
Just as the refractive index is
Typically reported
Point of this slide: polarizability of electrons in a molecule is related to the
Relative dielectric constant
180
165
900
800
150
2nd
-150
700
600
135
order
65
-135
500
400
120
-120
300
200
100
105
-105
0
-100
1st order
Grating
-200
90
- 90
- 75
75
- 60
60
- 45
45
-30
30
Angle of
reflection
i=45
15
0
-15
180
165
900
800
150
2nd
-150
700
600
135
order
65
-135
500
400
120
-120
300
200
100
105
-105
0
-100
1st order
-200
90
- 90
- 75
75
- 60
60
- 45
45
-30
30
Angle of
reflection
i=45
15
0
-15