6. Really Basic Optics
15000
10000
Amplitude
5000
0
0
0.2
0.4
0.6
0.8
-5000
-10000
-15000
Time (s)
Sample
Sample
Prep
Polychromatic light
Select
light
source
select
Instrument
Selected
light
Instrument
Out put
Signal (Data)
Turn off/diminish intensity
Sample
interaction
Turn on different wavelength
detect
1
1.2
Really Basic Optics
Key definitions
Phase angle
Atomic lines vs molecular bands
Atomic Line widths (effective; natural)
Doppler broadening
Molecular bands
Continuum sources Blackbody radiators
Coherent vs incoherent radiation
6. Really Basic Optics
A

Sin=opp/hyp
y
sin  

y
A
y  A sin 
1.5
 radians 

 t  t

s 
1
y  A sint 
Amplitude
0.5
/2
3/2

2
0
0
50
100
150
200
-0.5
-1
90o phase angle
/2 radian phase angle
-1.5
t (s)
250
300
350
400
y'  A sint   
Emission of Photons
Electromagnetic radiation is emitted when electrons relax from excited states.
A photon of the energy equivalent to the difference in electronic states
Is emitted
Ehi
Elo
  c
e
E  h 
hc

Frequency 1/s
Really Basic Optics
Key definitions
Phase angle
Atomic lines vs molecular bands
Atomic Line widths (effective; natural)
Doppler broadening
Molecular bands
Continuum sources Blackbody radiators
Coherent vs incoherent radiation
Theoretical width of an atomic spectral line
Natural Line Widths
Line broadens due
1. Uncertainty
2. Doppler effect
3. Pressure
4. Electric and magnetic fields
frequency
t  1
Lifetime of an excited state is typically 1x10-8 s
1
7 1
   8  5x10
s
10 s
  c1
d
 1c 2
d
d
 d
2
c

2
  
2
c
c

 5x10 7

1 2

 5x10 10  2
s
 


9

  nm 

8 m  10 nm
 3x10  


s m 
Example: 253.7 nm
 5x10 10 
2
5
  
 253.7nm  3.22 x10 nm
 nm 
Typical natural line widths are 10-5 nm
Line broadens due
1. Uncertainty
2. Doppler effect
3. Pressure
4. Electric and magnetic fields

0

 elocity
c
Line broadens due
1. Uncertainty
2. Doppler effect
3. Pressure
4. Electric and magnetic fields
The lifetime of a spectral event is 1x10-8 s
When an excited state atom is hit with another high energy atom
energy is transferred which changes the energy of the
excited state and, hence, the energy of the photon emitted.
This results in linewidth broadening.
The broadening is Lorentzian in shape.


FWHM


1

2
f ( )  
2

2 FWHM 
     o  
 


2  
FWHM = full width half maximum
o is the peak “center” in frequency units
We use pressure broadening
On purpose to get a large
Line width in AA for some
Forms of background
correction
Line spectra – occur when radiating species are atomic particles which
Experience no near neighbor interactions
Line broadens due
1. Uncertainty
2. Doppler effect
3. Pressure
4. Electric and magnetic fields
Line events
Can lie on top
Of band events
Overlapping line spectra lead to band emission
Continuum emission – an extreme example of electric and magnetic
effects on broadening of multiple wavelengths
High temperature solids emit Black Body Radiation
many over lapping line and band emissions influenced by
near neighbors
Planck’s Blackbody Law
Stefan-Boltzmann Law
I  T4

 8h 3 


3
 c

Intensity
exp
h
kT
1
= Energy density of radiation
h= Planck’s constant
C= speed of light
k= Boltzmann constant
T=Temperature in Kelvin
= frequency
0
500
1000
1500
2000
nm
 max
b

T
Wien’s Law
2500
3000
3500
 1
   3
 
8h
exp
hc
 kT
1
1. As  ↓(until effect of exp takes over)
2. As T,exp↓, 
Really Basic Optics
Key definitions
Phase angle
Atomic lines vs molecular bands
Atomic Line widths (effective; natural)
Doppler broadening
Molecular bands
Continuum sources Blackbody radiators
Coherent vs incoherent radiation
Incoherent radiation
The Multitude of emitters, even if they emit
The same frequency, do not emit at the
Same time
A
A  A 0 sin( t )
or
A  A 0 sin(2 ft )
B
0.25
0.2
0.15
B  B 0 sin( t   )
0.1
0.05
Frequency,, is the
Same but wave from particle
B lags behind A by the
Phase angle 
Series1
0
0
-0.05
-0.1
-0.15
-0.2
-0.25
0.05
0.1
0.15
0.2
0.25
0.3
0.35
Series2
END: Key Definitions
Begin
Using Constructive and Destructive
Interference patterns based on phase lag
By manipulating the path length can cause an originally coherent beam
(all in phase, same frequency) to come out of phase can accomplish
Many of the tasks we need to control light for our instruments
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
More Intense Radiation can be obtained by Coherent Radiation
Lasers
Beam exiting the cavity is in phase (Coherent) and therefore enhanced
In amplitude
Argument on the size of signals that follows is from Atkins, Phys. Chem. p. 459, 6th Ed
Stimulated Emission
Light Amplification by Stimulated Emission of Radiation
*
w  BN o
o
Photons can stimulate
Emission just as much
As they can stimulate
Absorption
(idea behind LASERs
Stimulated Emission)
w  BN *
*
The rate of stimulated event is described by : w  BN o w  BN
Where w =rate of stimulated emission or absorption

Is the energy density of radiation already present at the frequency of
The more perturbing photons the greater the
the transition
Stimulated emission
B= empirical constant known as the Einstein coefficient for stimulated
absorption or emission
N* and No
are the populations of upper state and lower states

can be described by the Planck equation for black body radiation at some T

 8h 3 


3
 c

exp
h
kT
frequency
1
In order to measure absorption it is required that the
Rate of stimulated absorption is greater than the
Rate of stimulated emission
wabsorption  BN o  wenussion  BN *
N o  N *
If the populations of * and o are the same the net absorption is zero as a photon is
Absorbed and one is emitted
Need to get a larger population in the excited state
Compared to the ground state (population inversion)
 N*0
 * g*

N 
N 
g0 0 

Degeneracies of the different energy levels
Special types of materials have larger excited state degeneracies
Which allow for the formation of the excited state population inversion
E
pump
Serves to “trap” electrons in the excited
State, which allows for a population
inversion
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.
Radiation not along the
Path is lost
mirror
Multiple directions,
Multiple phase lags
Incoherent radiation
mirror
Stimulated emission
1. Single phase
2. Along same path
=Constructive Interference
Coherent radiation
FTIR Instrument
1.5
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.
1
Voltage
0.5
0
0
0.2
0.4
0.6
0.8
-0.5
-1
-1.5
Time
Measurement  A1 sin2f s ,1t   As sin2f S ,2 t .... An sin2f S ,n t 
1
1.2
200000
180000
Time Domain: 2 frequencies
160000
140000
4000
amplitude
120000
3000
100000
80000
2000
60000
40000
1000
Amplitude
20000
0
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
frequency
1
60000
-1000
50000
-2000
40000
Amplitude
-3000
30000
-4000
20000
t (s)
10000
8000
1 “beat” cycle
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Frequency
6000
200000
4000
180000
160000
140000
120000
amplitude
Amplitude
2000
0
0
0.2
0.4
0.6
-2000
0.8
1
1.2
100000
80000
60000
40000
20000
-4000
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
frequency
-6000
-8000
t (s)
sin A  sin B  2 sin 21  A  B cos 21  A  B
B
A
Moving mirror
Fixed mirror
C
Beam
splitter
IR source
detector
Constructive interference occurs when
1
AC  BC  n
2
4000
3000
Light Intensity at Detector
2000
1000
0
-2
-1
0
-1000
-2000
-3000
-4000
Mirror distance
dis tan ce  f mirror velocity  mirror 
+1
INTERFEROGRAMS

n
2
  mirror 
 mirror  velocity mirror
n



2 mirror  mirror
f det ector
Remember that:
2 mirror
 

n
c

1

Frequency of light
 f det ector
2 mirror  light
cn
An interferometer detects a periodic wave with a
frequency of 1000 Hz when moving at a velocity of 1
mm/s. What is the frequency of light impinging on
the detector?
 f det ector
2 mirror  light
cn
No need to SELECT
Wavelength by using
Mirror, fiber optics,
Gratings, etc.
FOURIER TRANSFORMS
Advantages
1. Jaquinot or through-put
little photon loss; little loss of source intensity
2. Large number of wavelengths allows for ensemble averaging
(waveform averaging)
3. This leads to Fellget or multiplex advantage
multiple spectra in little time implies?
s population samples 
Signal
N oise

sinf inite population
N measurements
x sample  xblank
sblank
Signal
N oise

N measurements
sinf inite population
Signal
N oise

x
sample
N measurements
 xblank

DIFFRACTION
Huygen’s principle = individual propagating waves combine
to form a new wave front
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.
Can get coherent radiation if the slit is narrow enough.
Coherent = all in one phase
June 19, 2008, Iowa Flood
Katrina Levee break
Fraunhaufer diffraction at a single slit

B
W
 F’
O
C
1


L
D
d
9.
sin CBF  sin 
opposite
CF DE



hypotenuse BC OD
E
F
BD, OD, & CD, big  assume / /
10.
 DE 
 n  CF  BC sin   
 BC
 OD 
2.
if / /, then BCF  
3.
180  DEO  EOD    90  EOD  
OF ' B  90o
4.
5.
if / /, then CFB  90o
180  CFB  FBC  BCF  90    
6.
From which we conclude
7.
EOD  
8.
CF must be n for in
phase (constructive int erference)
11.
 DE 
 DE 
 n  
 BC  
 BC
 OD 
 OE 
d
12.
 n    W
 L
D
The complete equation for a slit is
d
B

W
 sin  
I  I0 

  
b
   sin  
2
E
L
b=W/2
2
Width of the line depends upon
The slit width!!
Therefore resolution depends
On slit width
1.2
1
Relative Intensity
0.8
0.6
0.4
0.2
0
-800
-600
-400
-200
0
Beta
200
400
600
800
Also “see”
This spectra
“leak” of
Our hard won
intensity
The base (I=0) occurs whenever
 sin  
I  I0 

  
2
sinβ =0
b
Which occurs when    sin     n
2
W
b
W
2
line width  2n  2 sin   2 sin   sin 
2
2
2
W
line width  sin 
2
1.2
1
Relative Intensity
0.8
0.6
0.4
0.2
0
-800
-600
-400
-200
0
Beta
200
400
600
800
The smaller the
Slit width the
Smaller
The line width,
Which leads
To greater
Spectral
Resolution
Remember R is
Inversely proportional
To the width of
The Gaussian base
SLIT IMAGE
Slit
A
1 2 3B
Image
1234 5
Position number
When edge AB at
Position 1
Position 2
Position 3
Position 4
Position 5
Detector output:
Detector Sees
0% power
50% power
100% power
50% power
0 % power
1.2
Triangle results when
Effective bandpass = image
Output of detector
1
eff
0.8
To resolve two images that are ∆ apart requires
0.6
0.4
0.2
0
0
1
2
3
4
5
6
  2eff
Image position
Implies want a narrower slit
Essentially,
Narrow slit widths
Are generally better
  2eff
GRATINGS
Gratings
UV/Vis
IR
Groves/mm
300/2000
10/20
Points:
1. Master grating formed by diamond tip under ground
1. Or more recently formed from holographic processes
2. Copy gratings formed from resins
+
-
opp DB
sinDAB 

hyp
d
r  DBA  90
q
DBA  BAD  90  180
DBA  BAD  90
r  BAD
q

sin r 

n  CB  BD

q  CAB  90

n  CB  BD
opp CB
sinCAB 

hyp
d
i  q  90
DB
d
CAB  i
CB
sini 
d
n  d sini   d sinr 
n  d sini   sinr 
EXAMPLE
Calculate  for a grating which has
i=45
2000 groves per mm
1) Get d
1mm
500nm

2000 groves grove
2) Use grating equation to solve for 
  
n  d sini   sinr 

n  500nm sin 45o  sinr 
Inputs that can be varied are in pink
d
grooves/mm
i in degrees
r in degrees
radians
order n
1
2
3
4
5
6
nlambda = d(sin+sinr)
Czerny-Turner
construction
2000
500
45 0.785398 i in radians
-40
-0.69813
-20
-0.34907
-10
-0.17453
32.15959
16.07979
10.71986
8.039896
6.431917
5.359931
182.5433
91.27166
60.84777
45.63583
36.50866
30.42389
266.7293
133.3647
88.90977
66.68233
53.34586
44.45488
0
10
20
40
0 0.174533 0.349066 0.698132
353.5534
176.7767
117.8511
88.38835
70.71068
58.92557
440.3775
220.1887
146.7925
110.0944
88.0755
73.39625
524.5635
262.2817
174.8545
131.1409
104.9127
87.42724
You get light of 674.9 nm
½; 1/3; 1/4; 1/5; etc.
440.3
220.1
146.8
88
73
All come
through
674.9472
337.4736
224.9824
168.7368
134.9894
112.4912
Multiple wavelengths
Are observed
At a single angle
Of reflection!!
Physical Dimensions:
89.1 mm x 63.3 mm x 34.4 mm
Weight:
190 grams
Detector:
Sony ILX511 linear silicon CCD array
Detector range:
200-1100 nm
Pixels:
2048 pixels
Pixel size:
14 μm x 200 μm
Pixel well depth: ~62,500 electrons
Sensitivity:
75 photons/count at 400 nm; 41 photons/count at 600 nm
Design:
f/4, Symmetrical crossed Czerny-Turner
Focal length:
42 mm input; 68 mm output
Entrance aperture:
Czerny-Turner
construction
5, 10, 25, 50, 100 or 200 µm wide slits or fiber (no slit)
Grating options: 14 different gratings, UV through Shortwave NIR
Detector collection lens option:
Yes, L2
OFLV filter options:
OFLV-200-850; OFLV-350-1000
Other bench filter options:
Longpass OF-1 filters
Collimating and focusing mirrors:
Standard or SAG+
UV enhanced window:
Yes, UV2
Fiber optic connector:
SMA 905 to 0.22 numerical aperture single-strand optical fiber
Spectroscopic Wavelength range:
Grating dependent
Optical resolution:
~0.3-10.0 nm FWHM
Signal-to-noise ratio:
250:1 (at full signal)
A/D resolution:
12 bit
Dark noise:
3.2 RMS counts
Dynamic range:
2 x 10^8 (system); 1300:1 for a single acquisition
Integration time: 3 ms to 65 seconds
Stray light:
<0.05% at 600 nm; <0.10% at 435 nm
Corrected linearity:
>99.8%
Electronics Power consumption:
90 mA @ 5 VDC
Data transfer speed:
Full scans to memory every 13 ms with USB 2.0 or 1.1 port, 300 ms with serial port
Ocean Optics
For fluorescence lab
440.3
220.1
146.8
88
73
All come
through
Monochromator we looked inside
440.3
Only
Hit grating first
Time to get 440.3
220.1
146.8
88
73
All come
through
Hit grating second time
220.1 nm
Another way to look at it is to say
We Lose some of the light
Not all of it ends up at the intended angle of reflection
Light of 100
Nm shows up
At -30.4
AND
-17.88
And
-6.1
And
5.3
Etc.
grooves/mm
i in degrees
2000
45
order
1
d
nlambda = d(sin+sinr)
500
0.785398 i in radians
2
3
4
5
Reflection Angle
wavelength
100
125
150
175
200
225
250
275
300
325
350
375
400
425
450
475
500
-30.47131
-27.20057
-24.02322
-20.92262
-17.88496
-14.89846
-11.95286
-9.039003
-6.148561
-3.273759
-0.407192
2.458355
5.330074
8.215299
11.12168
14.05736
17.03125
-17.88496
-11.95286
-6.148561
-0.407192
5.330074
11.12168
17.03125
23.13464
29.53092
36.36259
43.85957
52.45672
63.23909
83.16511
#NUM!
#NUM!
#NUM!
-6.148561
2.458355
11.12168
20.05324
29.53092
40.0079
52.45672
70.54325
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
5.330074 17.03125
17.03125 32.88081
29.53092 52.45672
43.85957 #NUM!
63.23909 #NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
GRATING DISPERSION
D-1 = Reciprocal linear dispersion
d dis tan ce between  d cos r
D 


dy dis tan ce on screen
nF
1
Where n= order
F = focal length
d= distance/groove
d
D r 0 
nF
1
POINT = linear dispersion
What is (are) the wavelength(s) transmitted at 45o
reflected AND incident light for a grating of 4000
groves/mm?
ave
1 2

RESOLUTION

 2 
ave
2
R

2  1

1
The larger R the greater the spread between the two wavelengths, normalized by
The wavelength region
R  nF
Where n = order and N = total grooves exposed to light
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
2
2




C
N

s
 12
7
 8.8552 x10
2   4  x10
2 
N m 
C 

c
1
2
111
. x10 17
s
m2
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 
 2   N A

 
 
2
3kT  3 0 M
  1 3 0 M 
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t 2
2 
n
2t 3
3 
n
2t 4
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
velocity1
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