6. Really Basic Optics.ppt

6. Really Basic Optics
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0
5000
10000
15000
0 0.2 0.4 0.6 0.8 1 1.2
Time (s)
Amplitude
Sample Sample
Prep
Instrument Instrument
Out put
Signal (Data)
Select
light
Sample
interaction
source select
Turn off/diminish intensity
detect
Polychromatic light Selected
light
Turn on different wavelength

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
y

Sin=opp/hyp
sin 
y
A
y A
 sin
A

 






 
radians
s
t t
 
y A t
 sin 
 
y A t
' sin
 
 
-1.5
-1
-0.5
0
0.5
1
1.5
0 50 100 150 200 250 300 350 400
t (s)
Amplitude
90o phase angle
/2 radian phase angle
/2  3/2 2

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
e
E h
hc
 


Ehi
Elo
Frequency 1/s
  c

Theoretical width of an atomic spectral line

Line broadens due
1. Uncertainty
2. Doppler effect
3. Pressure
4. Electric and magnetic fields
 
 t  1
Lifetime of an excited state is typically 1x10-8 s
  

1
10
5 10
1
8
7
s
x
s
 
 
c 1
d
d
c



  
1 2
d
c
d





2







c c

 
2
2
Natural Line Widths
frequency

 


























5 10
1
3 10
10
5 10
7 2
8
9
10
2
x
s
x
m
s
nm
m
x
nm


Example: 253.7 nm
 
 





 


5 10
2537 322 10
10
2 5
x
nm
nm x nm
. .
Typical natural line widths are 10-5 nm




0

elocity
c
Line broadens due
1. Uncertainty
2. Doppler effect
3. Pressure

Line broadens due
1. Uncertainty
2. Doppler effect
3. Pressure
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.
 
f
FWHM
FWHM
o
( )


  




















1 2
2
2
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
Overlapping line spectra lead to band emission
Line broadens due
1. Uncertainty
2. Doppler effect
3. Pressure
Line events
Can lie on top
Of band events

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

0 500 1000 1500 2000 2500 3000 3500
nm
Intensity
I T
  4
Wien’s Law
 max 
b
T
Stefan-Boltzmann Law

 









8
1
3
3
h
c
h
kT
exp
= Energy density of radiation
h= Planck’s constant
C= speed of light
k= Boltzmann constant
T=Temperature in Kelvin
= frequency












1 8
1
3
h
hc
kT
exp
1. As  ↓(until effect of exp takes over)
2. As T,exp↓, 
Planck’s Blackbody Law

A
B
The Multitude of emitters, even if they emit
The same frequency, do not emit at the
Same time
Incoherent radiation
-0.25
-0.2
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35
Series1
Series2
A A t
or
A A ft


0
0
2
sin( )
sin( )


B B t
 
0
sin( )
 
Frequency,, is the
Same but wave from particle
B lags behind A by the
Phase angle 

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
END: Key Definitions

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
Photons can stimulate
Emission just as much
As they can stimulate
Absorption
(idea behind LASERs
Stimulated Emission)
*
o
Stimulated Emission
The rate of stimulated event is described by :
w B N
  *
 Is the energy density of radiation already present at the frequency of
the transition
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
Where w =rate of stimulated emission or absorption
w B No
 
The more perturbing photons the greater the
Stimulated emission
Light Amplification by Stimulated Emission of Radiation
w B No
  w B N
  *


 









8
1
3
3
h
c
h
kT
exp
can be described by the Planck equation for black body radiation at some T

If the populations of * and o are the same the net absorption is zero as a photon is
Absorbed and one is emitted
In order to measure absorption it is required that the
Rate of stimulated absorption is greater than the
Rate of stimulated emission
w B N w B N
absorption o enussion
  
  *
N N
o  *
frequency

Need to get a larger population in the excited state
Compared to the ground state (population inversion)
N N
g
g
N
*0
*
*
 






0
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
Serves to “trap” electrons in the excited
State, which allows for a population
inversion
pump
E

Multiple directions,
Multiple phase lags
Stimulated emission
1. Single phase
2. Along same path
=Constructive Interference
Coherent radiation
Incoherent radiation
Radiation not along the
Path is lost
mirror
mirror
1. Laser
2. FT instrument
about distances
5. Holographic Interference filter.

-1.5
-1
-0.5
0
0.5
1
1.5
0 0.2 0.4 0.6 0.8 1 1.2
Time
Voltage
-1.5
-1
-0.5
0
0.5
1
1.5
0 0.2 0.4 0.6 0.8 1 1.2
Time
Voltage
-1.5
-1
-0.5
0
0.5
1
1.5
0 0.2 0.4 0.6 0.8 1 1.2
Time
Voltage
-1.5
-1
-0.5
0
0.5
1
1.5
0 0.2 0.4 0.6 0.8 1 1.2
Time
Voltage
-1.5
-1
-0.5
0
0.5
1
1.5
0 0.2 0.4 0.6 0.8 1 1.2
Time
Voltage
-1.5
-1
-0.5
0
0.5
1
1.5
0 0.2 0.4 0.6 0.8 1 1.2
Time
Voltage
-1.5
-1
-0.5
0
0.5
1
1.5
0 0.2 0.4 0.6 0.8 1 1.2
Time
Voltage FTIR Instrument
     
Measurement A f t A f t A f t
s s S n S n
  
1 1 2
2 2 2
sin sin .... sin
, , ,
  
1. Laser
2. FT instrument
about distances

-4000
-3000
-2000
-1000
0
1000
2000
3000
4000
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
t (s)
Amplitude
Time Domain: 2 frequencies
0
20000
40000
60000
80000
100000
120000
140000
160000
180000
200000
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
amplitude
0
10000
20000
30000
40000
50000
60000
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
Amplitude
-8000
-6000
-4000
-2000
0
2000
4000
6000
8000
0 0.2 0.4 0.6 0.8 1 1.2
t (s)
Amplitude
0
20000
40000
60000
80000
100000
120000
140000
160000
180000
200000
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
amplitude
1 “beat” cycle
     
   
 
sin sin sin cos
A B A B A B
   
2 1
2
1
2

Moving
mirror
IR source
Beam
splitter
Fixed mirror
B
C
A
detector
Constructive interference occurs when
AC BC n
 
1
2
 

-4000
-3000
-2000
-1000
0
1000
2000
3000
4000
Mirror distance
Light
Intensity
at
Detector
-2 -1 0 +1
 
dis ce f mirror velocity mirror
tan    

INTERFEROGRAMS


 
 
n
mirror
2 mirror mirror
velocity






 
n
mirror mirror
2
f
n
ector
mirror
det  
1 2



Remember that:



c
 f
cn
ector
mirror light
det
2 
Frequency of light

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
cn
ector
mirror light
det
2 

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
s
N
population samples
inite population
measurements

inf
S
N
x x
s
ignal
oise
sample blank
blank


 
S
N
N
s
x x
ignal
oise
measurements
inite population
sample blank
 
inf
S
N
N
ignal
oise
measurements


DIFFRACTION
Huygen’s principle = individual propagating waves combine
to form a new wave front
Can get coherent radiation if the slit is narrow enough.
Coherent = all in one phase
1. Laser
2. FT instrument
about distances

June 19, 2008, Iowa Flood
Katrina Levee break

Fraunhaufer diffraction at a single slit

1 BD OD CD big assume
, , & , / /

2. / /,
if then BCF
  
9. sin sin
 
  
CBF
opposite
hypotenuse
CF
BC
DE
OD

7.  
EOD 
8.
( int )
CF must be n for in
phase constructive erference

10. sin
   






n CF BC
DE
OD
BC
 
4 90
. '
 
OF B o
3 180 90
.         
DEO EOD EOD
 
6 180 90
.         
CFB FBC BCF  
From which we conclude
11.  





 






n
DE
OD
BC
DE
OE
BC

12.  






n
d
L
W

C
B
F
F’
D
E
d
W L


O
5 90
. / /,
if then CFB o
 



The complete equation for a slit is
I I










0
2
sin
 
 

b
2
sin
0
0.2
0.4
0.6
0.8
1
1.2
-800 -600 -400 -200 0 200 400 600 800
Beta
Relative
Intensity
Width of the line depends upon
The slit width!!
Therefore resolution depends
On slit width
Also “see”
This spectra
“leak” of
Our hard won
intensity
B
D
E
d
W L

b=W/2

The base (I=0) occurs whenever
 
  
  
b
n
2
sin
0
0.2
0.4
0.6
0.8
1
1.2
-800 -600 -400 -200 0 200 400 600 800
Beta
Relative
Intensity
I I










0
2
sin sinβ =0
Which occurs when
line width n
b
W
W
   
2 2
2
2 2
2 2
   
sin sin sin
line width
W

2
sin
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

0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5 6
Image position
Output
of
detector
SLIT IMAGE
1 2 3
Image
1 2 3 4 5 Position number
Slit
When edge AB at Detector Sees
Position 1 0% power
Position 2 50% power
Position 5 0 % power
Detector output:
Triangle results when
Effective bandpass = image
To resolve two images that are ∆ apart requires
 
 
 2 eff
eff
Implies want a narrower slit
A
B

Essentially,
Narrow slit widths
Are generally better
 
 
 2 eff

GRATINGS
Gratings Groves/mm
UV/Vis 300/2000
IR 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

 
n CB BD
  
 
sin   
CAB
opp
hyp
CB
d
i q
  90
q
q
q CAB
   90
 
CAB i
sini
CB
d

 
sin   
DAB
opp
hyp
DB
d
r DBA
DBA BAD
DBA BAD
r BAD
  
    
   
 
90
90 180
90
sinr
DB
d

 
n CB BD
  
   
 
n d i d r
  
sin sin
   
 
n d i r
  
sin sin
+ -

EXAMPLE
Calculate  for a grating which has
i=45
2000 groves per mm
1) Get d
1
2000
500
mm
groves
nm
grove

2) Use grating equation to solve for     
 
n d i r
  
sin sin
   
 
n nm r
o
  
500 45
sin sin

Inputs that can be varied are in pink
d nlambda = d(sin+sinr)
grooves/mm 2000 500
i in degrees 45 0.785398 i in radians
r in degrees -40 -20 -10 0 10 20 40
radians -0.69813 -0.34907 -0.17453 0 0.174533 0.349066 0.698132
order n
1 32.15959 182.5433 266.7293 353.5534 440.3775 524.5635 674.9472
2 16.07979 91.27166 133.3647 176.7767 220.1887 262.2817 337.4736
3 10.71986 60.84777 88.90977 117.8511 146.7925 174.8545 224.9824
4 8.039896 45.63583 66.68233 88.38835 110.0944 131.1409 168.7368
5 6.431917 36.50866 53.34586 70.71068 88.0755 104.9127 134.9894
6 5.359931 30.42389 44.45488 58.92557 73.39625 87.42724 112.4912
Multiple wavelengths
Are observed
At a single angle
Of reflection!!
You get light of 674.9 nm
½; 1/3; 1/4; 1/5; etc.
Czerny-Turner
construction
440.3
220.1
146.8
88
73
All come
through

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: 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
Czerny-Turner
construction

440.3
220.1
146.8
88
73
All come
through
Ocean Optics
For fluorescence lab
Monochromator we looked inside
440.3
220.1
146.8
88
73
All come
through
440.3
Only
Hit grating first
Time to get
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
d nlambda = d(sin+sinr)
grooves/mm 2000 500
i in degrees 45 0.785398 i in radians
order 1 2 3 4 5
Reflection Angle
wavelength
100 -30.47131 -17.88496 -6.148561 5.330074 17.03125
125 -27.20057 -11.95286 2.458355 17.03125 32.88081
150 -24.02322 -6.148561 11.12168 29.53092 52.45672
175 -20.92262 -0.407192 20.05324 43.85957 #NUM!
200 -17.88496 5.330074 29.53092 63.23909 #NUM!
225 -14.89846 11.12168 40.0079 #NUM! #NUM!
250 -11.95286 17.03125 52.45672 #NUM! #NUM!
275 -9.039003 23.13464 70.54325 #NUM! #NUM!
300 -6.148561 29.53092 #NUM! #NUM! #NUM!
325 -3.273759 36.36259 #NUM! #NUM! #NUM!
350 -0.407192 43.85957 #NUM! #NUM! #NUM!
375 2.458355 52.45672 #NUM! #NUM! #NUM!
400 5.330074 63.23909 #NUM! #NUM! #NUM!
425 8.215299 83.16511 #NUM! #NUM! #NUM!
450 11.12168 #NUM! #NUM! #NUM! #NUM!
475 14.05736 #NUM! #NUM! #NUM! #NUM!
500 17.03125 #NUM! #NUM! #NUM! #NUM!
Light of 100
Nm shows up
At -30.4
AND
-17.88
And
-6.1
And
5.3
Etc.

GRATING DISPERSION
D-1 = Reciprocal linear dispersion
D
d
dy
dis ce between
dis ce on screen
d r
nF

  
1  
tan
tan
cos
Where n= order
F = focal length
d= distance/groove
D
d
nF
r

 
1
0
POINT = linear dispersion

What is (are) the wavelength(s) transmitted at 45o
reflected AND incident light for a grating of 4000
groves/mm?

RESOLUTION
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
 
R ave




 
 


1 2
2 1
2


1 2
ave

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?
R nF

 
R ave




 
 


1 2
2 1
2


Change in path length results
In phase lag
     
 
E x y E x y ft x y
o
0 0 0
2
, , cos ,
 
 
The photo plate contains all the information
Necessary to give the depth perception when
decoded
1. Laser
2. FT instrument
about distances

Interference Filter
Holographic Notch Filter
1. Laser
2. FT instrument
about distances
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

End
Section on Using Constructive
and Destructive
Interference patterns based on
phase lags
1. Laser
2. FT instrument
about distances

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
12
2
2
88552 10



. x
C
N m
force N
kg m
s


capacitance
 0 Is the permeablity of free space and relates the current
In free space in response to a magnetic field and is defined as
It can be measured directly from capacitor measurements
 
0
7
2
2
4 10



x
N s
C

c 
1
0 0
 
c
x
C
N m
x
N s
C
















 
1
88552 10 4 10
12
2
2
7
2
2
. 
 
0
7
2
2
4 10



x
N s
C
 0
12
2
2
88552 10



. x
C
N m
c
x
s
m
x
s
m
 


1
111 10
1
333485 10
17
2
2
9
. .
c x
m
s
 2 9986 108
.

velocity 
1

c 
1
0 0
 


 


r
velocity
e
c
v
K
  
0 0 0
Dielectric constant
 
~ 0
Typically so
r
elocity
e
c
v
K
 
Maxwell’s relation
This works pretty well for gases (blue line)
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
1.51 4.63E+00 5.04 5.08 8.96E+00
sqrt dielectric
Index
of
refraction
0.9998
0.9999
1
1.0001
1.0002
1.0003
1.0004
1.0005
1.000034 1.000131 1.000294 1.00E+00
sqrt dielectric
Index
of
refraction
Says: refractive index is proportional to the dielectric
constant

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

Angle between incident and scattered
light
0
0.2
0.4
0.6
0.8
1
1.2
0 45 90 135 180 225 270 315 360
Angle of Scattered Light
Relative
Intensity
SCATTERING
Light in
 
I I
r
s o
 






8
1
4 2
4 2
2
 


cos
particle
 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
 
I I
vol molecule
s o









2
4

Most important parameter is the
relationship to wavelength

At sunset the shorter wavelength is
Scattered more efficiently, leaving the
Longer (red) light to be observed
Better sunsets in polluted regions
Long path allow more of the blue
light (short wavelength) to be
scattered
Blue is scattered
Red is observed

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)
 
I I
vol molecule
s o









2
4


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
elocity
c
v
 Refractive index = relative speed of radiation
Refractive index is related to the relative permittivity
(dielectric constant) at that Frequency



2
2
1
1



P
M
m
Where  is the mass density of the sample, M is the molar
mass of the molecules and Pm is the molar polarization
P
N
kT
m
A
o
 






3 3
2


 Where  is the electric dipole moment operator
 is the mean polarizabiltiy





  

2
2
0
2
0
1
1 3 3 3


 





 
N
M kT
N
M
A A


 

2
2
0
1
1 3



N
M
A
Clausius-Mossotti equation
 0 Is the permittivity of free space which describes the
Flux of the electric portion of the wave in vacuum and
Has the value
Point – refractive index
Is related to polarizability

 
2
3
2 2
e R
E

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
2
0
1
1 3



N
M
A




2
2
0
2 2
1
1 3
2
3















N
M
e R
E
A

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

An alternative expression for a single atom is

    
2
2
0
2 2
1
 
 

Ne
m
f
i
o e
j
j j j
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
Transition probability that
Interaction will occur
Molecules per
Unit volume
Each with
J oscillators
If you include the interactions between atoms and ignore absorbance you get




2
2
0
2 2
1
1 3
2
3















N
M
e R
E
A



   
2
2
2
0
2 2
1
2 3





Ne
m
f
o e
j
j j
j




2
2
0
2 2
1
1 3
2
3















N
M
e R
E
A

 
0
2 2
j j

when The refractive index is constant
when  
j j
2
0
2
 The refractive index depends on omega
And the difference  
0
2 2
j j
 Gets smaller so the
Refractive index rises
 r e
c
v
K
 

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
elocity
c
v

  
  


 
 
frequency
vacuum
frequency vacuum
frequency media elocity media
c
c
v


 ,

 
 


r
elocity
frequency vacuum
frequency media
vacuum
media
c
v
  

Filters can be constructed
By judicious combination of the
Principle of constructive and
Destructive interference and
Material of an appropriate refractive
index
t  1
2 '
t  '
t  3
2 '
t
n

2
'



 vacuum
'
t
n vacuum













2


2

t
n
vacuum

t
t
'
t
Wavelength
In media

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
n
vacuum


Holographic filters are better

INTERFERENCE WEDGES
AVAILABLE WEDGES
Vis 400-700 nm
Near IR 1000-2000 nm
IR 2.5 -14.5 m


1
1
2

t
n


2
2
2

t
n


3
3
2

t
n


4
4
2

t
n

The electromagnetic wave can be described in two components, xy, and
Xy - or as two polarizations of light.
Using constructive/destructive interference to select for polarized 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
Or transmitted
T
Air, n=1
Glass
n=1.5
The two polarization components are
reflected and transmitted with
Different amplitudes depending
Upon the angle of reflection, r,
And the angle of transmittence, t
Let’s start by examing
The Angle of transmittence

Snell’s Law
sin
sin




i
t
t
i

i
elocity
c
v

1
1
2
Less dense 1
Lower refractive index
Faster speed of light
More dense 1
Higher refractive index
Slower speed of light
sin
sin
sin
sin






1
2
  
i
t
i
t
t
i
velocity
velocity

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
sin




i
t
t
i


PRISM
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
1.51
1.515
1.52
1.525
1.53
1.535
0 100 200 300 400 500 600 700
wavelength nm
refractive
index
of
crown
glass
POINT, non-linear dispersive device
Reciprocal dispersion will vary with wavelength, since refractive index varies with
wavelength
Uneven spacing = nonlinear

T
Angle of transmittence
Is controlled by
The density of
Polarizable electrons
In the media as
Described by Snell’s Law
The intensity of light (including it’s component polarization) reflected as compared
to transmitted (refracted) can be described by the Fresnel Equations

 
T t
i i
i i t t
 
 











2
2
2 
   
cos
cos cos
 
T t
i i
i t t i
// //
cos
cos cos
 











2
2
2 
   
 
R r
i i t t
i i t t
 
 












2
2
   
   
cos cos
cos cos
 
R r
t i i t
i i t t
// //
cos cos
cos cos
 












2
2
   
   
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
sin




i
t
t
i

Since the total amount of light needs to remain constant we also know that
R T
R T
// //
 
 
 
1
1
Therefore, given the two refractive
Indices and the angle of incidence can
Calculate everything

Consider and air/glass interface
Here the transmitted parallel light is
Zero! – this is how we can select
For polarized light!
This is referred to as the polarization
angle
i
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0 10 20 30 40 50
Angle of incidence
Transmittance
Parallel
Perpendicular

Total Internal Reflection
T
Air, n=1
Glass
n=1.5
Here consider
Light propagating
In the DENSER
Medium and
Hitting a
Boundary with
The lighter
medium

0
0.2
0.4
0.6
0.8
1
1.2
0 20 40 60 80 100
Angle of incidence
Reflectance
and
Transmittence
Perpendicular
Parallel
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

 
R r
T
i i t t
i i t t
 













2
2
   
   
cos cos
cos cos
Set R to 1 &  to 90
The equation can be solved for the critical angle of incidence
t
ic
sin
,
,



c
transmitted less dense
incident dense

For glass/air
 
sin
.
.
sin( . ) .
.
.



c
c
o
a rads
rads
 
 

1
15
0 666
0 666 0 7297
0 7297 180
418
All of the light is reflected
internally

0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0 10 20 30 40 50 60 70 80 90
Angle of Incidence
%
Transmittence
The angle at which the discontinuity occurs:
1. 0% Transmittance=100% Reflectance
2. Total Internal Reflectance
3. Angle = Critical Angle – depends on refractive index
37
42
51
1.69/1
1.5/1
1.3/1
ni/nt Critical Angles
1.697 36.27
1.5 41.8
1.3 50.28

 
NA c incident dense transmitted less dense
  
sin , ,
  
2 2
The critical angle here is defined differently because we have to LAUNCH the
beam
Numerical Aperture

T
i=0
Using Snell’s Law the angle of transmittance is
0 









t
i
t
sin
sin
sin




i
t
t
i

sin
 
1
0 0
 t
Shining light directly through our sample
 
R r
T
i i t t
i i t t
 













2
2
   
   
cos cos
cos cos
 
R r
t i i t
i i t t
// //
cos cos
cos cos
 












2
2
   
   
cos0 1


 
R r
T
i t
i t
 









2
2
 
 
 
R r
t i
i t
// //
 








2
2
 
 
cos0 1

same
R
t i
i t
// 








 
 
2
The amount of light reflected depends
Upon the refractive indices of the medium

R
I
I
reflected
initial
t i
i t
//  








 
 
2
I I I
transmitted initial reflected
 
I I
reflected initial
t i
i t









 
 
2
I I I
transmitted initial initial
t i
i t
 








 
 
2
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?
I I
transmitted initial
t i
i t
 
















1
2
 
 

I I I
t o t
'' ''
. .
. .
'
. .
. .
 
















 
















1
15 133
15 133
1
15 133
15 133
2 2
I I
t o
 
















1
15 1
15 1
2
.
.
I I I
t o t
' '
. .
. .
. .
. .
 
















 
















1
133 15
133 15
1
133 15
133 15
2 2
Final exiting light
I I I
t o t
''' '''
.
.
.
.
''
 
















 
















1
15 1
15 1
1
15 1
15 1
2 2
Io It=I’o
It’ = I’’o
It’’ =
I’’’o
It’’’
Air
Glass, refractive index 1.5
Air, refractive index 1
Water,
refractive index
1.33
I I
t o
'
.
.
. .
. .
 

































1
15 1
15 1
1
133 15
133 15
2 2
I I
t o
''
.
.
. .
. .
 

































1
15 1
15 1
1
133 15
133 15
2 2 2
I I
t o
'''
.
.
. .
. .
 

































1
15 1
15 1
1
133 15
133 15
2 2 2 2

I I
glass air initial
/
.
.
.
.
2
2 2 2 2
1
05
2 5
1
017
2 83
 





























   
I I
glass air initial
/ . .
2
2 2
0 96 0 99

I I
glass air initial
/ .
2 0915

We lose nearly 10% of the light
I I
t o
'''
.
.
. .
. .
 

































1
15 1
15 1
1
133 15
133 15
2 2 2 2

Key Concepts
Light Scattering
I I
vol
s o







2
4

Refractive Index
Is wavelength dependent
Used to separate light by prisms
r
elocity
c
v


   
2
2
2
0
2 2
1
2 3





Ne
m
f
o e
j
j j
j




2
2
0
2 2
1
1 3
2
3















N
M
e R
E
A

2

t
n

Refractive index based
Interference filters

Key Concepts
Snell’s Law
sin
sin




i
t
t
i

Describes how light is bent based differing refractive indices
 
T t
i i
i i t t
 
 











2
2
2 
   
cos
cos cos
 
T t
i i
i t t i
// //
cos
cos cos
 











2
2
2 
   
 
R r
i i t t
i i t t
 
 












2
2
   
   
cos cos
cos cos
 
R r
t i i t
i i t t
// //
cos cos
cos cos
 












2
2
   
   
Fresnell’s Equations describe how polarized light is transmitted and/or reflected
at an interface
Used to create surfaces which select for polarized light

Key Concepts
Fresnell’s Laws collapse to
sin
,
,



c
transmitted less dense
incident dense

Which describes when you will get total internal reflection (fiber optics)
R
I
I
reflected
initial
t i
i t
//  








 
 
2
And
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 h
m  
 
Energy of
Ejected
electron
Frequency of impinging photon
(related to photon energy)
Work function=minimum energy binding an
Electron in the metal

KE h
m  
 
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
S kP k
electrical signal radiant power d
 
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

1. Capture all simultaneously
= multiplex advantage
2. Generally less sensitive
Rock to
Get different
wavelengths
Very sensitive detector

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

Photodiodes
Photomultiplier tube
The AA 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
+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”
1. Set device to accumulate
charge for some period of
time. (increase sensitivity)
2. Charge accumulated near
electrode
Requires special cooling, Why?
The fluorescence
experiment

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
o
r
r




3
1
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

-200
-100
0
100
200
300
400
500
600
700
800
900
65
-150
-135
-120
-105
-90
-75
-60
-45
-30
-15
0
15
30
45
60
75
90
105
120
135
150
165
180
Grating
2nd order
1st order
Angle of
reflection
i=45

-200
-100
0
100
200
300
400
500
600
700
800
900
65
-150
-135
-120
-105
-90
-75
-60
-45
-30
-15
0
15
30
45
60
75
90
105
120
135
150
165
180
2nd order
1st order
Angle of
reflection
i=45

6. Really Basic Optics.ppt

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6. Really Basic Optics.ppt