what is the resolution of a grating in the first order of 4000 groves/mm if 1 cm of the grating is...
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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 resultsIn phase lag
E x y E x y ft x yo0 0 02, , co s ,
The photo plate contains all the informationNecessary to give the depth perception whendecoded
Constructive/Destructive interference 1. Laser2. FT instrument3. Can be used to select wavelengths4. Can be used to obtain information
about distances5. Holographic Interference filter.
Interference Filter
Holographic Notch Filter
Constructive/Destructive interference 1. Laser2. FT instrument3. Can be used to obtain information
about distances4. Interference filter. 5. Can be used to select wavelengths
Can create a filter usingThe holographic principle To create a series of Groves on the surfaceOf the filter. The groovesAre very nearly perfectIn spacing
End Section on Using Constructive and DestructiveInterference patterns based on phase lags
Constructive/Destructive interference 1. Laser2. FT instrument3. Can be used to obtain information
about distances4. Interference filter. 5. Can be used to select wavelengths
Begin SectionInteraction 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 light3. 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 theFlux of the electric portion of the wave in vacuum andHas the value
012
2
28 8552 10
. xC
N m
force Nkg m
s
capacitance
0 Is the permeablity of free space and relates the currentIn free space in response to a magnetic field and is defined as
It can be measured directly from capacitor measurements
07
2
24 10x
N s
C
c 1
0 0
c
xC
N mx
N s
C
1
8 8552 10 4 10122
27
2
2.
07
2
24 10x
N s
C 0
122
28 8552 10
. xC
N m
c
xs
mx
s
m
1
1 11 10
1
3 33485 10172
2
9. .
c xm
s 2 9986 10 8.
v elocity 1
c 1
0 0
r
ve loc itye
c
vK
0 0 0
Dielectric constant
~ 0Typically so
re locity
e
c
vK
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
Ind
ex o
f re
frac
tio
n
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
Ind
ex o
f re
frac
tio
n
Says: refractive index is proportional to the dielectricconstant
Our image is of electrons perturbed by an electromagnetic field which causesThe 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 electromagneticRadiation at same frequency
Angle between incident and scatteredlight
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
Rel
ativ
e In
ten
sity
SCATTERING
Light in
I Irs o
81
4 2
4 22
cos
particle
Polarizability of electronsa) Number of electronsb) Bond lengthc) Volume of the molecule,
which depends upon the radius, r
= vacuum Io = incident intensity
I I
vo l m olecu les o
2
4
Most important parameter is the relationship to wavelength
At sunset the shorter wavelength is Scattered more efficiently, leaving theLonger (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
vo l m olecu les o
2
4
Our image is of electrons perturbed by an electromagnetic field which causesThe change in permittivity and permeability – and therefore, the speed of thePropagating electromagnetic wave.
It follows that the index of refraction should be related to the ability to Polarize the electron cloud
re locity
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
Mm Where is the mass density of the sample, M is the molar
mass of the molecules and Pm is the molar polarization
PN
kTmA
o
3 3
2
Where is the electric dipole moment operator
is the mean polarizabiltiy
2
20
2
0
1
1 3 3 3
N
M kT
N
MA A
2
20
1
1 3
N
MA
Clausius-Mossotti equation
0 Is the permittivity of free space which describes theFlux of the electric portion of the wave in vacuum andHas the value
Point – refractive indexIs related to polarizability
2
3
2 2e 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
20
1
1 3
N
MA
2
20
2 21
1 3
2
3
N
M
e R
EA
The change in the velocity of the electromagnetic radiation is a function of1.mass density (total number of possible interactions)2. the charge on the electron3. 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
22
02 21 Ne
m
f
io e
j
j j jj
A damping force term that account for Absorbance (related to delta E in priorExpression) Natural
Frequency ofThe oscillating electronsIn the single atom j
Frequency of incoming electromagneticwave
Transition probability thatInteraction will occur
Molecules perUnit volumeEach with J oscillators
If you include the interactions between atoms and ignore absorbance you get
2
20
2 21
1 3
2
3
N
M
e R
EA
2
2
2
02 2
1
2 3
Ne
m
f
o e
j
j jj
2
20
2 21
1 3
2
3
N
M
e R
EA
02 2j jwhen The refractive index is constant
when j j2
02 The refractive index depends on omega
And the difference 02 2j j Gets smaller so the
Refractive index rises r e
c
vK
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?
re locity
c
v
frequencyvacuum
frequency vacuum
frequency m ed ia eloc ity m ed ia
c
c
v
,
r
e locity
frequency vacuum
frequency m ed ia
vacuum
m edia
c
v
Filters can be constructedBy judicious combination of the Principle of constructive andDestructive interference andMaterial of an appropriate refractiveindext 1
2 '
t '
t 32 '
tn
2'
vacuum
'
tn 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 nmNear IR 1000-2000 nmIR 2.5 -14.5 m
112
t
n
222
t
n
332
t
n
442
t
n
The electromagnetic wave can be described in two components, xy, andXy - or as two polarizations of light.
Using constructive/destructive interference to select for polarized light
Refraction, Reflection, and Transmittance DefinedRelationship to polarization
The amplitude of the spherically oscillating electromagneticWave can be described mathematically by two componentsThe 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 toThe plane
It can be reflected
Or transmitted
T
R
Air, n=1
Glassn=1.5The two polarization components are
reflected and transmitted with Different amplitudes dependingUpon the angle of reflection, r,And the angle of transmittence, t
Let’s start by examingThe Angle of transmittence
Snell’s Law
sin
sin
i
t
t
i
ie loc ity
c
v
1
1Time=0
Time=1
Time=2
Time=3
Time=4
Time=0
Time=1
Time=2
Time=3
Time=4
2
Less dense 1Lower refractive indexFaster speed of light
More dense 1Higher refractive indexSlower 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 1.532450 nm 1.528550 nm 1.519590 nm 1.517620 nm 1.514650 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
refr
acti
ve i
nd
ex o
f cr
ow
n g
lass
POINT, non-linear dispersive deviceReciprocal dispersion will vary with wavelength, since refractive index varies with wavelength
Uneven spacing = nonlinear
T
R
Angle of transmittenceIs controlled byThe density of Polarizable electronsIn the media asDescribed 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 ti i
i i t t
2
22
cos
co s co s
T ti i
i t t i/ / / /
cos
cos cos
2
22
R ri i t t
i i t t
2
2 cos cos
cos cos
R rt i i t
i i t t/ / / /
co s co s
co s co s
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
1Therefore, given the two refractiveIndices and the angle of incidence canCalculate everything
Consider and air/glass interface
Here the transmitted parallel light isZero! – this is how we can selectFor polarized light!
This is referred to as the polarizationangle
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
Tra
nsm
itta
nce
Parallel
Perpendicular
Total Internal Reflection
T
R
Air, n=1
Glassn=1.5
Here considerLight propagatingIn the DENSERMedium andHitting aBoundary withThe lighter medium
0
0.2
0.4
0.6
0.8
1
1.2
0 20 40 60 80 100
Angle of incidence
Ref
lect
ance
an
d T
ran
smit
ten
ce
Perpendicular
Parallel
Same calculation but made the indicident medium denser so that wave isPropagating inside glass and is reflected at the air interface
Discontinuity at 42o signalsSomething unusual is happening
R rT
i i t t
i i t t
2
2
cos co s
co s co sSet R to 1 & to 90The equation can be solved for the critical angle of incidence
t
ic
sin,
,
c
transm itted less dense
inciden t dense
For glass/air
sin.
.
s in ( . ) .
..
c
c
o
a rads
rads
1
1 50 666
0 666 0 7297
0 7297 18041 8
All of the light is reflectedinternally
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
% T
ran
smit
ten
ceThe angle at which the discontinuity occurs:1. 0% Transmittance=100% Reflectance2. Total Internal Reflectance3. Angle = Critical Angle – depends on refractive index
3742
51
1.69/1
1.5/1
1.3/1
ni/nt Critical Angles
1.697 36.271.5 41.81.3 50.28
NA c inciden t dense transm itted less dense sin , , 2 2
The critical angle here is defined differently because we have to LAUNCH the beam
Numerical Aperture
R=?
T
i=0
Using Snell’s Law the angle of transmittance is
0
t
itsin
sin
sin
i
t
t
i
sin 1 0 0 t
Shining light directly through our sample
R rT
i i t t
i i t t
2
2
cos co s
co s co s
R rt i i t
i i t t/ / / /
co s co s
co s co s
2
2
cos 0 1
R rTi t
i t
22
R r t i
i t/ / / /
22
cos 0 1
same
R t i
i t/ /
2
The amount of light reflected depends Upon the refractive indices of the medium
RI
Ire flec ted
in itia l
t i
i t/ /
2
I I Itransm itted in itia l re flec ted
I Ire flec ted in itia lt i
i t
2
I I Itransm itted in itia l in itia lt 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 Itransm itted in itia lt i
i t
12
I I It o t' ' ' '. .
. .'
. .
. .
11 5 1 33
1 5 1 331
1 5 1 33
1 5 1 33
2 2
I It o
11 5 1
1 5 1
2.
.
I I It o t' '. .
. .
. .
. .
11 33 1 5
1 33 1 51
1 33 1 5
1 33 1 5
2 2
Final exiting light
I I It o t' ' ' ' ' '.
.
.
.' '
11 5 1
1 5 11
1 5 1
1 5 1
2 2
Io It=I’oIt’ = I’’o
It’’ = I’’’o
It’’’
AirGlass, refractive index 1.5
Air, refractive index 1
Water, refractive index 1.33
I It o'.
.
. .
. .
11 5 1
1 5 11
1 33 1 5
1 33 1 5
2 2
I It o' '.
.
. .
. .
11 5 1
1 5 11
1 33 1 5
1 33 1 5
2 2 2
I It o' ' '.
.
. .
. .
11 5 1
1 5 11
1 33 1 5
1 33 1 5
2 2 2 2
I Ig lass a ir in itia l/
.
.
.
.2
2 2 2 2
10 5
2 51
0 17
2 83
I Ig lass a ir in itia l/ . .2
2 20 96 0 99
I Ig lass a ir in itia l/ .2 0 915
We lose nearly 10% of the light
I It o' ' '.
.
. .
. .
11 5 1
1 5 11
1 33 1 5
1 33 1 5
2 2 2 2
Key Concepts
Interaction with Matter
Light Scattering I I
vo ls o
2
4
Refractive IndexIs wavelength dependentUsed to separate light by prisms
re loc ity
c
v
2
2
2
02 2
1
2 3
Ne
m
f
o e
j
j jj
2
20
2 21
1 3
2
3
N
M
e R
EA
2 t
nRefractive index based
Interference filters
Key Concepts
Interaction with Matter
Snell’s Lawsin
sin
i
t
t
i
Describes how light is bent based differing refractive indices
T ti i
i i t t
2
22
cos
co s co s
T ti i
i t t i/ / / /
cos
cos cos
2
22
R ri i t t
i i t t
2
2 cos cos
cos cos
R rt i i t
i i t t/ / / /
co s co s
co s co s
2
2
Fresnell’s Equations describe how polarized light is transmitted and/or reflectedat an interface
Used to create surfaces which select for polarized light
Key Concepts
Interaction with Matter
Fresnell’s Laws collapse to
sin,
,
c
transm itted less dense
inciden t dense
Which describes when you will get total internal reflection (fiber optics)
RI
Ire flec ted
in itia l
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 voltage4. 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 hm
Energy ofEjectedelectron
Frequency of impinging photon(related to photon energy)
Work function=minimum energy binding an Electron in the metal
KE hm
To convert photons to electrons that we can measure with an electrical circuit useA metal foil with a low work function (binding energy of electrons)
DETECTORSIdeal Properties1. High sensitivity2. Large S/N3. Constant parameters with wavelength
S kP kelec trica l signa l rad ian t pow er d
Where k is some large constantkd is the dark current
Classes of Detectors
Name commentPhotoemissive single photon eventsPhotoconductive “ (UV, Vis, near IR)
Heat average photon flux
Want low dark current
1. Capture all simultaneously = multiplex advantage2. Generally less sensitive
Rock toGet different wavelengths
Very sensitive detector
Sensitivity of photoemissiveSurface is variable
Ga/As is a good oneAs it is more or less consistentOver 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 acrossa 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 voltage4. Move charge to “gate” And Count, 5. move next “bin” of charge and keep on counting6. 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 fluorescenceexperiment
END6. Really Basic Optics
Since polarizability of the electrons in the material also controls the dielectricConstant 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 dependentJust as the refractive index isTypically reported
Point of this slide: polarizability of electrons in a molecule is related to theRelative dielectric constant
-200
-100
0
100
200
300
400
500
600
700
800
900-165
-150
-135
-120
-105
-90
-75
-60
-45
-30
-15015
30
45
60
75
90
105
120
135
150
165180
Grating
2nd order
1st order
Angle ofreflection
i=45
-200
-100
0
100
200
300
400
500
600
700
800
900-165
-150
-135
-120
-105
-90
-75
-60
-45
-30
-15015
30
45
60
75
90
105
120
135
150
165180
2nd order
1st order
Angle ofreflection
i=45