Download 3. Lab 1 Lecture Refractive index

Phys 570
Lab 1
Measurement of refractive index by
Brewster’s Angle method
Fresnel’s equations and Brewster’s angle
Electromagnetic wave
Linearly Polarized light
p-polarized light
E
E
Electric vector E
S-polarized light
S-polarized light
Perpendicular or vertical
p-polarized light
Horizontal
Laser Light – Linearly polarized.
45 degrees
Resultant
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Laser light is linearly polarized.
Use 45 degree polarizer in front of the laser light.
This make the relative intensities of the s and p components
the same, the light is polarized at 45 degrees.
Laser light – Linearly polarized
450 polarizer
Incident light
Reflected light
qi qr
S-polarization , coming out of page
qt
P-polarization , parallel to page
Transmitted light
Brewster’s angle setup
Photodetector
Analyzer
Sample
Slit
450 Polarizer
Laser
Pbo 50:50
Angle
75
70
65
64
63
62
61
60
59
58
57
56
55
50
45
40
Total Light Polarized light
4.54
35
2.51
40.4
0.52
26
0.34
20.6
0.26
19.7
0.23
19.1
0.15
12.9
0.15
14.5
0.16
14.9
0.15
13.1
0.18
12.9
0.21
12.3
0.24
11.7
0.37
6.75
0.59
5.68
0.87
5.32
Ratio= Polarized light/Total light
0.129714286
0.062128713
0.02
0.016504854
0.01319797
0.012041885
0.011627907
0.010344828
0.010738255
0.011450382
0.013953488
0.017073171
0.020512821
0.054814815
0.103873239
0.163533835
14
y = 0.0508x2 - 6.0695x + 182.09
R² = 0.9976
12
Ratio%
10
8
Series1
6
Poly. (Series1)
4
2
0
40
50
60
Angle (θ)
70
80
Ratio%
12.971
6.2129
2
1.6505
1.3198
1.2042
1.1628
1.0345
1.0738
1.145
1.3953
1.7073
2.0513
5.4815
10.387
16.353
Definitions: Planes of Incidence and the
Interface and the polarizations
Perpendicular (“S”)
Incident medium
polarization sticks out X
of or into
the plane of
incidence.
ki
polarization lies parallel
to the plane of
incidence.
ni
qi qr
Interface
Plane of the interface (here the yz
plane)y(perpendicular to page)
qt
z
Parallel (“P”)
Er
Ei
Plane of incidence
(here the xy plane) is the
plane that contains the
incident and reflected kvectors.
kr
nt
x
Et
kt
Transmitting medium
Fresnel’s Equations
ki
kr
Ei
Fraction of a light wave
reflected and transmitted by a
flat interface between two
media with different refractive
indices.
for the
perpendicular
polarization
qi qr
Plane of the interface (here
the yzy plane) (perpendicular
qt
to page)
x
z
Et k t
s-polarized
Eot
Eor
t 
r 
Eoi
Eoi
Er
ni ni
nt
nt
p-polarized
Eor
r 
Eoi
Eot
t 
Eoi
for the
parallel
polarization
where Eoi, Eor, and Eot are the field complex amplitudes.
We consider the boundary conditions at the interface for the electric and
magnetic fields of the light waves.
We’ll do the perpendicular polarization (s-polarized) first.
s-polarized
Eot
Eor
t 
r 
Eoi
Eoi
for the
perpendicular
polarization
Perpendicular polarization (s)
Boundary Condition for the Electric Field at an Interface
The Tangential Electric Field is Continuous
In other words:
The total E-field in
the plane of the
interface is
continuous.
Here, all E-fields are
in the z-direction,
which is in the plane
of the interface (xz)
ki
kr
Er
Ei
ni
qi qr
Br
Bi
Plane of the interface (here the yz
plane)y(perpendicular to page)
qt
z
x
Bt
Eoi  Eor  Eot
Et
nt
kt
Perpendicular polarization (s)
Boundary Condition for the Magnetic field at an Interface
The Tangential Magnetic Field is Continuous
ki
In other words:
kr
Er
Ei
The total B-field in
the plane of the
interface is
continuous.
ni
qi qr
Br
Bi
Plane of the interface (here the yz
plane)y(perpendicular to page)
qt
Here, all B-fields are
in the xy-plane, so
we take the
x-components:
x
z
Et
kt
Bt
 Boi
m
cos qi 
Bor
m
cos q r 
we're using non magnetic material m 1
nt
 Bot
m
cos qt
Reflection & Transmission Coefficients for Perpendicularly
(s) Polarized Light
E0i  E0 r  E0t
 B0i cos(qi )  B0 r cos(q r )   B0t cos(qt )
But B  nE and q r  qi :
ni ( E0r  E0i ) cos(qi )  nt E0t cos(qt )
E0i  E0 r  E0t
ni ( E0 r  E0i ) cos(qi )  nt E0t cos(qt )
A
Reflection & Transmission Coefficients for Perpendicularly
(s) Polarized Light
Solving for
E0 r
in Equation A yields the reflection coefficient :
E0i
E0 r  ni cos(q i )  nt cos(qt ) 
r 

E0i
 ni cos(qi )  nt cos(qt ) 
Analogously, the transmission coefficient,
Lab report
E0t
, is
E0i
E0t
2ni cos(qi )
t 

E0i  ni cos(qi )  nt cos(qt ) 
These equations are called the Fresnel Equations for
perpendicularly polarized light.
Lab report
Parallel polarization (p)
Boundary Conditions for the Electric Field and magnetic field at an
Interface
ki
Bi
kr
Br
ni
qi qr
Ei
Plane of the interface (here the yz
plane)y(perpendicular to page)
Er
qt
z
Bt
x
Et
nt
kt
 E0i cos(qi )  E0 r cos(q r )   E0t cos(qt )
B0i  B0 r  B0t
Reflection & Transmission Coefficients for Parallel (P) Polarized Light
But B  nE and q r  qi :
ni ( E0r  E0i ) cos(qi )  nt E0t cos(qt )
ni ( E0i  E0 r )  nt E0t
( E0 r  E0i ) cos(qi )   E0t cos(qt )
B
Solving for E0 r / E0i in Equation B yields the reflection coefficient :
E0 r  ni cos(qt )  nt cos(qi ) 
r|| 

E0i  ni cos(qt )  nt cos(qi ) 
Lab report
Analogously, the transmission coefficient, E0t / E0i , is
E0t
2ni cos(qi )
t|| 

E0i  ni cos(qt )  nt cos(qi ) 
Lab report
These equations are called the Fresnel Equations for parallel polarized light.
Reflection Coefficients for an Air-to-Glass Interface
E0 r  ni cos(qt )  nt cos(qi )
r|| 

E0i  ni cos(qt )  nt cos(qi )
tan(qi  qt )
r 
tan(qi  qt )
sin(qi  qt )
E0 r  ni cos(qi )  nt cos(qt )
r  
r 

sin(qi  qt )
E0i  ni cos(qi )  nt cos(qt )
Lab report
Reflectance
Reflectance (Intensity reflection coefficient ) is the square of the amplitude of
reflection coefficient
R = r2
2
tan
(qi  qt )
2
R r 
tan 2 (qi  qt )
2
sin
(qi  qt )
2
R  r  2
sin (qi  qt )
qi  qt  900
15%
4%
qi  q B  Brewster's angle
R  r 2  0 R  r2  sin 2 (qi  qt )
nt  tan q B
Lab report
Fluorescence of rare earth Sm3+ (Samarium)
ions in borate glasses.
Schematic representation of luminescence
Vibrational relaxation
Excited state
Triplet State
Ground state
Absorption
Fluorescence
Phosphorescence
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upport/tutorials.html
Your ion
Electronic Configuration of neutral rare earth atoms
Number
Element
Symbol
Electronic configuration
54
Xenon
Xe
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6
58
Cerium
Ce
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f1 5d1
59
Praseodymium
Pr
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f3
60
Neodymium
Nd
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f4
61
Promethium
Pm
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f5
62
Samarium
Sm
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f6
63
Europium
Eu
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f7
64
Gadolinium
Gd
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f7 5d1
65
Terbium
Tb
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f9
66
Dysprosium
Dy
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f10
67
Holmium
Ho
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f11
68
Erbium
Er
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f12
69
Thulium
Tm
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f13
70
Ytterbium
Yb
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14
71
Lutetium
Lu
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d1
Trivalent rare earth ion – remove 3 electrons
Atomic Term Symbols
Free ions can be described by Term Symbols:
2S + 1L
J
Ground-State Term Symbols for Any Electronic Configuration:
The Ground State has:
1. Choose Maximum S value
2. Choose Maximum L value
3. Choose J value:
(a) smallest J value if orbital sub-shell is J = L-S
less than half-filled
(b) largest J value if orbital sub-shell is J = L+S
greater than half-filled
22
Free ions can be described by Term Symbols:
2S + 1L
J
where:
• 2S + 1 is spin multiplicity
• L is Total orbital angular momentum
• J is the spin, orbit (L + S) coupling
Multiplicity:
2S + 1 is calculated from number of unpaired
electrons
S = 0 (all paired), S = 1/2 (one unpaired) , S = 1 (2 unpaired) , S = 3/2 (3 unpaired) ,
etc.
If total L = (Ltotal = l1 + l2 + … for each electron)
L = 0 -> S symbol
L = 1 -> P symbol
L = 2 -> D symbol
L = 3 -> F symbol
Russell-Saunders Coupling:
J ranges from L + S, …, L-S and all integers in between
23
Atomic Term Symbols
2S + 1L
J
Carbon atom with electronic configuration 1s 2 2 s 2 2 p 2
G!
6!
 15 states
N !(G  N )! 2!(6  2)!
ml= 0
ml= 0 ml= 1
0
-1
L  (l1  l2 )...........(l1  l2 )
L  2,1, 0
S  ( s1  s2 ).........( s1  s2 )
Ground state of Carbon
3
P0
S  0,1
Since both electrons are in the same subshell p, they have to satisfy Pauli's exclusion principle.
L
S
(2L+1)=M L and (2S+1)=M S
2
0
(2L+1)(2S+1)  5 1  5 states = 1 D = 1 D 2
1
1
(2L+1)(2S+1)  3  3 = 9 states = 3 P = 3 P2 , 3 P1 , 3 P0
0
0
(2L+1)(2S+1)  11 = 1state = 1S = 1S0
24
1
D2
J 2
M J  2, 1, 0, 1, 2
3
3
P2
J 2
M J  2, 1, 0, 1, 2
5
5 states
3
P1
J 1
M J  1, 0, 1
+
3
P0
J 0
MJ  0
+
1 = 9 states
1
S0
J 0
MJ  0
1 state
Total = 5+9+1 = 15 states
25
15 states
1
S0
Energy levels of Carbon atom
1
1
S
2
2s 2 p
1
2
D
3
H0
Central field
Hamiltonian
D2
3
P2
P
HC
3
Coulomb Interaction
Hamiltonian
3
P1
MJ
0
+2
+1
0
-1
-2
+2
+1
0
-1
-2
+1
0
-1
P0
HS-O Spin-orbit Interaction Hamiltonian 0
Hund’s Rules
1
1
S0
D2
3
P2
3
P1
3
P0
1.State with the largest value of S is most
stable and stability decreases with
decreasing S.
2.For states with same values of S, the
state with the largest value of L is the most
stable.
3.If states have same values of L and S
then, for a subshell that is less than half
filled, state with smallest J =L-S is most
stable; for subshells that are more than
half filled, state with largest value of J = L+S
is most stable.
27
Atomic Term Symbols
Praseodymium
Praseodymium neutral atom with electronic configuration
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f3
Praseodymium trivalent ion with electronic configuration
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 4f2
3
4f
2
1
0
-1
2
-2
-3
Ground state of trivalent
Praseodymium ion
3
G!
N !(G  N )!
H4
14!
Total States :
 91 states
2!(14  2)!
L  l1  l2 ...........l1  l2
l1  3 and l2  3
L  6, 5, 4
S  s1  s2 .........s1  s2
S  0,1
28
Trivalent Praseodymium ion
1
S0
3
P2
L
S
(2L+1)=M L and (2S+1)=M S
1
6
0
(2L+1)(2S+1)  13 1  13 states = 1 I = 1I 6
3
P1
1
(2L+1)(2S+1)  11 3  33 states = H = H 6 , H 5 , H 4
3
P0
0
(2L+1)(2S+1) = 9 1 = 9 states = G = G4
1
(2L+1)(2S+1)  7  3 = 21 states = F = F4 , F3 , F2
3
2
0
(2L+1)(2S+1)  5 1 = 5states = D = D2
F4
3
1
1
(2L+1)(2S+1) = 3  3 = 9 states = 3 P = 3 P2 , 3 P1 , 3 P0
F3
3
F2
0
0
(2L+1)(2S+1) = 11 =1 state = 1S = 1S0
3
H6
3
H5
3
H4
5
4
3
3
3
1
3
3
1
3
1
Total = 91 states
3
1
I6
D2
1
3
3
G4
1
29
Trivalent Praseodymium ion Energy levels
Trivalent Samarium ion Energy levels
62
Samarium
Excitation
Sm
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f6
Emission 4I13/2
Fluorescence
4
I
4 11/2
M
15/2
20
4
G5/2
4
G
5/2
707 nm
598 nm
12
646 nm
565 nm
6
F
11/2
6
F
9/2
6
F
6 7/2
F
5/2
6 6 6
F H F
5/2 15/2 3/2
6
H
11/2
8
4
6
H
488 nm
9/2
6
H
7/2
6
H
5/2
H7/2
598 nm
Intensity (a.u.)
Energy (103 cm-1)
16
6
4
6
G5/2
4
G5/2
H9/2
646 nm
6
H5/2
565 nm
4
G5/2
6
H11/2
707 nm
500
550
600
650
700
750
800
Wavelength (nm)
Write all the term symbols for Samarium
ion in your lab report