Phys 570. Lab 1: Refractive index and fluorescence of rare earth ions in borate glasses

Transcription

1 Phys 570 Lab 1: Refractive index and fluorescence of rare earth ions in borate glasses

2 Measurement of refractive index by Brewster s Angle method Fresnel s equations and Brewster s angle

3 Electromagnetic wave Linearly Polarized light p-polarized light E E Electric vector E S-polarized light

4 S-polarized light Perpendicular or vertical p-polarized light Horizontal Laser Light Linearly polarized. 45 degrees Resultant

5 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.

6 Laser light Linearly polarized 45 0 polarizer Incident light Reflected light q i q r q t S-polarization, coming out of page P-polarization, parallel to page Transmitted light

7 Brewster s angle setup Photodetector Analyzer Sample Slit 45 0 Polarizer Laser

8 Ratio% Pbo 50:50 Angle Polaized Light Total light Ratio= Polarized light/total light Ratio% y = x x R² = Series1 Poly. (Series1) Angle (θ)

9 Definitions: Planes of Incidence and the Interface and the polarizations Perpendicular ( S ) polarization sticks out X of or into the plane of incidence. ki Incident medium kr Plane of incidence (here the xy plane) is the plane that contains the incident and reflected k- vectors. E i B i q i q r Interface Plane of the interface (here the yz plane) y (perpendicular to page) q t E r B r n i Parallel ( P ) polarization lies parallel to the plane of incidence. z x E t B t Transmitting medium k t n t

10 Fresnel s Equations Fraction of a light wave reflected and transmitted by a flat interface between two media with different refractive indices. r E E or oi s-polarized t E E ot oi for the perpendicular polarization r q i q r Plane of the interface (here the yz y plane) (perpendicular to page) q t z x E E ki or oi E i E r E t kr k t p-polarized t E E ot oi n i n t n i n t for the parallel polarization where E oi, E or, and E ot 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.

11 r E E or oi s-polarized t E E ot oi 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: ki kr The total E-field in the plane of the interface is continuous. E i E r q i q B r i Plane of the interface (here the yz plane) y (perpendicular to page) B r n i q t Here, all E-fields are in the z-direction, which is in the plane of the interface (xz) z x B t E t k t n t E E E oi or ot

12 Boundary Condition for the Magnetic field at an Interface In other words: The total B-field in the plane of the interface is continuous. Here, all B-fields are in the xy-plane, so we take the x-components: Perpendicular polarization (s) The Tangential Magnetic Field is Continuous ki q i q B r i Plane of the interface (here the yz plane) y (perpendicular to page) z we're using non magnetic material m 1 x Boi Bor Bot cosqi cosqr cosqt m m m E i q t B t E r E t kr k t B r n i n t

13 Reflection & Transmission Coefficients for Perpendicularly (s) Polarized Light E E E 0i 0r 0t B cos( q ) B cos( q ) B cos( q ) 0i i 0r r 0t t But B ne and q q : r i n ( E E )cos( q ) n E cos( q ) i 0r 0i i t 0t t E E E 0i 0r 0t n ( E E )cos( q ) n E cos( q ) i 0r 0i i t 0t t A

14 Reflection & Transmission Coefficients for Perpendicularly (s) Polarized Light r E E 0r Solving for in Equation A yields the reflection coefficient : 0i E n cos( ) cos( ) 0 i qi n r t qt E n cos( q ) n cos( q ) 0i i i t t Lab report E Analogously, the transmission coefficient, E E0t ni cos( i ) t E q n cos( q ) n cos( q ) 0i i i t t 0t 0i, is Lab report These equations are called the Fresnel Equations perpendicularly polarized light. for

15 Parallel polarization (p) Boundary Conditions for the Electric Field and magnetic field at an Interface ki B i B r kr n i E i q i q r Plane of the interface (here the yz plane) y (perpendicular to page) q t E r z x B t n t E t k t E cos( q ) E cos( q ) E cos( q ) 0i i 0r r 0t t B B B 0i 0r 0t

16 Reflection & Transmission Coefficients for Parallel (P) Polarized Light But B ne and q q : r i n ( E E )cos( q ) n E cos( q ) i 0r 0i i t 0t t n ( E E ) n E i 0i 0r t 0t ( E E )cos( q ) E cos( q ) 0r 0i i 0t t B Solving for E / E in Equation B yields the reflection coefficient : 0r 0i r E n cos( ) cos( ) 0 i qt n r t qi E n cos( q ) n cos( q ) 0i i t t i Analogously, the transmission coefficient, E t / E 0t 0i E0t ni cos( qi ) E n cos( q ) n cos( q ) 0i i t t i, is Lab report Lab report These equations are called the Fresnel Equations for parallel polarized light.

17 Reflection Coefficients for an Air-to-Glass Interface r E n cos( ) cos( ) 0 i qt n r t qi E n cos( q ) n cos( q ) 0i i t t i r tan( q q ) i tan( q q ) i t t r E n cos( ) cos( ) 0 i qi n r t qt E n cos( q ) n cos( q ) 0i i i t t r sin( q q ) i sin( q q ) i t t Lab report

18 Reflectance (Intensity reflection coefficient ) is the square of the amplitude of reflection coefficient R = r R r tan ( qi qt) tan ( qi qt) Reflectance R r i q q sin ( qi qt) sin ( qi qt) t % 15% q q i B R Brewster's angle r 0 R r sin ( q q ) i t n t tanq B Lab report

19 Fluorescence of rare earth Sm + (Samarium) ions in borate glasses.

20 Schematic representation of luminescence Vibrational relaxation Excited state Triplet State Ground state Absorption Fluorescence Phosphorescence upport/tutorials.html

21 Electronic Configuration of hydrogen atom : 1s1

22 Energy level diagram of Hydrogen atom S, P, P, 1/ 1/ / 1 S 1/

23 Your ion

24 Electronic Configuration of neutral rare earth atoms Number Element Symbol Electronic configuration 54 Xenon Xe 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 58 Cerium Ce 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 6s 4f 1 5d 1 59 Praseodymium Pr 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 6s 4f 60 Neodymium Nd 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 6s 4f 4 61 Promethium Pm 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 6s 4f 5 6 Samarium Sm 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 6s 4f 6 6 Europium Eu 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 6s 4f 7 64 Gadolinium Gd 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 6s 4f 7 5d 1 65 Terbium Tb 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 6s 4f 9 66 Dysprosium Dy 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 6s 4f Holmium Ho 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 6s 4f Erbium Er 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 6s 4f 1 69 Thulium Tm 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 6s 4f 1 70 Ytterbium Yb 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 6s 4f Lutetium Lu 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 6s 4f 14 5d 1 For triply ionized rare earth ion you have to remove three electrons from the outermost orbit configuration. For example for Nd + ion the electronic configuration is 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 4f. Two electrons from 6s and one electron from 4f are removed.

25 60 Neodymium Nd 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 6s 4f 4 Nd + 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 4f.

26 Atomic Term Symbols Free ions can be described by Term Symbols: Ground-State Term Symbols for Any Electronic Configuration: The Ground State has: 1. Choose Maximum S value. Choose Maximum L value n S + 1 L J. 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 6

27 Free ions can be described by Term Symbols: S + 1 L J where: S + 1 is spin multiplicity L is Total orbital angular momentum J is the spin, orbit (L + S) coupling Multiplicity: S + 1 is calculated from number of unpaired electrons S = 0 (all paired), S = 1/ (one unpaired), S = 1 ( unpaired), S = / ( unpaired), etc. If total L = (L total = l 1 + l + for each electron) L = 0 -> S symbol L = 1 -> P symbol L = -> D symbol L = -> F symbol Russell-Saunders Coupling: J ranges from L + S,, L-S and all integers in between 7

28 Atomic Term Symbols S + 1 L J Carbon atom with electronic configuration 1s s p G! 6! m l = 0 15 states N!( G N)!!(6 )! m l = 0 m l = L ( l l )...( l l ) L 1 1,1,0 S ( s s )...( s s ) S 0,1 1 1 Ground state of Carbon P 0 Since both electrons are in the same subshell p, they have to satisfy Pauli's exclusion principle. L S (L+1)=M and (S+1)=M L (L+1)(S+1) 51 5 states = D = D 1 1 (L+1)(S+1) = 9 states = P = P, P, P (L+ 1)(S+1) 11 = 1state = S = S 0 S 1 0 8

29 9 1, 1, 0, 1, J D J M, 1, 0, 1, J P J M 1 1 1,0, 1 J P J M J P J M 5 states = 9 states J S J M 1 state Total = = 15 states

30 15 states Energy levels of Carbon atom 1 S 0 M J 0 s p H 0 Central field Hamiltonian 1 S 1 D H C P Coulomb Interaction Hamiltonian H S-O D P P P0 Spin-orbit Interaction Hamiltonian 0

31 1 1 S 0 D P P 1 P 0 Hund s Rules 1.State with the largest value of S is most stable and stability decreases with decreasing S..For states with same values of S, the state with the largest value of L is the most stable..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. 1

32 4 f Atomic Term Symbols Praseodymium neutral atom with electronic configuration 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 6s 4f Praseodymium trivalent ion with electronic configuration 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 4f G! N!( G N)! Total States : 14! 91 states!(14 )! Praseodymium Ground state of trivalent Praseodymium ion H 4 l and l 1 L l l... l l 1 1 L 6,5,4,,,1,0 S s s... s s S 0,1 1 1

33 L S Trivalent Praseodymium ion (L+1)=M and (S+1)=M L 6 0 (L+1)(S+1) 11 1 states = I = (L+1)(S+1) 11 states = H = H, H, H 4 0 (L+1)(S+1) = 91 = 9 states = G = S I (L+1)(S+1) 7 = 1 states = F = F, F, F 0 (L+1)(S+1) 51 = 5states = D = (L+1)( S+1) = = 9 states = P = P, P, P (L+1)(S+1) = 11 =1 state = S = S0 Total = 91 states D G S 0 P I 6 P 1 P 0 D G F F F 4 4 H H H 6 5 4

34 Trivalent Praseodymium ion Energy levels

35

36 Energy (10 cm -1 ) Intensity (a.u.) 4 G 5/ 6 H 7/ Trivalent Samarium ion Energy levels 6 Samarium Sm 1s s p 6 s p 6 4s d 10 4p 6 5s 4d 10 5p 6 6s 4f 6 0 Excitation Emission 4 I1/ 4 I11/ 4 M15/ Fluorescence 4 G5/ 598 nm nm nm 565 nm 646 nm 4 G 5/ 6 H 5/ 4 G 5/ 646 nm 6 H 9/ 6 F11/ 565 nm 6 F9/ 4 G 5/ F7/ 6F5/ F5/ H15/ F / 6 H11/ 6 H9/ 6 H 11/ 707 nm Wavelength (nm) 488 nm 6 H7/ 6 H5/ Write all the term symbols for Samarium ion in your lab report