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6. Really Basic Optics 15000 10000 5000 Amplitude 0 0 0.2 0.4 0.6 0.8 1 1.2 -5000 -10000 -15000 Time (s) Sample Instrument Instrument Sample Signal (Data) Prep Out put Polychromatic light Selected Turn off/diminish intensity light Sample Select interaction light select Turn on different wavelength source detect 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 y sin A y A sin 1.5 radians t t 1 s y A sint 0.5 /2 3/2 2 Amplitude 0 0 50 100 150 200 250 300 350 400 y ' A sint -0.5 90o phase angle /2 radian phase angle -1 -1.5 t (s) 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 e c hc Elo E h 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 frequency 1. Uncertainty 2. Doppler effect t 1 3. Pressure 4. Electric and magnetic fields Lifetime of an excited state is typically 1x10-8 s 1 7 1 8 5x10 10 s s c 1 d 1c 2 d d d c2 2 c2 c 1 2 5x10 7 s 5x10 10 2 8 m 10 nm 9 nm 3x10 s m Example: 253.7 nm 5x10 10 253.7nm 3.22 x10 nm 2 5 nm Typical natural line widths are 10-5 nm Line broadens due 1. Uncertainty 2. Doppler effect elocity 3. Pressure 4. Electric and magnetic fields 0 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 We use pressure broadening On purpose to get a large f ( ) 1 2 Line width in AA for some 2 o 2 FWHM Forms of background correction 2 FWHM = full width half maximum o is the peak “center” in frequency units 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 Stefan-Boltzmann Law Planck’s Blackbody Law I T4 8h 3 c 3 h exp kT 1 Intensity = 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 2500 3000 3500 nm 1 8h max b 3 T hc exp kT 1 Wien’s Law 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 0 Series1 Series2 Same but wave from particle -0.05 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 B lags behind A by the -0.1 Phase angle -0.15 -0.2 -0.25 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 Photons can stimulate * Emission just as much As they can stimulate Absorption w BN o w BN * (idea behind LASERs o Stimulated Emission) The rate of stimulated event is described by : w BN o w BN * 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 8h 3 frequency c 3 h exp kT 1 In order to measure absorption it is required that the Rate of stimulated absorption is greater than the Rate of stimulated emission wabsorption BN o wenussion BN * 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) * g* N*0 N N0 g0 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 mirror Stimulated emission 1. Single phase 2. Along same path =Constructive Interference Coherent radiation Multiple directions, Multiple phase lags Incoherent radiation 1.5 FTIR Instrument Constructive/Destructive interference 1. Laser 2. FT instrument 1 3. Can be used to select wavelengths 4. Can be used to obtain information about distances 0.5 5. Holographic Interference filter. Voltage 0 0 0.2 0.4 0.6 0.8 1 1.2 -0.5 -1 -1.5 Time Measurement A1 sin2f s,1t As sin2f S ,2 t .... An sin2f S ,n t 200000 180000 Time Domain: 2 frequencies 160000 140000 4000 120000 amplitude 100000 3000 80000 2000 60000 40000 1000 20000 Amplitude 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 frequency 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 60000 -1000 50000 -2000 40000 -3000 Amplitude 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 2000 140000 Amplitude 120000 amplitude 0 100000 0 0.2 0.4 0.6 0.8 1 1.2 80000 -2000 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 sin A sin B 2 sin 2 A B cos 2 A B -8000 1 1 t (s) B Fixed mirror Moving mirror A C Beam IR source splitter detector Constructive interference occurs when 1 AC BC n 2 4000 3000 2000 Light Intensity at Detector 1000 -2 -1 0 +1 0 -1000 -2000 -3000 -4000 Mirror distance dis tan ce f mirror velocity mirror INTERFEROGRAMS n mirror 2 mirror velocitymirror n 2 mirror mirror 1 2 mirror f det ector n Remember that: c Frequency of light 2 mirror light f det ector 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? 2 mirror light f det ector 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? x sinf inite population Signal N measurements s population samples sample xblank N measurements N oise sinf inite population Signal x sample xblank Signal N measurements N oise sblank N oise 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 D d 9. sin CBF sin B F’ opposite CF DE W O E L F hypotenuse BC OD C 1 BD, OD, & CD, big assume / / DE 10. n CF BC sin BC 2. if / /, then BCF OD 3. 180 DEO EOD 90 EOD 4. OF ' B 90 o 5. if / /, then CFB 90o 6. 180 CFB FBC BCF 90 From which we conclude DE DE 11. n BC BC 7. EOD OD OE 8. CF must be n for in 12. d n W phase (constructive int erference) L D The complete equation for a slit is d 2 sin B I I0 E W L b sin 2 b=W/2 Width of the line depends upon The slit width!! Therefore resolution depends 1.2 On slit width 1 0.8 Relative Intensity 0.6 0.4 0.2 Also “see” This spectra 0 “leak” of -800 -600 -400 -200 0 200 400 600 800 Beta Our hard won intensity The base (I=0) occurs whenever 2 sin sinβ =0 I I0 b Which occurs when sin n 2 W W b 2 sin W sin line width sin line width 2n 2 sin 2 2 2 2 2 1.2 The smaller the 1 Slit width the Smaller 0.8 The line width, Relative Intensity 0.6 Which leads To greater 0.4 Spectral Resolution 0.2 Remember R is 0 Inversely proportional -800 -600 -400 -200 0 200 400 600 800 Beta To the width of The Gaussian base SLIT IMAGE A Slit 1 2 3B Position number 1234 5 Image When edge AB at Detector Sees Position 1 0% power Position 2 50% power Position 3 100% power Position 4 50% power Position 5 0 % power Detector output: 1.2 Triangle results when 1 Effective bandpass = image eff 0.8 Output of detector 0.6 To resolve two images that are ∆ apart requires 0.4 0.2 0 2 eff 0 1 2 3 4 5 6 Image position Implies want a narrower slit 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 + - opp DB sinDAB hyp d q r DBA 90 DBA BAD 90 180 DBA BAD 90 q r BAD DB sin r d n CB BD n CB BD opp CB sinCAB hyp d n d sini d sinr i q 90 CAB i n d sini sinr CB q CAB 90 sini d 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 sini sinr n 500nm sin 45o sinr Inputs that can be varied are in pink Czerny-Turner 440.3 construction 220.1 d nlambda = d(sin+sinr) grooves/mm 2000 500 146.8 i in degrees 45 0.785398 i in radians 88 73 r in degrees -40 -20 -10 0 10 20 40 All come radians -0.69813 -0.34907 -0.17453 0 0.174533 0.349066 0.698132 through 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 You get light of 674.9 nm Are observed ½; 1/3; 1/4; 1/5; etc. 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 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) construction 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 Hit grating second time Time to get 440.3 220.1 146.8 88 73 220.1 nm All come through 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 Light of 100 Reflection Angle wavelength Nm shows up 100 -30.47131 -17.88496 -6.148561 5.330074 17.03125 At -30.4 125 -27.20057 -11.95286 2.458355 17.03125 32.88081 AND 150 -24.02322 -6.148561 11.12168 29.53092 52.45672 175 -20.92262 -0.407192 20.05324 43.85957 #NUM! -17.88 200 -17.88496 5.330074 29.53092 63.23909 #NUM! And 225 -14.89846 11.12168 40.0079 #NUM! #NUM! -6.1 250 -11.95286 17.03125 52.45672 #NUM! #NUM! 275 -9.039003 23.13464 70.54325 #NUM! #NUM! And 300 -6.148561 29.53092 #NUM! #NUM! #NUM! 5.3 325 -3.273759 36.36259 #NUM! #NUM! #NUM! Etc. 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! GRATING DISPERSION D-1 = Reciprocal linear dispersion 1 d dis tan ce between d cos r D dy dis tan ce on screen nF Where n= order F = focal length d= distance/groove 1 d D r 0 nF 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 1 2 2 ave R 2 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? 1 2 2 ave R R nF 2 1 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 E0 x, y Eo0 x, y cos 2ft 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 1 c is the velocity of the electromagnetic wave in free space c 0 0 0 Is the permittivity of free space which describes the Flux of the electric portion of the wave in vacuum and Has the value capacitance C2 0 8.8552 x10 12 N m2 kg m force 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 N s2 0 4 x10 7 C2 1 N s2 C2 c 0 4 x10 7 0 8.8552 x10 12 0 0 C2 N m2 1 c 12 C 2 7 N s 2 8.8552 x10 2 4 x10 2 N m C 1 1 c s 2 9 s 111x10 17 . 3.33485x10 m m2 m c 2.9986x10 8 s sqrt dielectric 1 1.000034 1.000131 1.000294 1.00E+00 c 1.8 1.0005 0 0 1.6 1.4 1.0004 1.0003 Index of refraction Index of refraction 1.2 1 velocity 1 1.0002 0.8 1.0001 0.6 1 0.4 c 0.9999 r Ke 0.2 vvelocity 0 0 0 0 1.51 4.63E+00 5.04 5.08 8.96E+00 0.9998 sqrt dielectric Dielectric constant Typically ~0 so This works pretty well for gases (blue line) c r Ke v elocity Says: refractive index is proportional to the dielectric Maxwell’s relation 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 vol molecule2 SCATTERING Is Io 4 Most important parameter is the Light in particle relationship to wavelength 8 4 2 I s I o 4 2 1 cos 2 Angle between incident and scattered r light 1.2 Polarizability of electrons a) Number of electrons 1 b) Bond length Relative Intensity 0.8 c) Volume of the molecule, which depends upon 0.6 the radius, r 0.4 = vacuum 0.2 Io = incident intensity 0 0 45 90 135 180 225 270 315 360 Angle of Scattered Light 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 molecule2 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 c r Refractive index = relative speed of radiation v elocity Refractive index is related to the relative permittivity (dielectric constant) at that Frequency 2 1 Pm Where is the mass density of the sample, M is the molar mass of the molecules and Pm is the molar polarization 1 2 M 0 Is the permittivity of free space which describes the Flux of the electric portion of the wave in vacuum and Has the value NA 2 Where is the electric dipole moment operator Pm 3 o 3kT is the mean polarizabiltiy Point – refractive index 1 N A 2 N A 2 Is related to polarizability 1 3 0 M 2 3kT 3 0 M 2 1 N A Clausius-Mossotti equation 1 3 0 M 2 1 N A 2 2e 2 R 2 2 1 3 0 M 3 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 1 N A 2e 2 R 2 1 3 0 M 3 E 2 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 N A 2e 2 R 2 1 2 3 0 M 3 E An alternative expression for a single atom is 2 Transition probability that Ne fj Interaction will occur 1 2 o me j 2 0j 2j i j A damping force term that account for Absorbance (related to delta E in prior Molecules per Natural Expression) Unit volume Frequency of Each with The oscillating electrons J oscillators In the single atom j Frequency of incoming electromagnetic wave If you include the interactions between atoms and ignore absorbance you get 2 1 N A 2e 2 R 2 1 2 3 0 M 3 E 2 1 Ne 2 fj 2 2 3 o me j 2 2j 0j when 2 j 2 0 j The refractive index is constant when 2 2j j 0 The refractive index depends on omega c r Ke And the difference 2j 2 0 j Gets smaller so the v 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? c r v elocity c frequency vacuum frequency c vacuum frequency v media elocity ,media c frequency vacuum vacuum r velocity frequency media media Filters can be constructed By judicious combination of the Principle of constructive and Destructive interference and t 2 ' 1 Material of an appropriate refractive index t n Wavelength In media ' t ' 2 t ' vacuum ' t n vacuum t 2 t 2 ' 3 2t vacuum t 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? 2t vacuum n Holographic filters are better INTERFERENCE WEDGES 2t1 1 n 2t2 2 n 2t3 3 n 2t4 4 n AVAILABLE WEDGES Vis 400-700 nm Near IR 1000-2000 nm IR 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 Glass The two polarization components are n=1.5 reflected and transmitted with Different amplitudes depending T Upon the angle of reflection, r, And the angle of transmittence, t Let’s start by examing The Angle of transmittence c i velocity 1 Snell’s Law sin i t 2 sin t i 1 Less dense 1 More dense 1 Lower refractive index Higher refractive index Faster speed of light 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 Crown Glass 1.53 (nm) refractive index of crown glass 400nm 1.532 1.525 450 nm 1.528 550 nm 1.519 590 nm 1.517 1.52 620 nm 1.514 650 nm 1.513 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 T In the media as Described by Snell’s Law i cosi t cost 2 2i cosi 2 R r cos cos T t 2 cos cos 2 i i t t i i t t 2i cosi 2 2 t cos i i cos t R/ / r/ / cos cos T/ / t / / cos cos 2 2 i i t t i t t 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 Therefore, given the two refractive R T 1 Indices and the angle of incidence can Calculate everything Consider and air/glass interface i 0.8 0.7 Perpendicular 0.6 Transmittance 0.5 0.4 0.3 Here the transmitted parallel light is 0.2 Zero! – this is how we can select For polarized light! 0.1 Parallel 0 0 10 20 30 40 50 Angle of incidence This is referred to as the polarization angle Total Internal Reflection Here consider Glass Light propagating n=1.5 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 Parallel 1 Reflectance and Transmittence Perpendicular 0.8 0.6 0.4 0.2 0 0 20 40 60 80 100 Angle of incidence t All of the light is reflected internally ic 2 i cos i t cos t RT r cos cos 2 i i t t Set R to 1 & to 90 The equation can be solved for the critical angle of incidence transmitted ,less dense sin c incident ,dense 1 For glass/air sin c 0.666 . 15 c a sin(0.666) 0.7297rads 0.7297rads180 418o . 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 1.3/1 0.6 1.5/1 ni/nt Critical Angles % Transmittence 0.5 1.697 36.27 0.4 1.5 41.8 1.3 50.28 0.3 0.2 0.1 0 0 10 20 30 40 50 60 70 80 90 37 Angle of Incidence 51 42 Numerical Aperture NA sinc 2 incident ,dense transmitted ,less dense 2 The critical angle here is defined differently because we have to LAUNCH the beam sin i t Shining light directly through our sample sin t i i=0 Using Snell’s Law the angle of transmittance is t 0 sin t i sin 1 0 t 0 cos0 1 2 i cos i t cos t RT r cos cos 2 i i t t 2 T t cos i i cos t R/ / r/ / cos cos 2 i i t t cos0 1 2 i t RT r 2 i t same 2 t i R/ / r/ / 2 i t 2 t i The amount of light reflected depends R/ / i t 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? 2 I reflected t i R/ / I initial i t 2 t i I reflected I initial i t I transmitted I initial I reflected 2 i I transmitted I initial I initial t i t 2 I transmitted I initial 1 t i i t Water, refractive index 1.33 It’ = I’’o It’’ = It’’’ Io It=I’o I’’’o Air, refractive index 1 Air Glass, refractive index 1.5 Final exiting light 15 1 2 133 15 2 . . . 15 1 2 . 15 1 2 . I t ' I o 1 1 I t ''' I ''' o 1 I t 1 '' 15 1 133 15 . . . 15 1 . 15 1 . 15 133 2 15 133 2 2 . . . . 15 1 2 133 15 2 . . . I t '' I o '' 1 I t ' 1 I t '' I o 1 1 15 133 . . 15 133 . . 15 1 133 15 . . . 133 15 2 . . 133 15 2 . . I t ' I ' o 1 I t 1 133 15 . . 133 15 . . 2 2 15 1 2 . 133 15 2 . . 15 1 2 I t ''' I o 1 1 I t I o 1 . 15 1 . 133 15 . . 15 1 . 2 2 15 1 2 . 133 15 2 . . I t ''' I o 1 1 15 1 . 133 15 . . 2 2 05 2 . 017 2 . I glass/ air 2 I initial 1 1 2.5 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 Key Concepts Interaction with Matter Light Scattering vol 2 I s Io 4 2 1 Ne 2 fj Refractive Index Is wavelength dependent r c v elocity 2 3 o me 2 j 2 2j 0j Used to separate light by prisms 2 1 N A 2e 2 R 2 1 3 0 M 3 E 2 2t Refractive index based Interference filters n Key Concepts Interaction with Matter sin i t Snell’s Law 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 i cosi t cost 2 2i cosi 2 R r cos cos T t 2 cos cos 2 i i t t i i t t 2i cosi 2 2 t cos i i cos t R/ / r/ / cos cos T/ / t / / cos cos 2 2 i i t t i t t i Used to create surfaces which select for polarized light Key Concepts Interaction with Matter Fresnell’s Laws collapse to transmitted ,less dense sin c incident ,dense Which describes when you will get total internal reflection (fiber optics) And 2 I reflected t i R/ / I initial i t 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 KEm h Work function=minimum energy binding an Electron in the metal Energy of Ejected Frequency of impinging photon electron (related to photon energy) KEm 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 Want low dark current kd is the dark current Classes of Detectors Name comment Photoemissive single photon events Photoconductive “ (UV, Vis, near IR) Heat average photon flux 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 3. Apply greater voltage 4. Move charge to “gate” +V 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 65 165 900 150 800 -150 700 135 600 -135 2nd order 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 -15 15 0 Angle of reflection i=45 180 65 165 900 150 800 -150 700 135 600 -135 2nd order 500 400 120 -120 300 200 100 -105 105 0 -100 1st order -200 - 90 90 - 75 75 - 60 60 45 - 45 30 -30 -15 15 0 Angle of reflection i=45