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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? 2 ave 2 R 2 1 1 R nF 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 E 0 x, y E o 0 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 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 8.8552 x10 12 C2 N m2 force capacitance kg m 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 0 4 x10 7 N s2 C2 c 1 0 0 c 0 4 x10 7 N s2 C2 0 8.8552 x10 12 1 C 12 7 N s 8.8552 x10 2 4 x10 2 N m C 2 c 1 2 111 . x10 17 s m2 2 1 3.33485x10 m c 2.9986 x10 s 8 9 s m C2 N m2 sqrt dielectric 1 0 0 c 1.000034 1.000131 1.000294 1.00E+00 1.8 1.0005 1.6 1.0004 1.4 1 velocity 1 1.0002 0.8 1.0001 0.6 1 r c vvelocity Ke 0 0 0 0.4 0.9999 0.2 0 0.9998 1.51 4.63E+00 5.04 5.08 8.96E+00 sqrt dielectric Dielectric constant Typically r ~ 0 so c velocity This works pretty well for gases (blue line) Ke Says: refractive index is proportional to the dielectric Maxwell’s relation constant Index of refraction Index of refraction 1.0003 1.2 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 Is Io 4 SCATTERING particle 8 4 2 2 I s I o 4 2 1 cos r 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 Angle between incident and scattered light 1.2 1 Relative Intensity Light in Most important parameter is the relationship to wavelength 0.8 0.6 0.4 0.2 0 0 45 90 135 180 225 Angle of Scattered Light 270 315 360 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 r c Refractive index = relative speed of radiation velocity Refractive index is related to the relative permittivity (dielectric constant) at that Frequency 2 1 Pm 2 M 1 NA 2 Pm 3o 3kT 0 Where is the mass density of the sample, M is the molar mass of the molecules and Pm is the molar polarization Is the permittivity of free space which describes the Flux of the electric portion of the wave in vacuum and Has the value Where is the electric dipole moment operator is the mean polarizabiltiy 1 N A N A 2 3kT 3 0 M 1 3 0 M 2 2 2 1 N A 2 1 3 0 M Point – refractive index Is related to polarizability Clausius-Mossotti equation 2e 2 R 2 3 E 1 N A 2 1 3 0 M 2 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 2 1 3 0 M 3 E 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 2 1 N A 2e 2 R 2 3 M 3 E 0 An alternative expression for a single atom is 2 Ne 1 o me 2 Molecules per Unit volume Each with J oscillators j Transition probability that Interaction will occur fj 2 0j 2j i 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 If you include the interactions between atoms and ignore absorbance you get 2 1 2 1 N A 2e 2 R 2 3 M 3 E 0 2 1 Ne 2 2 2 3 o me when when c r v Ke j fj 2 0j 2j 20 j 2j 2j 20 j The refractive index is constant The refractive index depends on omega And the difference 20 j 2j Gets smaller so the 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? r c velocity frequency frequency frequency r c vacuum c v c v elocity vacuum media elocity ,media frequency vacuum vacuum frequency media media t 21 ' n t ' 2 t Wavelength In media Filters can be constructed By judicious combination of the Principle of constructive and Destructive interference and Material of an appropriate refractive index ' t ' vacuum ' t t 23 ' t n vacuum t 2 2t vacuum 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 n 2t2 2 n 2t 3 3 n 2t4 4 n 1 AVAILABLE WEDGES Vis Near IR IR 400-700 nm 1000-2000 nm 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 The two polarization components are reflected and transmitted with Different amplitudes depending Upon the angle of reflection, r, And the angle of transmittence, t Glass n=1.5 T Let’s start by examing The Angle of transmittence i c velocity 1 Snell’s Law sin i t sin t i 2 1 Less dense 1 Lower refractive index Faster speed of light More dense 1 Higher refractive index 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 1.53 refractive index of crown glass 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 1.525 1.52 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 In the media as Described by Snell’s Law T R r R/ / r/ / 2 2 i cosi t cost i cosi t cost t cos i i cos t i cos i t cos t 2 T t 2 2 T/ / t / / 2 2i cosi i cosi t cost 2i cosi i cost t cosi 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 R T 1 2 Therefore, given the two refractive Indices and the angle of incidence can Calculate everything 2 Consider and air/glass interface i 0.8 0.7 Perpendicular Transmittance 0.6 0.5 0.4 0.3 Here the transmitted parallel light is Zero! – this is how we can select For polarized light! 0.2 Parallel 0.1 0 0 10 20 30 Angle of incidence This is referred to as the polarization angle 40 50 Total Internal Reflection Glass n=1.5 Here consider Light propagating 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 Reflectance and Transmittence Parallel 1 Perpendicular 0.8 0.6 0.4 0.2 0 0 20 40 60 Angle of incidence 80 100 t All of the light is reflected internally ic RT r 2 i cos i t cos t i cos i t cos t Set R to 1 & to 90 The equation can be solved for the critical angle of incidence sin c transmitted ,less dense For glass/air incident ,dense 1 sin c 0.666 15 . c a sin(0.666) 0.7297rads 0.7297rads180 418 . o 2 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 % Transmittence 0.6 1.3/1 1.5/1 ni/nt 0.5 0.4 Critical Angles 1.697 36.27 1.5 41.8 1.3 50.28 0.3 0.2 0.1 0 0 10 20 30 37 42 40 50 Angle of Incidence 60 70 51 80 90 Numerical Aperture NA sin c 2 incident ,dense 2 transmitted ,less dense The critical angle here is defined differently because we have to LAUNCH the beam sin i t sin t i Shining light directly through our sample i=0 Using Snell’s Law the angle of transmittance is t 0 sin t i sin 1 0 t 0 cos0 1 RT r R/ / r/ / 2 2 i cos i t cos t i cos i t cos t t cos i i cos t i cos i t cos t 2 2 T cos0 1 RT r R/ / r/ / 2 2 i t i t t i i t t i R/ / i t 2 2 same 2 The amount of light reflected depends 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? R/ / I reflected I initial I reflected t i i t 2 t i I initial i t I transmitted I initial I reflected I transmitted I initial I transmitted i I initial t i t 2 i I initial 1 t i t 2 2 Io It=I’o Water, refractive index 1.33 It’ = I’’o It’’ = I’’’o It’’’ Air, refractive index 1 Air Glass, refractive index 1.5 Final exiting light 2 2 15 15 . 1 . 1 '' I t ''' I ''' o 1 I t 1 15 . 1 15 . 1 2 2 15 15 . 133 . . 133 . I t '' I o ' ' 1 I t ' 1 15 . 133 . 15 . 133 . 2 2 133 133 . 15 . . 15 . I t ' I ' o 1 I t 1 133 . 15 . 133 . 15 . 2 15 . 1 I t I o 1 15 . 1 2 2 15 . 1 133 . 15 . I t ' I o 1 1 15 . 1 133 . 15 . 2 2 15 . 1 133 . 15 . I t ' ' I o 1 1 15 . 1 133 . 15 . 2 15 . 1 I t ''' I o 1 15 . 1 2 2 2 133 . 15 . 1 133 . 15 . 2 2 15 . 1 I t ''' I o 1 15 . 1 I glass / air 2 2 0.5 2 I initial 1 2.5 2 133 . 15 . 1 133 . 15 . 2 2 017 . 1 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 2 2 Key Concepts Interaction with Matter Light Scattering vol 2 I s Io 4 Refractive Index Is wavelength dependent Used to separate light by prisms r c velocity 2 1 Ne 2 2 2 3 o me j fj 2 0j 2j 2 1 N A 2e 2 R 2 2 1 3 0 M 3 E Refractive index based Interference filters 2t n Key Concepts Interaction with Matter Snell’s Law sin i t 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 R r R/ / r/ / 2 2 i cosi t cost i cosi t cost t cos i i cos t i cos i t cos t 2 T t 2 2 T/ / t / / 2 2i cosi i cosi t cost 2 2i cosi i cost t cosi Used to create surfaces which select for polarized light 2 Key Concepts Interaction with Matter Fresnell’s Laws collapse to sin c transmitted ,less dense incident ,dense Which describes when you will get total internal reflection (fiber optics) And R/ / I reflected I initial t i i t 2 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 m h Energy of Ejected electron Work function=minimum energy binding an Electron in the metal Frequency of impinging photon (related to photon energy) KE m 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 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 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 +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” 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 180 165 900 800 150 2nd -150 700 600 135 order 65 -135 500 400 120 -120 300 200 100 105 -105 0 -100 1st order Grating -200 90 - 90 - 75 75 - 60 60 - 45 45 -30 30 Angle of reflection i=45 15 0 -15 180 165 900 800 150 2nd -150 700 600 135 order 65 -135 500 400 120 -120 300 200 100 105 -105 0 -100 1st order -200 90 - 90 - 75 75 - 60 60 - 45 45 -30 30 Angle of reflection i=45 15 0 -15