Introduction to Basics of Raman Spectroscopy Chandrabhas Narayana

Introduction to Basics of Raman Spectroscopy Chandrabhas Narayana Chemistry and Physics of Materials Jawaharlal Nehru Centre for Advanced Scientific Research, Jakkur P. O. , Bangalore 560064, India cbhas@jncasr. ac. in http: //www. jncasr. ac. in/cbhas Lecture at MASTANI Summer School, IISER, Pune, June 30, 2014 to July 12, 2014 July 11, 2014

What happens when light falls on a material? Transmission Reflection Absorption Luminescence Elastic Scattering Inelastic Scattering

Raman Spectroscopy 1 in 107 photons is scattered inelastically Scattered Excitation virtual state Rotational Raman Vibrational Raman Electronic Raman v” = 1 v” = 0 Infrared Raman (absorption) (scattering)

Raman visible through unaided eye

Raman, Fluorescence and IR Scattering Absorption and emission Absorption

Concept of normal modes in a molecule • There are 3 N possible movements in a molecule made of N atoms, each of which moving in one of three directions, x, y and z. – There are three transitional movements: all atoms in the molecule moving in x, y or z direction at the same time. – There are three rotational movements around x, y or z-axis • Linear molecules are exceptions because two axes that are perpendicular to the molecular axis are identical. – The rest of movements are vibrational movements • For linear molecules, 3 N – 5 movements • For non-linear molecules, 3 N – 6 movements – All vibrational movements of the sample can be described as linear combinations of vibrational normal modes.

Vibrations in Molecules Sym. Stretching HCl n = 2991 cm-1 H 2 O n 1 = 3835 cm-1 n 1 = 774. 55 cm-1 HF Asym. Stretching Sym. Bending n 3 = 3939 cm-1 n 2 = 1648 cm-1 NH 3 n 1 = 3505. 7 cm-1 SF 6 8086 cm-1 = 1 e. V n = 4139 cm-1 Asym. Bending n 3 = 3573. 1 cm-1 n 5 = 643. 35 cm-1 n 3 = 947. 98 cm-1 n 2 = 1022 cm-1 n 2 = 615. 02 cm-1 n 4 = 1689. 7 cm-1 n 6 = 348. 08 cm-1 n 4 = 523. 56 cm-1

Vibrational Spectroscopy DAB For a diatomic molecule (A-B), the bond between the two atoms can be approximated by a spring that restores the distance between A and B to its equilibrium value. The bond can be assigned a force constant, k (in Nm-1; the stronger the bond, the larger k) and the relationship between the frequency of the vibration, , is given by the relationship: or, more typically where , c is the speed of light, is the frequency in “wave numbers” (cm-1) and is the reduced mass (in amu) of A and B given by the equation: 0 r. AB re re = equilibrium distance between A and B DAB = energy required to dissociate into A and B atoms

Vibrational Spectroscopy can be rearranged to solve for k (in N/m): Molecule (cm-1) k (N/m) (amu) HF 3962 878 19/20 HCl 2886 477 35/36 or 37/38 HBr 2558 390 79/80 or 81/82 HI 2230 290 127/128 Cl 2 557 320 17. 5 Br 2 321 246 39. 5 CO 2143 1855 6. 9 NO 1876 1548 7. 5 2331 2240 7 N 2 For a vibration to be active (observable) in an infrared (IR) spectrum, the vibration must change the dipole moment of the molecule. (the vibrations for Cl 2, Br 2, and N 2 will not be observed in an IR experiment) For a vibration to be active in a Raman spectrum, the vibration must change the polarizability of the molecule.

Classical Picture of Raman Induced Polarization Stokes Raman Polarizability Anti-Stokes Raman

Mutually exclusive principle

Raman Scattering Selection rule: v = ± 1 Overtones: v = ± 2, ± 3, … Must also have a change in polarizability Classical Description does not suggest any difference between Stokes and Anti-Stokes intensities www. andor. com

Are you getting the concept? Calculate the ratio of Anti-Stokes to Stokes scattering intensity when T = 300 K and the vibrational frequency is 1440 cm-1. -34 h = 6. 63 x 10 -34 Js k = 1. 38 x 10 -23 J/K ~ 0. 5

Energy diagram and Quantum picture Virtual states Raman cross section S photon <eg, p 2|Her|p 2, eb> <eb, p 2|Hep|p 1, ea> <ea, p 1|Her|p 1, eg> |Es-Eb|x|Ei-Ea| a, b ex g Electronic states Vibrational states If Ei = Ea or Es = Eb We have Resonance Raman effect

• • • Intensity of Normal Raman Peaks The intensity or power of a normal Raman peak depends in a complex way upon the polarizability of the molecule, the intensity of the source, and the concentration of the active group. The power of Raman emission increases with the fourth power of the frequency of the source; - photodecomposition is a problem. Raman intensities are usually directly proportional to the concentration of the active species.

Raman Depolarization Ratios Polarization is a property of a beam of radiation and describes the plane in which the radiation vibrates. Raman spectra are excited by plane-polarized radiation. The scattered radiation is found to be polarized to various degrees depending upon the type of vibration responsible for the scattering.

Raman Depolarization Ratios The depolarization ratio p is defined as Experimentally, the depolarization ratio may be obtained by inserting a polarizer between the sample and the monochromator. The depolarization ratio is dependent upon the symmetry of the vibrations responsible for scattering.

Raman Depolarization Ratios Polarized band: p = < 0. 76 for totally symmetric modes (A 1 g) Depolarized band: p = 0. 76 for B 1 g and B 2 g nonsymmetrical vibrational modes Anomalously polarized band: p = > 0. 76 for A 2 g vibrational modes

Raman spectra of CCl 4 Isotope effect Cl has two isotopes 35 Cl and 37 Cl Relative abundance is 3: 1

CCl 4 Spectra • 461. 5 cm-1 is due to 35 Cl 4 C • 458. 4 cm-1 is due to 35 Cl 337 Cl. C • 455. 1 cm-1 is due to 35 Cl 237 Cl 2 C • What are the two question marks? • Why are these bands weak?

Raman Spectra of Methanol and Ethanol CH stretching Raman Intensity (arbitrary unit) CCO stretching CH 3 and CH 2 deformation OH stretching CO stretching CH 3 deformation Raman Shift (cm-1) CASR Significant identification of alcohols which differ just in one CH 2 -g

Peak position – Chemical identity – Similar Structures CASR 3, 4 -Methylenedioxymethamphetamine (MDMA) Methamphetamine ecstasy Raman Intensity (arbitrary unit) 500 1000 1500 2000 Raman Shift (cm-1) 2500 3000 3500

The Mass Effect on Raman Spectra Significant identification of salts (SO 42 -) which differ just in the metal ion employed Mg - SO 4 Na 2 - SO 4 CASR

CASR Peak positions – Chemical identity Diasteromers Ephedrine Pseudoephedrine Raman Intensity (arbitrary unit) 500 1000 1500 2000 Raman Shift (cm-1) 2500 3000 3500

Peak Position – Crystal Phases – Polymorphs Both Anatase and Rutile are Ti. O 2 but of different polymorphic forms - identical chemical composition, different crystalline structures. Rutile Anatase

CASR Peak Shift – Stress and Strain Larnite (b – Ca 2 Si. O 4) inclusion in Diamond Nasdala, L. , Harris, J. W. & Hofmeister, W. (2007): Micro-spectroscopy of diamond. Asia Oceania Geosciences Society, 4 th Annual Meeting, Bangkok, Thailand, August, 2007. Nasdala, L. , Raman barometry of mineral inclusions in diamond crystals. Mitt. Österr. Miner. Ges. 149 (2004)

CASR Bandwidth – Crystallinity – Structural order/disorder Raman spectra of zircon, showing typical amorphous (blue) and crystalline (red) spectra.

CASR Intensity – Concentration 4 -Nitrophenol dissolved in CH 2 Cl 2

CASR Raman technique – what requirements are needed? Requirements for Raman technique to determine peak position, peak shift, bandwidth and intensity - Laser Excitation - Reduction of stray light - Collecting Optics - Spectral resolution and spectral coverage - Spatial resolution and confocality - Sensitivity: subject to detector - Light flux: subject to dispersion

CASR What do we need to make a Raman measurement ? Detector Light source (A way of focusing the laser onto the sample and then collecting the Raman scatter. ) • Sampling optics Grating • Rejection filter (A way of removing the scattered light that is not shifted( changed in colour). Filter Laser • Monochromatic typically a laser Sample • Spectrometer and detector (often a single grating spectrometer and CCD detector. )

Demonstration of the very high CASR spectral resolution obtained in the triple additive mode Rotation-Vibration Spectrum of O 2 Triple additive mode Slit widths= 30 mm Triple subtractive mode. Slit=30 mm

CASR Laser excitation – Laser Selection to avoid fluorescence Laser wavelength, 3 Laser wavelength, 1 Fluorescence Raman shift, 1 -1+ Laser wavelength: 3 < 2 < 1 Raman shift, 3 -1+

CASR Laser excitation – Laser selection to avoid fluorescence Green spectrum: 532 nm laser Red spectrum: 633 nm laser Dark red spectrum: 785 nm laser Fluorescence is wavelength dependent Ordinary Raman is wavelength independent

CASR Laser excitation – Laser selection to avoid fluorescence Commercial Hand Cream 785 nm – 633 nm – 473 nm ÚReduction of Fluorescence

CASR Laser excitation – laser radiation power Laser wavelength: 473 nm Laser power at sample: 25. 5 m. W Laser wavelength: 633 nm Laser power at sample: 12. 6 m. W Objective N. A. Laser spot size (µm) Radiatio n power (k. W/cm 2) 100× 0. 90 0. 64 ~7900 100× 0. 90 0. 85 ~2200 50× 0. 75 0. 77 ~5400 50× 0. 75 1. 03 ~1500 10× 0. 25 2. 31 ~600 10× 0. 25 3. 09 ~200 Laser Radiation spot size power (µm) (k. W/cm 2)

CASR Laser excitation – laser radiation power • Keep in mind: the usage of high numerical objective lenses causes a very small spot size of the laser which results in a high power density • To avoid sample burning radiation power has to be adapted INDIVIDUALY to the sample

Collection solid angle Large for high N. A. lens Small for low N. A. lens Low N. A. lens Working distance θ High N. A. lens Sampling volume Small for high N. A. lens Large for low N. A. lens Laser spot size Small for high N. A. lens Large for low N. A. lens NA = n · sin (Q) n: refraction index Q: aperture angle θ Working distance Collecting Optics CASR

CASR Collecting Optics – Overview on common objectives Objective N. A. Working distance [mm] x 100 0. 90 0. 21 x 50 0. 75 0. 38 x 10 0. 25 10. 6 x 100 LWD 0. 80 3. 4 x 50 LWD 0. 50 10. 6

CASR Collecting Optics – what objective should be used? A distinction between opaque and transparent samples has to be made For opaque samples, high N. A. lens works better because there is almost no penetration of the laser into the sample. High N. A. lens enables - High laser power density (m. W/ m 3) increases sensitivity - Wide collection solid angle increases sensitivity Example for an opaque sample: Silicon wafer 100 % 70 % 5% Silicon x 100 – NA = 0. 9 – 31. 350 C/s x 50 – NA = 0. 75 - 21. 995 C/s x 10 – NA = 0. 25 - 1. 462 C/s

CASR Collecting Optics – what objective should be used? A distinction between opaque and transparent samples has to be made For transparent samples, low N. A. lens works better because of penetration of the laser into the sample. Low N. A. lens enables - Large sampling volume increases sensitivity Sample: Cyclohexane Instrument: ARAMIS Red: x 100 LWD, 7, 000 cts/s Blue: Macro lens, 14, 500 cts/s 48 % 100 %

Spectral resolution and spectral coverage CASR Schematic diagram of a Czerny-Turner spectrograph Slit Collimating mirror Grating Detector Focusing mirror Focal Length

CASR Spectral resolution and spectral coverage • Spectral resolution is a function of 1. dispersion, 2. widths of entrance slit and 3. pixel size of the CCD • Dispersion is the relation between refraction of light according to the wavelength of light • Dispersion is a function of the 1. focal length of spectrograph the 2. groove density of the grating and 3. the excitation wavelength • In general, long focal length and high groove density grating offer high spectral resolution.

CCD Detector Short focal length Same grating Same excitation wavelength Long focal length CCD Detector CASR Dispersion as a function of the focal length

CASR Dispersion as a function of the focal length vis-a vis wavelength Dispersion in cm-1 / pixel 1800 gr/mm Grating Lab. RAM (F = 300 mm) Lab. RAM HR (F = 800 mm)

CASR Dispersion as a function of excitation wavelength Short wavelength CCD Detector Same focal length Same grating Long wavelength

CASR Spectral coverage - dependence from excitation wavelength Length of CCD Chip 473 nm – 633 nm – 785 nm Same focal length Same grating Length of CCD Chip Relative Raman shift of 3100 cm-1 corresponds to 81 nm Length of CCD Chip Relative Raman shift of 3100 cm-1 corresponds to 154 nm Relative Raman shift of 3100 cm-1 corresponds to 252 nm

Low density groove grating Same focal length Same excitation wavelength High density groove grating CCD Detector Dispersion as a function of groove density CCD Detector CASR

CASR Spectral resolution as a function slit width Slit One of parameters that determines the spectral resolution is the entrance slit width. The narrower the slit, the narrower the FWHM (full width at half maximum), and higher the spectral resolution. When recording a line whose natural width is smaller than the monochromator’s resolution, the measured width will reflect the spectrograph’s resolution.

Spectral resolution as a function of pixel CASR size Detector • Because a CCD detector is made of very small pixels, each pixel serves as an exit slit (pixel size = exit slit width) • For the same size CCDs, the CCD with a larger number of smaller pixels produces a larger number of spectral points closer to each other increasing the limiting spectral resolution and the sampling frequency • 26 m pixel vs. 52 m pixel (simulation)

CASR Choice of Laser for Raman The choice of laser will influence different parameters: • Signal Intensity: (1/ )4 rule applies to Raman intensity. • Probing volume: spot size and material penetration depth. • Fluorescence: may overflow Raman signal. • Enhancement: some bounds only react strongly at a certain wavelength. • Coverage range and Resolution: grating dispersion varies along the spectral range. Silicium

CASR Spatial resolution: penetration depth Penetration depth in Silicon General: The larger the excitation wavelength, the deeper the penetration. The exact values depend on material.

CASR Spatial resolution: penetration depth 325 nm Strained Si of top layer 325 nm Si of Si. Ge layer ~nm 488 nm 785 nm Strained silicon layer Uniform layer of Si. Ge ~µm 488 nm Gradient Si. Ge layer Si of silicon substrate 785 nm Pure Si substrate The higher the excitation wavelength, the deeper the penetration.

Spatial resolution: penetration depth CASR EXAMPLE 325 nm laser results The strain is not homogenous. A characteristic, cross-hatch pattern is observed.

CASR Spatial resolution and spot size Laser spot size D is defined by the Rayleigh criterion: excitation wavelength ( ) D = 1. 22 / NA objective numerical aperture (NA) With NA=n sinα Spatial resolution is half of the laser spot diameter

CASR Confocality and Spatial Resolution Nearly closed confocal apertur Sample P P' P' P' Image P ' of laser focus P matches exactly the confocal hole. Length of Laser Focus

Axial resolution of a Confocal Raman Microprobe CASR

Confocality and spatial Resolution Wide Hole Laser focus waist diameter Depth of laser focus (d. o. f) Sampling Volume Narrow Hole Confocal z-scan against silicon with different hole apertures exc = 633 nm CASR Narrow Hole: Collecting Raman radiation that originates only from within a diffraction limited laser focal volume with a dimension of: Focus waist diameter ~ 1. 22 / NA Depth of laser focus ~ 4 / (NA)2

Example of application using the confocality principle : depth profile on a multilayer polymer sample x Polyethylene nylon Polyethylene z CASR 75 m

Thank you for your attention!

Symmetry – Identity (E) Identity operation (E) This is a very important symmetry operation which is where the molecule is rotated by 360º. In otherwords a full rotation or doing nothing at all. This appears in all molecules!!!

Symmetry – Rotation (Cn) Rotations (Cn) These are rotations about the axes of symmetry. n denotes 360º divided by the number for the rotation. C 2 = 180º C 3 = 120º C 4 = 90º C 5 = 72º C 6 = 60º

Symmetry – Reflections (s) Reflections (sh, sd and sv) These are reflections in a symmetry planes (x, y and z). h - Horizontal Plane (y) perpendicular to the highest rotation axis v - Vertical Plane (z) parallel to the highest rotation axis d - Diagonal (dihedral) Plane (x) the Diagonal Plane that bisects two axes

Symmetry – Inversion (i) Inversion centre (i) These are where the molecule can be inverted through the centre of the molecule (atom or space).

Symmetry – Improper Rotation (Sn) Improper rotations (Sn) These are rotations about the axes of symmetry followed by reflections.

Vibrational Spectroscopy For polyatomic molecules, the situation is more complicated because there are more possible types of motion. Each set of possible atomic motions is known as a mode. There a total of 3 N possible motions for a molecule containing N atoms because each atom can move in one of the three orthogonal directions (i. e. in the x, y, or z direction). A mode in which all the atoms are moving in the same direction is called a translational mode because it is equivalent to moving the molecule - there are three translational modes for any molecule. A mode in which the atoms move to rotate (change the orientation) the molecule called a rotational mode - there are three rotational modes for any non-linear molecule and only two for linear molecules. Translation al modes Rotational modes The other 3 N-6 modes (or 3 N-5 modes for a linear molecule) for a molecule correspond to vibrations that we might be able to observe experimentally. We must use symmetry to figure out what how many signals we expect to see and what atomic motions contribute to the particular vibrational modes.

Vibrational Spectroscopy and Symmetry We must use character tables to determine how many signals we will see in a vibrational spectrum (IR or Raman) of a molecule. This process is done a few easy steps that are similar to those used to determine the bonding in molecules. 1. Determine the point group of the molecule. 2. Determine the Reducible Representation, tot, for all possible motions of the atoms in the molecule. 3. Identify the Irreducible Representation that provides the Reducible Representation. 4. Identify the representations corresponding to translation (3) and rotation (2 if linear, 3 otherwise) of the molecule. Those that are left correspond to the vibrational modes of the molecule. 5. Determine which of the vibrational modes will be visible in an IR or Raman experiment.

Finding the Point Group

Vibrational Spectroscopy and Symmetry Example, the vibrational modes in water. The point group is C 2 v so we must use the appropriate character table for the reducible representation of all possible atomic motions, tot. To determine tot we have to determine how each symmetry operation affects the displacement of each atom the molecule – this is done by placing vectors parallel to the x, y and z axes on each atom and applying the symmetry operations. As with the bonds in the previous examples, if an atom changes position, each of its vectors is given a value of 0; if an atom stays in the same place, we have to determine the effect of the symmetry operation of the signs of all three vectors. The sum for the vectors on all atoms is placed into the reducible representation. Make a drawing of the molecule and add in vectors on each of the atoms. Make the vectors point in the same direction as the x (shown in blue), the y (shown in black) and the z (shown in red) axes. We will treat all vectors at the same time when we are analyzing for molecular motions. top view

Vibrational Spectroscopy and Symmetry Example, the vibrational modes in water. The E operation leaves everything where it is so all nine vectors stay in the same place and the character is 9. z y x The C 2 operation moves both H atoms so we can ignore the vectors on those atoms, but we have to look at the vectors on the oxygen atom, because it is still in the same place. The vector in the z direction does not change (+1) but the vectors in the x, and y directions are reversed (-1 and -1) so the character for C 2 is -1. The v (xz) operation leaves each atom where it was so we have to look at the vectors on each atom. The vectors in the z and x directions do not move (+3 and +3) but the vectors in the y direction are reversed (-3) so the character is 3. The ’v (yz) operation moves both H atoms so we can ignore the vectors on those atoms, but we have to look at the vectors on the oxygen atom, because it is still in the same place. The vectors in the z and y directions do not move (+1 and +1) but the vectors in the x direction is reversed (-1) so the character is 1. C 2 V E C 2 v (xz) ’v (yz) tot 9 -1 3 1 C 2 v (xz) ’v (yz)

Vibrational Spectroscopy and Symmetry C 2 V E C 2 v (xz) ’v (yz) tot 9 -1 3 1 C 2 V E C 2 v (xz) ’v (yz) A 1 1 1 z x 2, y 2, z 2 A 2 1 1 -1 -1 Rz xy B 1 1 -1 x, Ry xz B 2 1 -1 -1 1 y, Rx yz From the tot and the character table, we can figure out the number and types of modes using the same equation that we used for bonding: This gives: Which gives: 3 A 1’s, 1 A 2, 3 B 1’s and 2 B 2’s or a total of 9 modes, which is what we needed to find because water has three atoms so 3 N = 3(3) =9.

Vibrational Spectroscopy and Symmetry Now that we have found that the irreducible representation for tot is (3 A 1 + A 2 + 3 B 1+ 2 B 2), the next step is to identify the translational and rotational modes - this can be done by reading them off the character table! The three translational modes have the symmetry of the functions x, y, and z (B 1, B 2, A 1) and the three rotational modes have the symmetry of the functions Rx, Ry and Rz (B 2, B 1, A 2). Translation Rotational al modes C 2 V E C 2 v (xz) ’v (yz) A 1 1 1 z x 2, y 2, z 2 A 2 1 1 -1 -1 Rz xy B 1 1 -1 x, Ry xz B 2 1 -1 -1 1 y, Rx yz The other three modes (3(3)-6 = 3) that are left over for water (2 A 1 + B 1) are the vibrational modes that we might be able to observe experimentally. Next we have to figure out if we should expect to see these modes in an IR or Raman vibrational spectrum.

Vibrational Spectroscopy and Symmetry Remember that for a vibration to be observable in an IR spectrum, the vibration must change the dipole moment of the molecule. In the character table, representations that change the dipole of the molecule are those that have the same symmetry as translations. Since the irreducible representation of the vibrational modes is (2 A 1 + B 1) all three vibrations for water will be IR active (in red) and we expect to see three signals in the spectrum. The three vibrational modes for water. Each mode is listed with a (Greek letter ‘nu’) and a subscript and the energy of the vibration is given in parentheses. 1 is called the “symmetric stretch”, 3 is called the “anti-symmetric stretch” and 2 is called the “symmetric bend”. For a vibration to be active in a Raman spectrum, the vibration must change the polarizability of the molecule. In the character table, representations that change the polarizability of the molecule are those that have the same symmetry as the second order functions (with squared and multiplied variables). Thus all three modes will also be Raman active (in blue) and we will see three signals in the Raman spectrum. C 2 V E C 2 v (xz) ’v (yz) A 1 1 1 z x 2, y 2, z 2 A 2 1 1 -1 -1 Rz xy B 1 1 -1 x, Ry xz B 2 1 -1 -1 1 y, Rx yz

The Geometry of the Sulfur Dioxide Molecule

Cs structure: 3 normal modes, all having A' symmetry The Cs structure should have 3 IR active fundamental transitions. These three fundamental transitions also should be Raman active. We would expect to observe three strong peaks in the IR and three strong peaks in the Raman at the same frequency as in the IR. All of the Raman lines would be polarized because they are totally symmetric (A' symmetry).

C 2 v structure: 3 normal modes, two with A 1 symmetry, one with B 2 The C 2 v structure should have 3 IR active fundamental transitions. These three fundamental transitions also should be Raman active. We would expect to observe three strong peaks in the IR and three strong peaks in the Raman at the same frequency as in the IR. Two of the Raman lines are totally symmetric (A 1 symmetry) and would be polarized. One Raman line would be depolarized.

The Dooh structure should have two IR active fundamental transitions. It will have one Raman active fundamental transition at a different frequency than either of the IR peaks. . The Raman line will be polarized.

Experimental Observation The experimental infrared and Raman bands of liquid and gaseous sulfur dioxide have been reported in a book by Herzberg 7. Only the strong bands corresponding to fundamental transitions are shown below. The polarized Raman bands are in red. Fundamental 2 1 3 IR (cm-1) 519 1151 1336 Raman (cm-1) 524 1151 1336

Conclusion The existence of three experimental bands in the IR and Raman corresponding to fundamental transitions weighs strongly against the symmetrical linear (Dooh) structure. We usually do not expect more strong bands to exist than are predicted by symmetry. Group theory predicts that both bent structures would have three fundamental transitions that are active in both the IR and Raman. However all three of the Raman lines would be polarized if the structure were unsymmetrical (Cs symmetry). The fact that one Raman line is depolarized indicates that the structure must be bent and symmetrical (C 2 v symmetry). The sulfur dioxide molecule has C 2 v symmetry.

Problems with Raman: a) Very Weak – for every 106 photons only 1 photon Raman a) Resonant Raman not feasible with every sample. b) Absorption a better process than scattering

International and National Patent (2007), G. V. Pavan Kumar et al Current Science (2007) 93, 778. Micro–Raman setup Raman Spectrometers Optical fiber CCD Monochromator Focusing lens Edge filter Camera Computer Dichroic Mirror LASER Objective lens Stage