Lecture Notes on Scattering and Refraction - Optical Spectroscopy | PHYS 552, Assignments of Optics

Download Lecture Notes on Scattering and Refraction - Optical Spectroscopy | PHYS 552 and more Assignments Optics in PDF only on Docsity! 1 Scattering and refraction We have presented the classical theory of spectroscopy and concentrated on its application to absorption and emission of energy, and polarization. This section will concentrate on the use of this theory to describe light scattering and optical phenomena, such as the index of refraction. Scattering from an isolated particle We remember that the light radiated from an oscillating dipole is: ( )( ) 2 0 2 sin coss m rE t crc ω θ ω= − − , where m0 is the amplitude of the oscillating dipole. If an atom or molecule is placed in an oscillating electric field ( )0 cosE tω , and if the frequency of the oscillating field does not correspond to a resonance frequency of the molecule (actually, is far away) then the amplitude of the oscillating dipole is equal to 0Eα , and the equation for the field of the radiation becomes ( )( ) 2 0 2 sin coss rE E t crc ωα θ ω= − − . The intensity of this radiated field is: ( ) ( ) ( )2 2 2 4 4 200 4 sincRs sI E r drd c I dπ α ω θ= Ω = Ω∫ . Rc is the distance light travels in one second in a vacuum. This is the intensity traveling in a certain direction (within a solid angle). To get the total intensity scattered in all directions from this single particle: ( ) 2 2 4 2 4 4 3 0 04 0 0 0 8sin 2 sin 3st s s I I d I d d c I d I c π π π πα ωθ θ ϕ π α ω θ θ= Ω = = =∫ ∫ ∫ ∫ . Homework: This result has assumed that the incoming light is polarized – as homework show that this (the total scattered intensity) is the same if the incoming light is non- polarized. As a first step show that the energy scattered into a solid angle dΩ making an angle χ with the incoming beam is ( ) ( )( )2 4 4 20 1 cosc I dα ω χ+ Ω . We will use this equation later today. 2 Scattering by more than one particle If there are more than one particle, then the fields due to all the oscillating dipoles must be added vectorially, and of course this depends on the geometrical arrangement of the particles (whether this is random, in a crystal, in an amorphously organized sample). So we cannot give a general solution for this. However, there are a couple of important special cases and general characteristics that are useful (we will use these below): 1) The light scattered in the forward direction will always result in constructive interference. This is the beam that is related to refraction (see later). 2) The scattered intensity is always proportional to 4 4or ω λ − . Thus blue light is scattered more than red light (blue color of the sky, and red color of a sunset – Lord Rayleigh – Rayleigh scattering). This is also important to take into account when making absorption measurements in a cuvette with highly scattering samples (macromolecules). Of course in a highly concentrated “microscope sample”, or in biological tissue, this is of prime importance. One has to correct the absorption spectrum for this effect. This will be part of a lab exercise. 3) What happens in a perfect crystal when the wavelength of light is long compared to the molecular (atomic) spacing. If the crystal is homogeneous, and we do not consider fluctuations in spatial locations of the atoms, then the waves of the sidewise scattered light will all be out of phase, and the result is complete cancellation of the scattered fields, except for the forward scattering (think of crystal optics). 4) If there is a random arrangement of the atoms (molecules) such as in a gas, then the intensity of the detected scattered light will be proportional to the number of scattering centers. Heuristically, the scattered light will be in phase half the time and out of phase half the time, so the average of the energy from two particles (which is proportional to E2) is (0 + 4) / 2 = 2. And so on for all the particles. This the reason for the scattering from a fog. The classical theory as presented above can be used to explain a number of different phenomenon that are related to (or part of – depending on your point of view) optical spectroscopy. We will discuss some of these shortly: 1) scattering by density fluctuations in pure liquids and solids 2) concentration fluctuations from solutions of particles (molecules) 3) Small-angle scattering by particles 4) Refraction of light due to scattering 5) X-ray diffraction – we will not discuss this. 5 Concentration fluctuations from solutions of particles (molecules) This is done with the same method as the last section. We just have to change the expression for the ( )2εΔ . This is now the mean square fluctuation due to a fluctuation of the particle concentration in a selected cell of volume V. ( ) ( )( )d dN N d dC dC dN Nε ε εΔ = Δ = Δ , where N is the number of particles in V, C is the concentration in grams/cm3, and ΔN is the fluctuation of the number of particles coming into the volume V from the bulk solution. If m is the mass of an individual particle, then ( )dC dN m V= . Now we use a statistical mechanical argument, calculating the free energy fluctuation due to the fluctuation in the number of particles. We use the chemical potentials (μ) to calculate the change in free energy change (ΔA) per change in the number of particles, and using ( )2A kT NαΔ = Δ , where ( ) ( )2m d dC VkTα μ= , we arrive at: ( ) ( ) ( ) ( ) ( ) 2 2 22 2 2 12 2 2 2 2 1 2 N N N e d Nd m d m m d dkT dC V dC V V dC dCe d N α α ε ε ε με α ∞ − Δ − −∞ ∞ − Δ −∞ Δ Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞Δ = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠Δ ∫ ∫ . We can substitute this into ( ) ( ) 2 22 04 1 cos2s I I d Vπ χ ε λ = + Ω Δ to give: ( ) ( ) 2 1 2 12 2 2 2 0 04 41 cos 1 cos2 2s m d d d dI I d kT V I d mkT V dC dC dC dC π ε μ π ε μχ χ λ λ − − ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + Ω = + Ω⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ This is the scattering of a solution of particles in the direction χ and in the solid angle dΩ. We have, at these frequencies, ( )2d dC n dn dCε = ; if the solution is dilute enough to be ideal, we also have 0 lnkT Cμ μ= + , giving d dC kT Cμ = . Therefore the expression for the scattering from a solution of dilute particles is: ( ) 22 2 2 04 2 1 coss dnI I d n mC dC π χ λ ⎛ ⎞= + Ω ⎜ ⎟ ⎝ ⎠ . Peter Debye noted that the scattering is proportional to the molecular mass of the solute, and can be used for determining the molecular weight. All you have to do is to determine a value for dn dC , and the energy scattered into an angle (usually 90 degrees) relative to the incoming light beam (or the integrated scattered energy, if you can). The value for dn dC is determined with a simple refractometer. sI is proportional to 41 λ . You must correct this in UV absorbance measurements of large biological molecules! Homework: 0.1 gm of a polymer is dissolved into 100 cc of solution (water). The refractive index of the water is changed by 0.00019. The wavelength of the light is 4360 Angstrøm. The total energy of light that is scattered passing through 1 cc of the solution is 0.001% of the incident light. What is the molecular weight of the polymer? 6 Small angle scattering by particles: This will be a qualitative discussion. The light scattered at angles other than the forward direction is small if the wavelength of the light is large compared to the particle size. But the forward scattering will interfere with itself and with the incoming light. The atoms (molecules) at a particular plane perpendicular to the incoming plane wave will all vibrate electronically at the same phase, therefore leading to interference. Consider a cube that is divided in stripes the width of the wavelength of light. The lens will gather the light and focus the light that is scattered from planes perpendicular to the direction of the light travel. So the focused light from a cube where the incoming planar light wave differs by one wavelength will cancel due to the differences in phase (covering all the phases between 0 and 2π ). This light is at an angle of ( )11 tan Lθ λ−= , where L is the width of the cube. However, the light perpendicular to a plane intersecting the wave fronts at a smaller angle will only partially interfere with each other, and this will result in light spots where the focused light does not interfere destructively. This will have a maximum at an angle ( )12 tan 2Lθ λ−= . The cube has to be much larger that the wavelength of the light, and then there will be a diffraction pattern at the screen where the light is focused, and this will have the highest intensity in the center spot, where the dark ring around this center spot has an angle of 2 2 Lϕ θ λ= . This will of course only happen if the wavelength of the light is on the order of the diameter of the particles. So, for proteins we are talking about X-rays and this is small angle x-ray scattering. Provided that we are talking about an ensemble of particles all with the same size and structure, we will see these peaks, and by analyzing the shape of the wings of the focused scattered light we can estimate the size of particles (molecules) and even the internal structure sometimes. We will not go into the details (see J.T. Edsall, in The Proteins, H. Neurath and K. Bailey, eds., Vol I, part B, Academic Press, New York, 1953). 7 Refraction of light from light scattering (using the above formalism) Consider one layer of molecules in a plane perpendicular to the direction of the light plane wave. If the incident light has a field of ( )( )0 cos rE E t cω= − then the dipole of one molecule is ( )( )0 cos rm E t cα ω= − . This dipole scatters as: ( )( ) ( )( ) 2 2 0 02 2 4sin cos sin coss r rE t E t Ec crc r αω πθ ω θ ω λ ⎛ ⎞ ⎛ ⎞ = − − = − −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ . Without going into the derivation, we can write the scattered wave of the complete array of atoms (molecules) as ( )( )0' sinsE zE t z cγ ω= − Δ − , where 24 Nγ π α λ= . Note the cosine wave of each particle has become a sine wave – that is, there is a shift of 90 degrees. Now the total field of both the incident and scattered wave is just their sum: ( )( ) ( )( )0' cos sintot sE E E E t z c z t z cω γ ω⎡ ⎤= + = − − Δ −⎣ ⎦ Then, defining tan zωτ γ= Δ (see why below) we have ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) 0 0 cos tan sin cos cos sin sin / cos totE E t z c t z c E t z c t z c ω ωτ ω ωτ ω ωτ ω ωτ ⎡ ⎤= − − −⎣ ⎦ ⎡ ⎤= − − −⎣ ⎦ . Now using a simple trigonometric identity, ( )( ) ( ) ( )( ) ( ) 0 0 1 cos cos cos cos tantot E t z c E t z c E z ω τ ω τ ωτ γ− ⎡ ⎤+ − + −⎣ ⎦= = Δ . When zγΔ is very small, 1tan zγ− Δ is very small, and ( )1cos tan 1zγ−⎡ ⎤Δ ≈⎣ ⎦ . This gives us: ( )( )0 costotE E t z cω τ+ − This equation says that the arrival of the electric field at position z is delayed by τ from that if there were only a vacuum present. The delay is dependent on the thickness of the layer (remember tan zωτ γ= Δ , or ( )1tan /z zτ γ ω γ ω−= Δ ≈ Δ ).