The Interaction of Light and Matter:  and n The interaction of light and matter is what makes life interesting. Everything we see is the result of this interaction. Why is light absorbed or transmitted by a particular medium? Light causes matter to vibrate. Matter in turn emits light, which interferes with the original light. Destructive interference means absorption. Mere out-of-phase interference changes the phase velocity of light, or refractive index. Excited atoms & the forced oscillator Complex Lorentzian: 1/(  -i  ) Absorption coefficient, , and refractive index, n. 0  Absorption coefficient Refractive index 0 n–1 Frequency, 

Waves using complex amplitudes The resulting "complex amplitude" is: As written, this entire field is complex!

Complex numbers simplify optics! This isn't so obvious using trigonometric functions, but it's easy with complex exponentials: where all initial phases are lumped into E 1, E 2, and E 3. Adding waves of the same frequency, but different initial phase, yields a wave of the same frequency.

When two waves add together with the same complex exponentials, we add the complex amplitudes, E 0 + E 0 '. Adding complex amplitudes Slower phase velocity LaserAbsorption + = time Destructive interference: = time Constructive interference: + = time "Quadrature phase" ±90° interference: i 1-0.2i

Light excites atoms, which emit light that adds (or subtracts) with the input light. When light of frequency  excites an atom with resonant frequency  0 : An excited atom vibrates at the frequency of the light that excited it and re-emits the energy as light of that frequency. The crucial issue is the relative phase of the incident light and this re-emitted light. For example, if these two waves are ~180° out of phase, the beam will be attenuated. We call this absorption. Electric field at atom Electron cloud Emitted field On resonance (  =  0 ) + = Incident light Emitted light Transmitted light

The Forced Oscillator When we apply a periodic force to a natural oscillator (such as a pendulum, spring, swing, or atom), the result is a forced oscillator. Examples: Child on a swing being pushed Pushed pendulum Bridge in wind or an earthquake Electron in a light wave Nucleus in a light wave The forced oscillator is one of the most important problems in physics. It is the concept of resonance. Tacoma Narrows Bridge collapsing because oscillatory winds blew at its resonance frequency.

The Forced Oscillator: The relative phase of the oscillator motion with respect to the input force depends on the frequencies. Below resonance  <<  0 ForceOscillator On resonance  =  0 Above resonance  >>  0 Let the oscillator’s resonant frequency be  0, and the forcing frequency be . We could think of the forcing function as a light electric field and the oscillator as a nucleus of an atom in a molecule. Weak vibration. In phase. Strong vibration. 90° out of phase. Weak vibration. 180° out of phase.

The relative phase of an electron cloud’s motion with respect to input light depends on the frequency. Below resonance  <<  0 Electric field at atom Electron cloud On resonance  =  0 Above resonance  >>  0 Let the atom’s resonant frequency be  0, and the light frequency be . The electron charge is negative, so there’s a 180° phase shift in all cases (compared to the previous slide’s plots). Weak vibration. 180° out of phase. Strong vibration. -90° out of phase. Weak vibration. In phase.

The relative phase of emitted light with respect to the input light depends on the frequency. Below resonance  <<  0 Electric field at atom Electron cloud On resonance  =  0 Above resonance  >>  0 The emitted light is 90° phase-shifted with respect to the atom’s motion. Emitted field Weak emission. 90° out of phase. Strong emission. 180° out of phase. Weak emission. -90° out of phase.

The Forced Oscillator: Math Consider an electron on a spring with position x e (t), and driven by a light wave, E 0 exp(-i  t) : So the electron oscillates at the incident light wave frequency (  ), but with an amplitude that depends on the difference between the frequencies. The solution is:

Checking our solution Substitute the solution for x e (t) into the forced oscillator equation to see if it works. QED

The problem with this model Exactly on resonance, when  =  0, x e goes to infinity. This is unrealistic. We’ll need to fix this.

The Damped Forced Oscillator Our solution has infinite amplitude on resonance, which is unrealistic. We fix this by using a damped forced oscillator: a harmonic oscillator experiencing a sinusoidal force and viscous drag. The electron still oscillates at the light frequency and with a potential phase shift, but now with a finite amplitude for all . The solution is now: We must add a viscous drag term:

Why we included the damping factor,  Atoms spontaneously decay to the ground state after a time. Also, the vibration of a medium is the sum of the vibrations of all the atoms in the medium, and collisions cause the sum to cancel. Collisions "dephase" the vibrations, causing cancellation of the total medium vibration, typically exponentially. (We can use the same argument for the emitted light, too.) time

Assuming, this becomes: The atom’s response is approximately a Complex Lorentzian. Consider: In terms of the variables    and  = , the function 1/(  – i  ), is called a Complex Lorentzian. Its real and imaginary parts are:

Complex Lorentzian

Damped forced oscillator solution for light-driven atoms The forced-oscillator response is sinusoidal, with a frequency- dependent strength that's approximately a complex Lorentzian: Here, e < 0. When  <<  0, the electron vibrates 180° out of phase with the light wave: When  =  0, the electron vibrates -90° out of phase with the light wave: When  >>  0, the electron vibrates in phase with the light wave:

The relative phase of an electron cloud’s motion with respect to input light depends on the frequency. Below resonance  <<  0 Electric field at atom Electron cloud On resonance  =  0 Above resonance  >>  0 Let the atom’s resonant frequency be  0, and the light frequency be . The electron charge is negative, so there’s a 180° phase shift in all cases (compared to the previous slide’s plots). Weak vibration. 180° out of phase. Strong vibration. -90° out of phase. Weak vibration. In phase.

Okay, so we’ve determined what the light wave does to the atom. Now, what does the atom do to the light wave?

Re-emitted light from an excited atom interferes with original light beam The re-emitted light may interfere constructively, destructively, or, more generally, somewhat out of phase with the original light wave. We model this process by considering the total electric field, Maxwell's Equations will allow us to solve for the total field, E(z,t). The input field will be the initial condition. E(z,t) = E original ( z,t ) + E re-emitted ( z,t ) + = Incident light Emitted light Transmitted light z

The Inhomogeneous Wave Equation The induced polarization,, is due to the medium: where: and e is the electron charge, and N = the electron number density. For our vibrating electrons: E(z,t)E(z,t) x We often also write: where: or:

The Slowly Varying Envelope Approximation For now, we'll assume that the wave won't be dramatically affected by the induced polarization, which itself will not change in time. Let Specifically, the envelope, E 0 (z), is assumed to vary slowly; the fast variations will all be in the complex exponential, The time derivatives are easy (as before, they just bring down a –  2 ) because the envelopes are independent of t. The z -derivatives: Constant in time Because variations of the envelope, E 0 (z), in space will be slow, we’ll neglect the 2 nd z -derivative.

The Slowly Varying Envelope Approximation Substituting the derivatives into the inhomogeneous wave equation: Now, because k =  /c, the 2nd and 3rd terms cancel. And canceling the complex exponentials leaves: If P 0 is constant, the integration is trivial. Usually, however, P 0 = P 0 (E 0 ) and hence P(z), too. Or: Converting to finite differences, the re-emitted field at a given z will be: Note the i, which means that the re-emitted field has a 90° phase shift with respect to the polarization and hence a -90° phase shift with respect to the electron cloud motion.

The re-emitted wave leads the electron cloud motion by 90° in phase. Below resonance  <<  0 Electric field at atom Electron cloud On resonance  =  0 Above resonance  >>  0 This phase shift adds to the potential phase shift of the electron cloud motion with respect to the input light. Emitted field Weak emission. 90° out of phase. Strong emission. 180° out of phase. Weak emission. -90° out of phase.

Solving for the slowly varying envelope Substituting for P 0, the SVEA becomes: This differential equation has the solution: where  is the absorption coefficient and n is the refractive index. so that: Define new quantities for the real and imaginary parts of  :

The complete electric field in a medium: The electromagnetic wave in the medium becomes (combining the slowly varying envelope with the complex exponential): Simplifying: Absorption Refractive index attenuates the field changes the k-vector E0(z)E0(z)

A light wave entering a medium Typically, the speed of light, the wavelength, and the amplitude decrease, but the frequency, , doesn’t change. n = 1n = 2 k0k0 nk 0 Vacuum (or air)Medium Absorption depth = 1/     n Wavelength decreases

The irradiance is proportional to the (average) square of the field. Since E(z)  exp(-  z/2), the irradiance is then: Absorption Coefficient and the Irradiance where I(0) is the irradiance at z = 0, and I(z) is the irradiance at z. Thus, due to absorption, a beam’s irradiance exponentially decreases as it propagates through a medium. The 1/e distance, 1/ , is a rough measure of the distance light can propagate into a medium. I(z) = I(0) exp(-  z)

Refractive index and Absorption coefficient n comes from the real part of  : Simplifying:  comes from the imaginary part of  : These results are valid for small values of these quantities.

Refractive index and Absorption coefficient Frequency,  00

Different atoms have different resonance frequencies,  0, and widths, .

Molecules have a higher density of states than atoms and so have much more complex absorption spectra. Due to the finite width of absorptions, these levels can overlap. In any case, a batch of identical molecules can be highly absorbing over a large spectral range. Ground electronic state 1 st excited electronic state 2 nd excited electronic state Energy Actually, it’s not correct to broaden a single level; it’s really each pair of levels that broadens.

The absorption of glass Transmission ranges for various glasses It’s difficult to find transparent materials below 100 nm and above 70  m.

Absorbing glass filters A wide range of absorbing glass filters allow us to manipulate the spectrum of a light beam. A red filter transmits red light, etc.

What determines the color of an object? The wavelengths that are not absorbed will be reflected and/or scattered into our eyes.

Why are most plants green and then red or yellow in the fall? Chlorophyll absorbs in the red and blue, and hence reflects in the green. But it breaks down in the fall. In the fall, trees produce carotenoids, which reflect yellow, and anthocyanins, which reflect orange and red. Biologists believe that these molecules ward off insects.

Materials that are transparent in the IR can be opaque in the visible. And materials that are opaque in the visible are often transparent in the IR. (You can try this at home with your IR remote control!)

By choosing a high- power laser that is absorbed by a material, we can laser-weld or cut. Here, a CO 2 laser, which lases at 10.6  m, cuts metal. Laser surgery works on the same principle.

The sun’s emission spectrum in the visible The dark lines below represent absorption due to elements on the surface of the sun from its various constituents.

Absorption spectrum of air Air consists of numerous molecules that are non-absorbing in the visible, but which can absorb very strongly elsewhere. Note that the vertical axis is a log of a log scale!

Water’s absorption distance vs. wavelength Visible spectrum Penetration depth into water (1/  ) Wavelength 1 km1 m1 mm1 µm1 nm UV X-ray Radio Microwave IR 1 mm 1 km 1 µm 1 m Penetration depth into water vs. wavelength Notice that the penetration depth varies by over ten orders of magnitude! Water is clear in the visible, but not elsewhere.

Measuring Lake Purity Secchi Disk Measure the absorption vs. depth using reflection off a white disk lowered into the lake.

Absorption Spectrum of Human Tissue 1.3  m 650 nm

Illuminating a water droplet with different wavelengths Visible light: 2.78-µm light (ErCr:YSGG): Light passes through droplet; no energy is absorbed. Nothing happens to droplet. Light is absorbed in several microns. Water is vaporized, yielding massive droplet acceleration in the forward direction. Water droplet Excited region Excited region vaporizes and expands rapidly. Expansion accelerates rest of droplet forward rapidly. Vapor Liquid Laser-accelerated water droplets are now being used by dentists to cut teeth!

Refractive Index vs. Wavelength Since resonance frequencies exist in many spectral ranges, the refractive index varies in a complex manner. Electronic resonances usually occur in the UV; vibrational and rotational resonances occur in the IR, and inner-shell electronic resonances occur in the x-ray region. n increases with frequency, except in anomalous dispersion regions.

Refractive indices for glasses in the UV, visible, and IR We’ll use n = 1.5 for the refractive index of the glass we usually encounter.

Refractive Index and the Speed of Light The speed of light is   /k. Since k 0 becomes k = nk 0 in a medium, where c 0 is the speed of light in vacuum. The refractive index, n, of a medium is thus the ratio of the speed of light in vacuum to the speed of light in the medium. It can be defined as the ratio: The refractive index is usually > 1. But it can be < 1.

The Irradiance in a medium The irradiance: Substitute for c in the medium: c = c 0 /n Now we need  n = c 0 /c, but c 0 = 1/ √  0  0 and c = 1/ √  and  0 so: n = √  /√  0  =  0 n 2 So the irradiance in the medium becomes or:

Refraction and Snell's Law The electric field (and its wave-fronts) are continuous at a boundary. But the speed of light will be different in the two media. AD = BD/sin(  i ) AD = AE/sin (  t ) So: BD/sin (  i ) = AE/sin (  t ) But: BD = v i  t = (c 0 /n i )  t & AE = v t  t = (c 0 /n t )  t So: (c 0 /n i )  t / sin (  i ) = (c 0 /n t )  t / sin (  t ) Or: n i sin (  i ) = n t sin (  t ) nini ntnt ii tt D B A E v i  t v t  t ii tt

Snell’s Law causes things to look bent in water.

Snell's Law explains many everyday effects The refractive index increases with density (and hence decreases with temperature). Occasionally, astronomers speculate whether light can orbit a planet somewhere in the universe… This effect may play a role in mirages.

Snell’s Law explains why stars twinkle. Star Cooler regions of air (with higher refractive index) The atmosphere has non-uniform temperature and hence non- uniform refractive index. And these regions move about in time. As the air masses move about, the amount of light reaching our eyes from the star varies.

Snell's Law for many parallel layers So we can ignore the intermediate layers if we’re only interested in the output angle! If the layers are parallel, then these angles are always equal. n1n1 n2n2 n3n3 n4n4 n5n5 11 55 22 22

Refraction allows prisms to expand or compress a beam. Entering at an oblique angle and exiting at near-normal incidence yields magnification. d int d out d in Magnification (or demagnification) occurs at both surfaces.

Dispersion of the refractive index allows prisms to separate white light into its components and to measure the wavelength of light. Dispersion is the tendency of optical properties to depend on frequency. Dispersion can be good or bad, depending on what you’d like to do. Dispersive element White light Dispersed beam n( )

The Sellmeier Equation approximates the refractive-index curve for most materials. CoefficientValue B B x10 -1 B C x10 -3 C x10 -2 C x10 2 These values are obtained by measuring n for numerous wavelengths and then curve-fitting.

Practical numbers for material dispersion dn/d  m 

Dispersion is undesirable in lenses. Unfortunately, dispersion also causes lenses to focus different colors at different distances, making lens design difficult. This is called chromatic aberration. Parallel red and blue input rays The lens refractive index is higher for blue, so f is smaller for blue. f( ) Lens designers go a to great lengths to compensate dispersion.

Because the refractive index depends on wavelength, the refraction angle also depends on wavelength: this is called angular dispersion. Snell’s Law: Prisms disperse light. Prism Input white beam Dispersed beam n( ) Suppose light enters the prism at 0°, and  t int is its incidence angle at the exit face. Differentiating implicitly w.r.t. : We obtain the prism dispersion: Because n generally decreases with wave- length ( dn/d < 0 ), the shorter the wavelength, the greater the refraction angle.

More generally, prisms disperse light at both surfaces. The dispersion of a general prism is: Dispersion at exit face Magnification at exit face Dispersion at entrance face

The dispersion of a series of prisms Let D i be the dispersion of the i th prism ( D i is positive for upward-pointing prisms and negative for downward-pointing prisms) and M i be the magnification of the i th prism. The total dispersion of all the prisms will be: The contribution of each prism’s dispersion is reduced by the total magnification after it! D2 M2D2 M2 D3 M3D3 M3 D4 M4D4 M4 D5 M5D5 M5 D1 M1D1 M1

Rainbows result from dispersion in the refraction of sunlight in water droplets. Note that there can be two rainbows, and the top one is inverted. And the sky is much brighter below the bottom one.

Rainbow explanation: Light in a spherical droplet Light can enter a droplet at different distances from its edge. We must compute the angle of the emerging light as a function of the incident position. Light paths Water droplet 180° 165° 150° 138° 150° 160° Path leading to mini- mum deflec- tion Minimum deflection angle (~138°) For a cool java applet showing this, try

Plotting deflection angle vs. wavelength is the key. Lots of violet deflected at this angle Lots of red deflected at this angle Lots of light of all colors is deflected by >138°, so the region below rainbow is bright and white. Because n varies with wavelength, the minimum deflection angle varies with color.

Explanation of 2nd rainbow The 2nd (upper) rainbow results from light entering the droplet in its lower half and making 2 internal reflections in the droplet. Minimum deflection angle (~232.5°) yielding a rainbow radius of 52.5°. Water droplet Because energy is lost at each reflection, the 2nd rainbow is weaker. 3rd and 4th rainbows are weaker, more spread out, and toward the sun. 5th rainbow overlaps 2nd, and 6th is below the 1st, but too weak to see.