General

Question 1

What factors influence light-solid interaction

Answer 1

Light matter interactions are fundamentally a function of the material and the light source where the optical properties are described through the refractive index

Question 2

Light (light-matter interactions)

Answer 2

This is a transverse electromagnetic wave with an electric and magnetic field which oscilate perpindicular to the direction of propogation at an optical frequency of 10¹³-10¹⁷ hz which includes UV, visible, and IR.

Fundamentally light is EMR governed by Maxwell’s equations and produced by the acceleration of electric charges.

Question 3

Solids (light-matter interactions)

Answer 3

These are considered to be materials of positive and negative charged particles with inducible dipoles

Question 4

Interaction (light-matter)

Answer 4

The fundamental interaction of light and matter is that light polarizes the atoms by nature of its transverse electromagnetism to produce oscillating dipole moments. The interaction produces either reflection, transmission, absorption, or scattering.

Question 5

Dipole moments

Answer 5

This is defined as qx where q is the partial charge and x is the distance of the partial charge from equilibrium

Question 6

Electronic oscillations

Answer 6

This is the concept of electrons surrounding the nucleas behaving like springs where they oscillate about the positively charged protons. In this model there are several electrons attached to springs surrounding the nucleas that oscillates like a hula hoop.

Because mass of electrons is so much less than that of the nucleas the shifting electronic field of EMR makes the electrons shift with respect to the nucleas and not vice versa

Question 7

Resonant frequency

Answer 7

This describes the frequency of light that coincides with the frequency of oscillations in order to create constructive interference.

Question 8

Resonant frequency of electronic oscillations

Answer 8

ω is proportional to (1/m_e + 1/m_n)^.5 = (1/μ)^.5 where m=mass and μ = reduced mass. Because me<

Question 9

Resonant frequency of electronic oscillations speed

Answer 9

These are the easiest to induce and therefore and the fastest and have frequencies in the UV and visible spectra. They are a function of collisions, electronegativity, and orbital arrangements

Question 10

Vibrational oscillators

Answer 10

This is a type of dipole based on charged atoms that are bonded to one another vibrating from their equilibrium position. It is dependent on the rigidity of the bond and the degree of covalent nature.

Because the speed of this type of action is slower the resonant frequency is in the IR range. The oscillating atoms create phonons.

Question 11

Quantum Theory version of vibrational oscillators

Answer 11

Quantum theory says that electrons do not vibrate in a bond but are elevated from the inner orbitals into the conductance bands at the resonant frequency where they emit photons when returning to the ground state

Question 12

Free electron oscillator’s

Answer 12

These are electrons that are within the outer orbits of large atoms (like metals) that can oscillate from light interactions without any true restoring force. These oscillations have a resonant frequency of 0.

In metals these are known as the conductance electrons and are responsible for the reflective nature of metals.

Question 13

Polarization of a Medium Equation

Answer 13

P = ε_o χ E where

P=Polarizability of the medium (dependent variable)

ε_o = permittivity of a vacuum =

χ = Dialectric susceptibility of the medium

E = electric field that the material is within

This is the basic way of describing how a material behaves but is best applied to net effects

Question 14

Diaeletric

Answer 14

A dielectric is an insulator that can be polarized in an electric field not because of electron flow but electron drift. The negative charges in atoms align with the negative end of the electric field and the positive charges to the positive side of the field.

These are effectively capacitors

Question 15

Electric Susceptibility

Answer 15

χ represents the electric susceptibility of a dielectric material. It is a property of the material that represents the ease at which is polarizes in response to an electric field.

It directly influences the electric permittivity of the material.

Question 16

Vaccum permittivity

Answer 16

This represents the ability for an electric field to exist in a vacuum. It is given in terms of F/m which represents the capacitance per meter of space.

ε_o = 8.854 * 10^-12

Question 17

Polarizability of a medium meaning

Answer 17

This represents the density of moment dipoles in the medium. Thus, it is given in C/m² It represents how easily a material accumulates charge at a point that is not equilibrium.

Moment dipole = qx

Question 18

3 levels of polarizability with EMR

Answer 18

At low frequency (radio/microwave) χ (magnetic susceptibility) is highest because the wavelength is long enough for materials to reorient to constructively interfere with the EMR. It is a rotational shift.

In the IR/Vis there is vibrational and electronic dipoles. It slowly decreases with frequency in this space. In the lower frequencies (IR) the vibrational dipoles create heat through distortion of the bonds. This is phonons

In the visible and higher frequencies only, electronic vibrations occur because these are the fastest so they can occur rapidly. They resonate with electronic vibrations to create photons.

Question 19

polarizability with consideration of induced dipoles

Answer 19

P = ε_o ( χ *E + χ²*EE + χ³*EEE+…)

This equation fundamentally says that the various oscillations of the atoms and charged particles induce impacts onto the polarizability of a material

In most optics (except lasers nothing above the second order term matters.

This represents that the polarizability of the material in electronic oscillations is influenced by the induced electric fields produced within oscillations.

Question 20

The Dipole Oscillator Model

Answer 20

This says that within electronic vibrations we consider the electron to fluctuate with their natural resonant frequency (ω_o) with dampening. Their position is given by

m_e ( d²x/dt² + γ (dx/dt) + (ω_o)²x) =qE

where γ = dampening constant and x is given by the motion of the electron from the equilibrium point along the vector of E

This fundamentally says that F = ma where a = d²x/dt²

Question 21

Dipole moment of a Lorentz oscillator (e with springs)

Answer 21

For N = #of charges/volume

P = Σ qx = E*N [q²/m /(ω_o² - ω² + iγω)] where ω = the frequency of light and ω_o =natural frequency of the oscillator

This is found by using E = E_o e^iωt as an induced oscillation created by the EMR with the dipole oscillator equation. It represents how polarizable a material is within an EMR interaction

Question 22

Real Refractive index of a material with 1+ oscillator

Answer 22

n² = 1 + (q²/ε_om) (Σ_iⁿ N_i/(ω_i² -ω + iγ_i ω)

where n = refractive index

ε_o = permittivity of an electric field in a vacuum

N_i = number of i atoms

ω_i = resonant frequency of atom i

γ_i = dampening of the atom i

ω = frequency of light

Question 23

The Complex Refractive index

Answer 23

n = n + ik

n = real refractive index = c/v

k = extinction coefficient which is proportional to the absorption coefficient (α) which is related to the dissapation/attenuation of light

Question 24

How does n vary with ω

Answer 24

IF ω i then n is ~constant and low

IF ω = ω_i then n is increases rapidly because iγ_i ω is ~0. This produces a resonance line.

IF ω > ω_i then n is ~ 1

Question 25

How does n vary with frequency in visible light?

Answer 25

n is proportional to c* ω so at higher ω n increases which changes the angle of refraction and creates dispersion

Question 26

When are materials transparent?

Answer 26

This occurs between resonant frequencies where n ~n and thus almost no dampening occurs because dampening is given by the term iγ_i ω

Question 27

Electric field for a light wave

Answer 27

E = E_o e^{i(kz - ωt)} where

k = the wave vector = nω/c if transparent = (n + iK)(ω/c) with absorption

Plugging this in

E = E_o e^-(ωKz)/c e^{i(ωnz/c - ωt)}

Where E_o e^-(ωKz)/c is the amplitude and e^{i(ωnz/c - ωt)} represents the real refractive index of a wave

The important part about this function is that it exponentially decays with z

Question 28

Absorption coefficient

Answer 28

α = 2Kω/c where K = extinction coefficient ω = angular frequency of light

Question 29

Claussis Masetti equation for isotropic media

Answer 29

n² - 1 / n² + 2 = χ/3 This says that for sufficiently dense media that the dipoles of nearby atoms influence the polarization of each dipole. χ is the electric susceptibility which is directly related to P

Question 30

Four Measurable Optical Parameters

Answer 30

You can measure reflection, absorption, scattering, and transmission

Question 31

Reflection

Answer 31

R = I_R/I_o = [(n-1)² + k²]/ [(n+1)² + k²]

where: I = beam intensity (lumens)
* n*=real refractive index of the material

k = extinction coefficient ~0 for transparent materials

Question 32

Absorption

Answer 32

This occurs when the angular frequency of light = resonant frequency and vibrational oscillations lead to changes in kinetic energy due to attenuation of the wave (friction)

α = attenuation efficiency

Question 33

Absorption Equation

Answer 33

This is given by Beer’s law which says

I(z) = I_o e^-αz so it decays exponentially with α being the absorption efficiency coefficient.

Question 34

Scattering

Answer 34

This is the idea that certain light is not attenuated in a solid but the waves are split and change direction which lowers E

Question 35

Scattering Equation

Answer 35

I(z) = I_o e^-Sz Where: S = scattering coefficient

S is proportional to 1/λ⁴

Question 36

Scattering and attenuation

Answer 36

Because we cannot measure the difference between S and α we say α_T = S + α but S is ~0

Question 37

Conservation of Energy when measuring light

Answer 37

For light with an intensity of I_o upon entering a material

1 = (I_R/I_o) + (I_{<span>T</span>}/I_o) + (I_A/I_o) = R + T + A

Where R = reflection, T = transmission, and A = absorption

Question 38

Transmission equation

Answer 38

T = (1-R)² e^-αl

Where: R = reflection coefficient

α = attenuation efficienvy

l = thickness of the material

T = proportion of light transmitted

Question 39

Tool to measure light-matter interactions

Answer 39

A spectrometer is used to measure light-matter interactions by measuring I_o , I_R and I_T as a function of wavelength

Question 40

Fourier Transform Infrared (FTIR)

Answer 40

This uses a glow bar to create a light source over IR and NIR for spectrometry

Question 41

UV-VIS double beam spectrometer

Answer 41

This uses a tungsten and deuterium lamp for analyzing UV to NIR

Question 42

For dialectrics and semiconductors how does A change with wavelength?

Answer 42

At the lower bound of transparency the short wavelengths are absorbed by electron oscillators which are excited to the conductance band.

At the upper bound is when atomic vibrations begin to occur which create heat in the IR range. Thus a higher atomic mass leads to a greater maximum wavelength

Question 43

Transparency

Answer 43

When materials are transparent (when resonant frequency matches the angular frequency of light) k (extinction coefficient) is very small which means that n ~ n

n is the real refractive index and is a function of N, the number of electronic oscillators within a volume of material where an increase in N leads to greater n and greater R. Thus transparent materials with a higher electron density (like semiconductors) are more reflective

Question 44

Ways to find n

Answer 44

n is principally found via snell’s law that says n₁ sin (θ₁) = n₂ sin (θ₂​) where monochromatic light needs to be used to find n because n = f(λ)

There is also ellipsometry which uses the properties of a thin film and changing angles of incidence to derive n

Question 45

Refractometers: Use and types

Answer 45

These are used to find the angle of refraction for an incident ray to find the real refractive index.

Abbe refractometers use a glass hemisphere

V-block refractometers use a V-shaped prism

Prism gonimeter uses a prism specimin and a prism of glass

Question 46

When is ellipsometry preferred for finding n

Answer 46

n is better found using ellipsometry when investigating light that is at a frequency above the transparency frequency (where electrons are excited across the conductance band and absorption dominates)

This would be most useful for semiconductors that have smaller e_g gaps so they become opaque at lower frequencies

Question 47

Plasma frequency

Answer 47

ω_p² = plasma frequency = Nq²/ε_om_e and n² = 1-ω_p²/ω²

This is the frequency that metals stop reflecting light. It is usually near UV.

When ω< ωp n² is imaginary so it is opaque

When ω> ω_p n² is real so it transmits light

Question 48

n² formula for metals

Answer 48

n² = 1-q²N/ε_o m ω

Question 49

Assumptions for metals in the lorentz model of e^- oscillators

Answer 49

This assumes that there is not a dampener because electrons flow and ω_o = 0

It also assumes that all electrons are equal

Question 50

Plasma frequency formula

Answer 50

1-(q²N/ε m)= ω_p²

Question 51

Classes of Diffraction Gratings

Answer 51

Diffraction Grating are either reflective or transmissive

Question 52

Diffraction Gratings

Answer 52

These are optical elements with repetitive parallel patterns embedded into a grating that create dispersion which is used to separate wavelengths of light.

Question 53

Transmissive Diffraction Gratings

Answer 53

These function by having many tiny slits which influence the output waves to create interference

Question 54

Diffraction Grating Formula

Answer 54

a[sin(θ_m -+sin θ_i] = +-mλ

Where λ = wavelength of interest

θ_m = angle from the surface normal to the transferred ray

θ_i = angle to normal of the incident ray

Question 55

Sign convention for θ

Answer 55

If both rays exit the transmission grating on the same side of the surface normal then they are both positive

Question 56

Sign convention for grating formula

Answer 56

Both signs are negative for transmission and positive for reflection

Question 57

Order of Principle Maxima

Answer 57

Diffraction grating create diffraction patterns where m = the order of principle maxima which refers to which “hump” it is on the diffraction pattern. At every m not equal to 0 the light wavelength is a function of θm

Question 58

Reflective gratings

Answer 58

This usually describes a metal deposited onto an optic with parallel groves ruled onto the surface. This has a specular reflection given by θ_r

Question 59

Blazed Gratings

Answer 59

These are gratings that use a step-like pattern to, in theory, maximize efficiency at one value of m for the blaze wavelength (design wavelength). Instead of the light energy being dispersed over many humps, it is concentrated on one m. They do this by having each reflective surface be individually tapered.

Question 60

Blaze grating anatomy

Answer 60

a = length of one step

blaze arrow = the “up” direction where the non-lee side faces

γ = the angle from the surface normal to the step normal.

Question 61

Blazed Grating zero order geometry

Answer 61

For m=0 -θr = θi

Question 62

Blazed grating general relation between θr and θi

Answer 62

If theta is on the same side of the normal then θr <0 and θi-θr = 2*γ

Question 63

Blazed Grating θi =0

Answer 63

When the angle of incidence = 0 then a * sin (-2*γ ) = mλ

Question 64

Littrow Configuration Equation

Answer 64

2a sin(θL) = mλ_D

θL = γ

where θL = Littrow configuration angle for the design wavelength λ_D

Question 65

Angular separation formula

Answer 65

ΔλFSR = λ/m

ΔλFSR = the free spectral range which is the distance of spread for the specta which decreases as m increases because spectra begin to overlap. At higher m the efficiency decreases too.

Question 66

Littrow Configuration

Answer 66

This is a configuration for maximizing the intensity of one order (m=1) of light. It is used in monochromoters and spectrometers

Question 67

Echelle Grating

Answer 67

This is a ruled diffraction grating with extremely high blaze angles and large a to concentrate higher modes of principle diffraction. It is mainly used for producing very high resolution monochromatic light

Question 68

Ruled Gratings

Answer 68

These are mechanically cut and polished gratings like blaze gratings

Question 69

Holographic gratings

Answer 69

These are etched by lasers and have a sinusoidal pattern. They have less error at a reduced efficiency where dispersion = f(grooves/mm)

Question 70

Gratings being used in monochromators

Answer 70

The idea is that you input polychromatic light which colimmeates (becomes parallel rays) in a focusing mirror which interacts with a grating and the spectrum is then mechanically selected

Question 71

Diffraction Efficiency

Answer 71

This is defined as the incident optical power (W/m²)/entrant power

Question 72

Pattern density influence on pattern

Answer 72

With increased density of pattern there are more reflectors and there are more “small humps”

Question 73

Diffraction gratings: missing orders

Answer 73

With transmissive gratings if the distance between slits is given by w=a/m where a is the size of the opening then the m’th order will not appear in the dispersion

Question 74

Spectrogrpahs

Answer 74

These are things that use gratings to disperse light which is then measured using an electric analyzer. They take a broadband source and disperse it so that way each wavelength is in a different location to be read by the signal detector.

Question 75

Gratings use with laser pulses

Answer 75

Gratings are used to disperse the laser pulse to make the energy/area less by effectively lengthening the pulse time. This is helpful for protecting sensitive instruments/objects. The lasers can be reconstructed to the original amplitude too.

Question 76

What is light fundamentally?

Answer 76

Light is generated as the result of the acceleration of electric charges

Question 77

Maxwell’s equation for light propagation

Answer 77

E(z,t) = x E_o cos( ωt-kz+ ϕ) for monochromatic light propagating in the z direction

ω = 2pi v (v=frequency)

k = 2pi/λ = wavenumber

ϕ = arbitrary phase correction

E = the electric field vector (perpendicular to the z direction)

Question 78

Types of Light Polarization

Answer 78

Light is polarized when it is constrained to oscillate within specific planes

Linearly polarized light is light that is polarized within a single plane

Circular polarization is when the oscilations are in a perfect circle

Elliptical polarization: This describes when light’s E and h fields are not equal creating an elliptical cross section

Question 79

How does polarization influence reflection

Answer 79

s-polarized light reflects better than p-polarized light

Question 80

Types of reflectance polarization

Answer 80

s-polarization refers to when light oscillates parallel to the surface it is reflecting on.

p-polarization is when the light is oscillating normal to the propagation direction and thus intersects the surface.

Question 81

Brewster angle

Answer 81

this is the incident angle where p-polarization has zero reflectance. It is the minimum reflectance in the reflectance vs incident angle curve

Question 82

angular dispersion

Answer 82

This is a a measure of how wide the wavelength is displayed for a diffraction grating

dθ/dλ = L/Acosθ_i

where A = spacing =a and L = order = m

This says that with lower spacing, higher order, and higher incidence angles that the radians/wavelength increases

Question 83

Monochromator

Answer 83

This takes polychromatic light and reflects it on a colimeating light then a diffraction grating. This seperates the wavelengths by distance. The specified wavelength of interest is the only one that is allowed through a slit based on where it appears in the diffraction.

Question 84

Thermal polychromatic light sources

Answer 84

These work by heating a material above some critical temperature or by passing an electric current through a high MP metal. These are good sources because they are characteristically broad and continuous.

Question 85

Blackbodies

Answer 85

These are materials that absorb all incident radiation independent of wavelength and conversely emits all wavelength as a function of temperature

Question 86

Radiant Emittance Formula for blackbodies

Answer 86

M(T) = C₁ λ^-5(exp (C₂/λT) -1)^-1

Where M = radiant emmittence (W/cm²*unit wavelength)

λ = emitted wavelength

This says that as wavelength increases the power decreases

Question 87

Weins Displacement Law

Answer 87

This says that the maximum power of a blackbody as a function of temperature plots along .289878 cmK

Question 88

Real Refractive index of a material derivation

Answer 88

Using maxwell’s displacement vector which describes how an electric field permeates through a material (D = ε_o E + P = εE)

and n² = ε_r = ε/ε_o and P = Eε_oχ where χ = f(oscilations within E) to get that

n² = 1 +χ

n² = 1 + (q²/ε_om) (Σ_iⁿ N_i/(ω_i2 -ω + iγ_i ω)

Question 89

Transparency of glass with wavelength curve

Answer 89

Glass become transparent in the near vis range until the UV range because the conductance band gap for SiO₂ is relatively large compared to a semiconductor which set a low upper limit for wavelength (high upper limit for omega). Additionally because the atoms in glass are smaller than those of a semidconductor which usually have metals, phonons perterbate at lower IR wavelengths.

Question 90

Transparancy of a semiconductor with wavelength

Answer 90

Semiconductors are typically doped glasses with heavier metals (EX: Ge)

This means that on the low wavelength (high frequency) bound of transparency it is lower than dielectrics because the conductance band is smaller so less energy is needed to excite electrons.

On the high wavelength (low frequency) bound of transparency, the heavier metallic atoms oscillate slower and therefore require longer wavelengths to resonate. This effectively shifts the transparency curve to higher wavelengths.

Question 91

Plasma frequency meaning

Answer 91

ω_p² = plasma frequency = Nq²/ε_om_e and n² = 1-ω_p²/ω²

n>=1 hence,

When ω< ωp n² is imaginary because n is less than one. This means that kappa (extinction coefficient) is large so R approaches unity so the metal is opaque. This is at high wavelengths.

When ω> ω_p n² is real so R becomes a function of n and not kappa. This means that it decreases significantly

Question 92

Reflection Equation meaning

Answer 92

R = IR/Io = [(n-1)² + k²]/ [(n+1)² + k²]

IF the material is transparent then k ~0 and n dominate the equation. This reduces to R = [(n-1)² ]/ [(n+1)²] where R ~ 0 when n = 1 and increases quasi linearly as n increases to approach 1.

When n is complex k>>n and R approaches 1 (R = k²/k²) This is why metals are reflective at low frequencies where n is complex

Question 93

Reflectance of a metal as a function of wavelength

Answer 93

At low wavelengths (high frequencies) usually in the UV region, metal is non-reflective (n is real and R~0). At the plasma frequency n becomes imaginary and metals become highly reflective. This is because the band gap for metals is very small and can be excited at a variety of wavelengths.

Question 94

Plasma frequency derivation

Answer 94

ω_p² = plasma frequency = Nq²/ε_om_e and n² = 1-ω_p²/ω²

Basically, because we assume metals are homogenous with little variation in their electrons, which are in the conductance band and have a natural frequency of zero we assume that ω_i = 0; γ=0;

This means that the big long equation for the complex refractive index reduces and we can take out a constant and call it the plasma frequency.

Question 95

Why is the reflectance term in the transparency equation squared?

Answer 95

This is because reflectance (R) occurs twice when light transmits through a material (front and back).