Week 2

Question 1

What is light?

Answer 1

Light is made of an oscillating magnetic field and an electric field. They are coupled together and propagate as a electromagnetic wave. They are well described by the MaxWell equations.
From the MaxWell equations one can work out the so-called wave equations and plane waves are solution of these equations. E(r,t)=E_0cos(kr-omegat)
B(r,t)=B_0cos(kr-omegat)
The corrsponding complex quantities are:
E(r,t)= E_0e^(i(kr-wt)) B(r,t)= B_0e^(i(kr-wt)) but remember that one measure onely the real part of these quantities.
The important physical quantities that enter these expressions are: wave vector taht points toward the propagation direction (K=2pi/lambda), the angular frequency (omega=2pinu), they are related by: omega/k=c=(mu_0e_0)^(-1/2) speed of ligth

Question 2

How are wavelength, frequency and energy related with each other?

Answer 2

lambda=c/nu (c=310^8 ms^(-1))
E=hnu
(h=610^(-34) Js = 410^(-15) eVs)

Question 3

How can ligth interacti with a particle?

Answer 3

(reflection, refraction, transmission, absorption, scattering, Raman scattering, fluorescence, thermal emission)
The incident ligth hits the particle, at the interface between particle and surrounding medium some portion of it will be reflected also refraction happens according to Snell’s law (then you’ll have again refraction at the second interface particle-surrounding medium).
Another portion of incident ligth will be transmitted through the particle itself while another portion is going to be absorbed.
Some of the incident beams are going to be scattered in all the directions, if the scattering is isotropic, you also have Raman scattering.
You also have fluorescence and thermal emission.

Question 4

Polarization of ligth?

Answer 4

We talk about linear polarization if the electric field propagates keeping the same orientation (e.g., vertical).
A generic polarization state is a superposition of 2 orthogonal linear polarization states (linear combination of the y - polarization and x - polarization).
You can select a certain polarization from an unpolarized source by using a polarizer.
Ligth reflected from many surfaces is partially polarized, so these reflections can be removed with a polarizer oriented in the orthogonal direction. This mechanism is used in sun glasses.

Question 5

How does light propagate in a uniform medium?

Answer 5

We make the hypothesis of homogeneous and isotropic dielectric.
Homogeneous: single phase so a block of meatl or metal oxide or a dielectric.
Isotropic: same properties in all the directions.
Dielectric: very big band gap, larger than 2 eV.
The medium responds to the application of these oscillating electric field and magnetic field with an induced polarization and an induced magnetization. Polarization and Magnetization influence the propagation of the wave, their average can be measured macroscopically.
Influencing the propagation of the ligth means that these new quantities enter the MaxWell equations within the medium. We will define the auxiliary fields D (electric displacement) and H (magnetic field): D=e_0E+P; H=(1/mu_0)B-M

Question 6

How do you define the polarization and the magnetization?

Answer 6

Constitutive relations (linear medium)
P=e_0Chi_eE, Chi_e is the electric susceptibility
M=Chi_mH, Chi_m is the magnetic susceptibility
So you can rewrite the electric displacement and the applied field as follows:
D=e_0e_rE e_r=(1-Chi_e)
B=mu_0mu_rH mu_r=(1-Chi_m)
Result: by deriving the wave equations foe E and B within the medium you get that the propagation speed id lower than in vacuum v=c/n where
c=(mu_0e_0)^0.5
n=(mu_r*e_r)^0.5
Light is slowed down in a medium (the wavelength decreases = lambda/n)

Question 7

What is the refractive index?

Answer 7

For a given homogeneous and hysotropic medium we can define the refractive index as n= (e_r*mu_r)^(0.5). It is a function of the wavelength (draw how different wavelenghts are refracted at a certain interface, slide 32/105).
Therefore all quantites that depend on the refractive index n, also depends on the wavelenght lambda.
It describes how light travels through a certain medium.
For example, to analyze biological samples (living cells) using microscopy, you use glycerol to match the refractive index of the glass so that you don’t have air between the glass and the sample, in this way you don’t see the glass at all.
For lambda=532 nm (green)
n=1 vacuum
n=1.0003 air
n=1.33 water
n=2.67 silicon carbide
n=2.67 titanium dioxide
n=1.46 glass
n=1.47 glycerol
In absence of absorption the refractive index is a real number.

Question 8

Why is the refractive index real in absence of absorption?

Answer 8

?

Question 9

What is reflection?

Answer 9

Whenever light hits a surface, some of it will be reflected at the interface according to Snell’s law.
The angle between the normal to the interface and the incident beam is the same as the angle between the normal and the reflected beam.

Question 10

What is refraction?

Answer 10

Refraction of ligth occurs whenever the light travels across an interface between 2 media with different refractive index.
The incident ligth is hereby bent according to the Snell’s law: n_1 sin (theta_1)=n_2 sin (theta_2)
Where we define theta as the angle between the normal to the interface and the direction of propagation of light.

Question 11

Explain what happens when white light travels through a prism.

Answer 11

We can visualize the white light as the overlap of the different wavelenghts taht belong to the visible portion of the electromagnetic spectrum. So from 380 nm up to 700 nm.
The refractive index, which naively tells us how light travels through a medium, is wevelenght dependent, therefore, we will observe the splitting into the components of this incident white light through the medium.
Therefore, the light that is transmitted from the prism to the surrounding medium (air) is made up of all the different wavelenghts in the visible region of the spectrum.
Snell’s law: n_airsin(theta_incident)=n_glass (lamda)sin(theta_refraction). (slide 34/105)

Question 12

Explain what happens when white light travels through a lens.

Answer 12

Chromatic aberration is a physical phenomenum that occurs whenever light travels across a lens. Due to the wavelenght dependence of the refractive index of the glass and therefore on refraction, we will observe that the different focal points will be at different areas.
This leads to missing information in the observation of the sample (spectroscopy with a microscope).
This type of systematic error can be corrected, at least partially.
Whenever you are performing this type of spectroscopy it is important to be aware of chromatic aberration and which kind of aberration is corrected and which is not.

Question 13

What happens if the dimension of the medium is comparable to the waveleght of the incident light?

Answer 13

2 things can happen, mostly: scattering and transmission of the residual incident light.

Question 14

What is scattering?

Answer 14

When the wavelenght of the incident beam is comparable or smaller than the characteristic size of the particle, scattering happens. Scattering happens in all directions if the particle is small. To quantify the scattering you can define the scattering cross-section.

Question 15

What is the scattering cross-section?

Answer 15

The scattering cross-section quantifies the scattering for a certain particle or physical system. It can be measured by looking in the far field regime (at distances very large compared to the wavelength), how the different wavelengths are scattered at different angles.

Question 16

What is the optical cross-section?

Answer 16

It is the power removed from an incident plane wave and normalized by the incident intensity.

Question 17

What is the differential (scattering) cross-section?

Answer 17

It is an area, it is a function of both the radial and azhimuthal angle.
It is defined as follows: sigma_sca(theta, phi)=dP_sca(theta, phi)/d omega *1/I_in

Question 18

What is the total (scattering) cross-section?

Answer 18

It is the integral over the entire solid angle 4pi of the differential scattering cross-section. sigma_tot=P_sca/I_in, it is an area.

Question 19

what is the absorption cross-section?

Answer 19

absorbed power/incident intensity, it is an area

Question 20

what is the extinction cross section? why is it useful?

Answer 20

it is useful because it corresponds to what we usually measure. the extinction cross section is the sum of the scattering cross section and the sbsorption cross section. it quantifies the loss of the incident power due to both the scattering and the absorption
note: measuring absolute cross section is very challenging, it is easier to measure cross sections relative to an arbitrary reference (i.e., measuring at 2 or several wavelenghts)

Question 21

which parameters does the cross section depend on?

Answer 21

it depends on both the refractive index n (so also on the wavelength) and the size D of the particle. The joint effect of n and D determines the spectral (i.e., wavelenght) dependence of the scattering.

Question 22

which are the different scattering regimes? for simplicity we can neglect absorption

Answer 22

raileigh scattering D«lambda (sigma_tot=lambda^(-4))
mie scattering D=lambda

Question 23

Why is blue light scattered 9 times more than red light?

Answer 23

Because of Rayleigh law: sigma_scatt_tot = lambda^(-4).
draw graph

Question 24

what is mie scattering?

Answer 24

it is a scattering regime where the incident wavelenght is comparable with the size of the system. There’s no general rule to determine the relation between the total scattering cross-section and the wavelenght. This behaviour strongly depends on the size of the particle. It can either be measured or simulated with the knoowledge of these information about the particle.
in correspondence of the peaks a strong interaction between the particle and the electromagnetic field id observed.
Mie scattering is anisotropic (forward) scattering
highly resonant peaks creates selective scattering and strong field confinement within the structure.

Question 25

what is rayleight scattering?

Answer 25

it is a scattering regime, D<

Question 26

draw the plot of scattering cross section as a function of the wavelenght

Answer 26

plot

Question 27

what is the effect of the refractive index in scattering?

Answer 27

hogh refractive index (n>2) nanomaterials produce strong scattering even for small sizes (<100 nm) e.g. D=50 nm Si NPs (n=4, in air) do not absorb in the VIS. Si is a semiconductor but it behaves as a dielectric in the VIS range of the EM spectrum. Sometimes the silicon nanobeads are called old dielectric material (because of no absorption, beacuse the energy of photons in the vis range is smaller then the band gap)

Question 28

how do colors arise from scattering?

Answer 28

sharp mie resonances enable color formation by using spectrally selective scattering, without light generation. as you increase the size of the particles, you get redshift of the wavelenght. as you increase the size of the particle from 400 nm to 500 nm, you get access to higher orders: electric dipole, magnetic quadripole...

Question 29

why is the sky red-orange during sunrise/sunset and why is it blue at midday? why do clouds appear white?

Answer 29

There are 2 factors to take into account 1. thickness of the atmosphere that light travels from the sun to the observer 2. wavelenght dependence of scattering (rayleigh scattering, these particles are much smaller than the wavelenght) red light is 9 times less scattered than blue light. The scattering of blue light wins for short distances, while for longer distances the red light wins because it's less scattered. when talking about clouds the scattering centers are now of the order of the wavelenght of the incident light, they might be even larger (microns) so you are in the Mie regime. you have multiple scattering: that's why the clouds are white and sometimes even dark gray: because they have even a stronger scattering and the light is totally lost within the cloud.

Question 30

what is the condition to have total internal reflection? What is the critical angle?

Answer 30

n_inside > n_outside theta_c=arccos(n_out/n_in)

Question 31

How can you reach 1D confinement in a 3D structure?

Answer 31

Via the so-called whispering gallery modes. Where you can trap the light inside a spherical structure of higher refractive index. The light is just trapped close to the interface between air and glass fiber in the form of an evanescent wave. The name whispering gallery modes is in analogy with the acoustic waves in the St Paul's cathedral in London.

Question 32

What are evanescent waves?

Answer 32

This is a way that is exponentially decaying with imaginary k vector. We can associate to this negative exponential wave a penetration depht that is of the order of lambda / 10. For moderate refractive index (n<2) the confinement is not perfect. it is an exponentially decaying wave with an imaginary k vector. the penetration depth is of 50-100 nm for visible light. evanescent means tending to vanish which is appropriate because the intensity of evanescent waves decays exponentially rather than sinusoidally.

Question 33

How can you apply whispering gallery modes and evanescent waves for biosensing?

Answer 33

Biomolecues on the surface air-glass fiber, determine a shift of the whispering gallery mode resonance (dip in the intensity vs. Wavelength)

Question 34

How is the field inside the high refractive index particles?

Answer 34

we know that in the case of metallic NPs the electric field is at the interface. conversely, for dielectric (n=3.5) spherical NPs the electric field is almost zero inside and it's strong on the surface. while the magnetic field id strong in the inside of the particle. efficiently confined modes correspond to resonant scattering peaks. as you change the incident wavelenght you are exciting different resonances of different modes (electric quandripole, magnetic dipole...)

Question 35

name an application of dielectric nanoresonator for biosensing

Answer 35

on a 2*2 cm chip you have a quartz substrate and on top of it silicon pillars (silicon nanodisk), you can functionalize only the silicon pillars when the proteins are captured, the refractive index of the exterior medium changes, shifting the nanodisk resonance: you will observe some resonance shift vs. time. the shift in resonance is of the order of the nanometer so you really need a good resolution to measure it.

Question 36

what is the expression for the electromagnetic density?

Answer 36

u=0.5 *(epsilon*E^2 + (1/mu)*B^2) energy per unit volume

Question 37

what is the relation between the light wavelenght and the corresponding photon energy in the quantum picture?

Answer 37

E (eV) * lambda (microns) = 1.24

Question 38

how is the absorption described in the expression of the refractive index?

Answer 38

macroscopically, the presence of the absorption is described by an imaginary part of the dielectric constant, or equivalently by an imaginary part of the refractive index n. the absorption mechanism depends on the material. an incident plain wave is attenuated by a negative exponential.

Question 39

how do the real part and complex part of the refractive index behave as functions of the wavelenght?

Answer 39

the complex part has some kind of resonance the absorption is very stron at about 1 micron (1.24 micron corresponds to 1 eV) this is where you have the band gap of Silicon. if you use silicon with electromagnetic waves in the IR region, larger than 1 micron, then it is not transparent anymore.