What is the resolution of a grating in the first order of 4000 groves/mm if 1 cm of the grating is illuminated?

Are 489 and 489.2 nm resolved?

R nF

R ave

1 2

2 1

2

Change in path length resultsIn phase lag

E x y E x y ft x yo0 0 02, , co s ,

The photo plate contains all the informationNecessary to give the depth perception whendecoded

Constructive/Destructive interference 1. Laser2. FT instrument3. Can be used to select wavelengths4. Can be used to obtain information

about distances5. Holographic Interference filter.

Interference Filter

Holographic Notch Filter

Constructive/Destructive interference 1. Laser2. FT instrument3. Can be used to obtain information

about distances4. Interference filter. 5. Can be used to select wavelengths

Can create a filter usingThe holographic principle To create a series of Groves on the surfaceOf the filter. The groovesAre very nearly perfectIn spacing

End Section on Using Constructive and DestructiveInterference patterns based on phase lags

Constructive/Destructive interference 1. Laser2. FT instrument3. Can be used to obtain information

about distances4. Interference filter. 5. Can be used to select wavelengths

Begin SectionInteraction with Matter

In the examples above have assumed that there is no interaction with Matter – all light that impinges on an object is re-radiated with it’s Original intensity

Move electrons around (polarize)Re-radiate

“virtual state”Lasts ~10-14s

Move electrons around (polarize)Re-radiate

This phenomena causes:1. scattering 2. change in the velocity of light3. absorption

First consider propagation of light in a vacuum

c 1

0 0 c is the velocity of the electromagnetic wave in free space

0 Is the permittivity of free space which describes theFlux of the electric portion of the wave in vacuum andHas the value

012

2

28 8552 10

. xC

N m

force Nkg m

s

capacitance

0 Is the permeablity of free space and relates the currentIn free space in response to a magnetic field and is defined as

It can be measured directly from capacitor measurements

07

2

24 10x

N s

C

c 1

0 0

c

xC

N mx

N s

C

1

8 8552 10 4 10122

27

2

2.

07

2

24 10x

N s

C 0

122

28 8552 10

. xC

N m

c

xs

mx

s

m

1

1 11 10

1

3 33485 10172

2

9. .

c xm

s 2 9986 10 8.

v elocity 1

c 1

0 0

r

ve loc itye

c

vK

0 0 0

Dielectric constant

~ 0Typically so

re locity

e

c

vK

Maxwell’s relation

This works pretty well for gases (blue line)

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

1.51 4.63E+00 5.04 5.08 8.96E+00

sqrt dielectric

Ind

ex o

f re

frac

tio

n

0.9998

0.9999

1

1.0001

1.0002

1.0003

1.0004

1.0005

1.000034 1.000131 1.000294 1.00E+00

sqrt dielectric

Ind

ex o

f re

frac

tio

n

Says: refractive index is proportional to the dielectricconstant

Our image is of electrons perturbed by an electromagnetic field which causesThe change in permittivity and permeability – that is there is a “virtual” Absorption event and re-radiation causing the change

It follows that the re-radiation event should be be related to the ability to Polarize the electron cloud

10-14 s to polarize the electron cloud and re-release electromagneticRadiation at same frequency

Angle between incident and scatteredlight

0

0.2

0.4

0.6

0.8

1

1.2

0 45 90 135 180 225 270 315 360

Angle of Scattered Light

Rel

ativ

e In

ten

sity

SCATTERING

Light in

I Irs o

81

4 2

4 22

cos

particle

Polarizability of electronsa) Number of electronsb) Bond lengthc) Volume of the molecule,

which depends upon the radius, r

= vacuum Io = incident intensity

I I

vo l m olecu les o

2

4

Most important parameter is the relationship to wavelength

At sunset the shorter wavelength is Scattered more efficiently, leaving theLonger (red) light to be observed

Better sunsets in polluted regions

Long path allow more of the blue light (short wavelength) to be scattered

Blue is scattered

Red is observed

What is the relative intensity of scattered light for 480 vs 240 nm?

What is the relative intensity of scattered light as one goes from Cl2 to Br2? (Guess)

I I

vo l m olecu les o

2

4

Our image is of electrons perturbed by an electromagnetic field which causesThe change in permittivity and permeability – and therefore, the speed of thePropagating electromagnetic wave.

It follows that the index of refraction should be related to the ability to Polarize the electron cloud

re locity

c

v

Refractive index = relative speed of radiation

Refractive index is related to the relative permittivity (dielectric constant) at that Frequency

2

2

1

1

P

Mm Where is the mass density of the sample, M is the molar

mass of the molecules and Pm is the molar polarization

PN

kTmA

o

3 3

2

Where is the electric dipole moment operator

is the mean polarizabiltiy

2

20

2

0

1

1 3 3 3

N

M kT

N

MA A

2

20

1

1 3

N

MA

Clausius-Mossotti equation

0 Is the permittivity of free space which describes theFlux of the electric portion of the wave in vacuum andHas the value

Point – refractive indexIs related to polarizability

2

3

2 2e R

E

Where e is the charge on an electron, R is the radius of the molecule and ∆E is the mean energy to excite an electron between the HOMO-LUMO

2

20

1

1 3

N

MA

2

20

2 21

1 3

2

3

N

M

e R

EA

The change in the velocity of the electromagnetic radiation is a function of1.mass density (total number of possible interactions)2. the charge on the electron3. The radius (essentially how far away the electron is from the nucleus)4. The Molar Mass (essentially how many electrons there are)5. The difference in energy between HOMO and LUMO

An alternative expression for a single atom is

22

02 21 Ne

m

f

io e

j

j j jj

A damping force term that account for Absorbance (related to delta E in priorExpression) Natural

Frequency ofThe oscillating electronsIn the single atom j

Frequency of incoming electromagneticwave

Transition probability thatInteraction will occur

Molecules perUnit volumeEach with J oscillators

If you include the interactions between atoms and ignore absorbance you get

2

20

2 21

1 3

2

3

N

M

e R

EA

2

2

2

02 2

1

2 3

Ne

m

f

o e

j

j jj

2

20

2 21

1 3

2

3

N

M

e R

EA

02 2j jwhen The refractive index is constant

when j j2

02 The refractive index depends on omega

And the difference 02 2j j Gets smaller so the

Refractive index rises r e

c

vK

REFRACTIVE INDEX VS

Anomalous dispersion near absorption bands which occur at natural harmonic frequency of material

Normal dispersion is required for lensing materials

What is the wavelength of a beam of light that is 480 nm in a vacuum if it travels in a solid with a refractive index of 2?

re locity

c

v

frequencyvacuum

frequency vacuum

frequency m ed ia eloc ity m ed ia

c

c

v

,

r

e locity

frequency vacuum

frequency m ed ia

vacuum

m edia

c

v

Filters can be constructedBy judicious combination of the Principle of constructive andDestructive interference andMaterial of an appropriate refractiveindext 1

2 '

t '

t 32 '

tn

2'

vacuum

'

tn vacuum

2

2 t

n vacuum

t

t

'

t

Wavelength In media

What is (are) the wavelength(s) selected from an interference filter which has a base width of 1.694 m and a refractive index of 1.34?

2 t

n vacuum

Holographic filters are better

INTERFERENCE WEDGES

AVAILABLE WEDGES

Vis 400-700 nmNear IR 1000-2000 nmIR 2.5 -14.5 m

112

t

n

222

t

n

332

t

n

442

t

n

The electromagnetic wave can be described in two components, xy, andXy - or as two polarizations of light.

Using constructive/destructive interference to select for polarized light

Refraction, Reflection, and Transmittance DefinedRelationship to polarization

The amplitude of the spherically oscillating electromagneticWave can be described mathematically by two componentsThe perpendicular and parallel to a plane that described the advance of The waveform. These two components reflect the polarization of the wave

When this incident, i, wave plane strikes a denser surface with polarizable electrons at an angle, i, described by a perpendicular toThe plane

It can be reflected

Or transmitted

T

R

Air, n=1

Glassn=1.5The two polarization components are

reflected and transmitted with Different amplitudes dependingUpon the angle of reflection, r,And the angle of transmittence, t

Let’s start by examingThe Angle of transmittence

Snell’s Law

sin

sin

i

t

t

i

ie loc ity

c

v

1

1Time=0

Time=1

Time=2

Time=3

Time=4

Time=0

Time=1

Time=2

Time=3

Time=4

2

Less dense 1Lower refractive indexFaster speed of light

More dense 1Higher refractive indexSlower speed of light

sin

sin

sin

sin

1

2

i

t

i

t

t

i

velocity

velocity

What is the angle of refraction, 2, for a beam of light that impinges on a surface at 45o, from air, refractive index of 1, to a solid with a refractive index of 2?

sin

sin

i

t

t

i

PRISM

Crown Glass(nm) 400nm 1.532450 nm 1.528550 nm 1.519590 nm 1.517620 nm 1.514650 nm 1.513

1.51

1.515

1.52

1.525

1.53

1.535

0 100 200 300 400 500 600 700

wavelength nm

refr

acti

ve i

nd

ex o

f cr

ow

n g

lass

POINT, non-linear dispersive deviceReciprocal dispersion will vary with wavelength, since refractive index varies with wavelength

Uneven spacing = nonlinear

T

R

Angle of transmittenceIs controlled byThe density of Polarizable electronsIn the media asDescribed by Snell’s Law

The intensity of light (including it’s component polarization) reflected as compared to transmitted (refracted) can be described by the Fresnel Equations

T ti i

i i t t

2

22

cos

co s co s

T ti i

i t t i/ / / /

cos

cos cos

2

22

R ri i t t

i i t t

2

2 cos cos

cos cos

R rt i i t

i i t t/ / / /

co s co s

co s co s

2

2

The amount of light reflected depends upon the Refractive indices and the angle of incidence.

We can get Rid of the angle of transmittence using Snell’s Law

sin

sin

i

t

t

i

Since the total amount of light needs to remain constant we also know that

R T

R T/ / / /

1

1Therefore, given the two refractiveIndices and the angle of incidence canCalculate everything

Consider and air/glass interface

Here the transmitted parallel light isZero! – this is how we can selectFor polarized light!

This is referred to as the polarizationangle

i

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 10 20 30 40 50

Angle of incidence

Tra

nsm

itta

nce

Parallel

Perpendicular

Total Internal Reflection

T

R

Air, n=1

Glassn=1.5

Here considerLight propagatingIn the DENSERMedium andHitting aBoundary withThe lighter medium

0

0.2

0.4

0.6

0.8

1

1.2

0 20 40 60 80 100

Angle of incidence

Ref

lect

ance

an

d T

ran

smit

ten

ce

Perpendicular

Parallel

Same calculation but made the indicident medium denser so that wave isPropagating inside glass and is reflected at the air interface

Discontinuity at 42o signalsSomething unusual is happening

R rT

i i t t

i i t t

2

2

cos co s

co s co sSet R to 1 & to 90The equation can be solved for the critical angle of incidence

t

ic

sin,

,

c

transm itted less dense

inciden t dense

For glass/air

sin.

.

s in ( . ) .

..

c

c

o

a rads

rads

1

1 50 666

0 666 0 7297

0 7297 18041 8

All of the light is reflectedinternally

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 10 20 30 40 50 60 70 80 90

Angle of Incidence

% T

ran

smit

ten

ceThe angle at which the discontinuity occurs:1. 0% Transmittance=100% Reflectance2. Total Internal Reflectance3. Angle = Critical Angle – depends on refractive index

3742

51

1.69/1

1.5/1

1.3/1

ni/nt Critical Angles

1.697 36.271.5 41.81.3 50.28

NA c inciden t dense transm itted less dense sin , , 2 2

The critical angle here is defined differently because we have to LAUNCH the beam

Numerical Aperture

R=?

T

i=0

Using Snell’s Law the angle of transmittance is

0

t

itsin

sin

sin

i

t

t

i

sin 1 0 0 t

Shining light directly through our sample

R rT

i i t t

i i t t

2

2

cos co s

co s co s

R rt i i t

i i t t/ / / /

co s co s

co s co s

2

2

cos 0 1

R rTi t

i t

22

R r t i

i t/ / / /

22

cos 0 1

same

R t i

i t/ /

2

The amount of light reflected depends Upon the refractive indices of the medium

RI

Ire flec ted

in itia l

t i

i t/ /

2

I I Itransm itted in itia l re flec ted

I Ire flec ted in itia lt i

i t

2

I I Itransm itted in itia l in itia lt i

i t

2

For a typical Absorption Experiment,How much light will we lose from the cuvette?Or another way to put it is how much light will get transmitted?

I Itransm itted in itia lt i

i t

12

I I It o t' ' ' '. .

. .'

. .

. .

11 5 1 33

1 5 1 331

1 5 1 33

1 5 1 33

2 2

I It o

11 5 1

1 5 1

2.

.

I I It o t' '. .

. .

. .

. .

11 33 1 5

1 33 1 51

1 33 1 5

1 33 1 5

2 2

Final exiting light

I I It o t' ' ' ' ' '.

.

.

.' '

11 5 1

1 5 11

1 5 1

1 5 1

2 2

Io It=I’oIt’ = I’’o

It’’ = I’’’o

It’’’

AirGlass, refractive index 1.5

Air, refractive index 1

Water, refractive index 1.33

I It o'.

.

. .

. .

11 5 1

1 5 11

1 33 1 5

1 33 1 5

2 2

I It o' '.

.

. .

. .

11 5 1

1 5 11

1 33 1 5

1 33 1 5

2 2 2

I It o' ' '.

.

. .

. .

11 5 1

1 5 11

1 33 1 5

1 33 1 5

2 2 2 2

I Ig lass a ir in itia l/

.

.

.

.2

2 2 2 2

10 5

2 51

0 17

2 83

I Ig lass a ir in itia l/ . .2

2 20 96 0 99

I Ig lass a ir in itia l/ .2 0 915

We lose nearly 10% of the light

I It o' ' '.

.

. .

. .

11 5 1

1 5 11

1 33 1 5

1 33 1 5

2 2 2 2

Key Concepts

Interaction with Matter

Light Scattering I I

vo ls o

2

4

Refractive IndexIs wavelength dependentUsed to separate light by prisms

re loc ity

c

v

2

2

2

02 2

1

2 3

Ne

m

f

o e

j

j jj

2

20

2 21

1 3

2

3

N

M

e R

EA

2 t

nRefractive index based

Interference filters

Key Concepts

Interaction with Matter

Snell’s Lawsin

sin

i

t

t

i

Describes how light is bent based differing refractive indices

T ti i

i i t t

2

22

cos

co s co s

T ti i

i t t i/ / / /

cos

cos cos

2

22

R ri i t t

i i t t

2

2 cos cos

cos cos

R rt i i t

i i t t/ / / /

co s co s

co s co s

2

2

Fresnell’s Equations describe how polarized light is transmitted and/or reflectedat an interface

Used to create surfaces which select for polarized light

Key Concepts

Interaction with Matter

Fresnell’s Laws collapse to

sin,

,

c

transm itted less dense

inciden t dense

Which describes when you will get total internal reflection (fiber optics)

RI

Ire flec ted

in itia l

t i

i t/ /

2

And

Which describes how much light is reflected at an interface

PHOTONS AS PARTICLES

The photoelectric effect:

The experiment:1. Current, I, flows when Ekinetic > Erepulsive

2. E repulsive is proportional to the applied voltage, V

3. Therefore the photocurrent, I, is proportional to the applied voltage4. Define Vo as the voltage at which the photocurrent goes to zero = measure of the maximum kinetic energy of the electrons

5. Vary the frequency of the photons, measure Vo, = Ekinetic,max

KE hm

Energy ofEjectedelectron

Frequency of impinging photon(related to photon energy)

Work function=minimum energy binding an Electron in the metal

KE hm

To convert photons to electrons that we can measure with an electrical circuit useA metal foil with a low work function (binding energy of electrons)

DETECTORSIdeal Properties1. High sensitivity2. Large S/N3. Constant parameters with wavelength

S kP kelec trica l signa l rad ian t pow er d

Where k is some large constantkd is the dark current

Classes of Detectors

Name commentPhotoemissive single photon eventsPhotoconductive “ (UV, Vis, near IR)

Heat average photon flux

Want low dark current

1. Capture all simultaneously = multiplex advantage2. Generally less sensitive

Rock toGet different wavelengths

Very sensitive detector

Sensitivity of photoemissiveSurface is variable

Ga/As is a good oneAs it is more or less consistentOver the full spectral range

Diode array detectors-Great in getting-A spectra all at once!

Background current(Noise) comes from?

One major problem-Not very sensitive-So must be used-With methods in -Which there is a large-signal

Photodiodes

Photomultiplier tube

The AA experiment

Charge-Coupled Device (CCD detectors)

1. Are miniature – therefore do not need to “slide” the image acrossa single detector (can be used in arrays to get a Fellget advantage)

2. Are nearly as sensitive as a photomultiplier tube

+V

3. Apply greater voltage4. Move charge to “gate” And Count, 5. move next “bin” of charge and keep on counting6. Difference is charge in One “bin”

1. Set device to accumulate charge for some period of

time. (increase sensitivity)2. Charge accumulated near

electrode

Requires special cooling, Why?

The fluorescenceexperiment

END6. Really Basic Optics

Since polarizability of the electrons in the material also controls the dielectricConstant you can find a form of the C-M equation with allows you to compute The dielectric constant from the polarizability of electrons in any atom/bond

N

o

r

r

3

1

2

N = density of dipoles= polarizability (microscopic (chemical) property)r = relative dielectric constant

Frequency dependentJust as the refractive index isTypically reported

Point of this slide: polarizability of electrons in a molecule is related to theRelative dielectric constant

-200

-100

0

100

200

300

400

500

600

700

800

900-165

-150

-135

-120

-105

-90

-75

-60

-45

-30

-15015

30

45

60

75

90

105

120

135

150

165180

Grating

2nd order

1st order

Angle ofreflection

i=45

-200

-100

0

100

200

300

400

500

600

700

800

900-165

-150

-135

-120

-105

-90

-75

-60

-45

-30

-15015

30

45

60

75

90

105

120

135

150

165180

2nd order

1st order

Angle ofreflection

i=45