Part One Optics(1)

8/14/2019 Part One Optics(1)

1/13

Part One

Geometrical Optics1-1why optics?

Why study optics? Lasers and fiber optics will soon replace most wires.

1-2Light rays:Electromagnetic SpectrumWe can see from about 400 nm to 700 nm

This is known as the visible spectrum


2/13


3/13


4/13

Ray

Wavelength()

Frequency () Hz

Energy (E) ev

Gamma 10-2

10-6

nm 3x1019

-3x1022

19.86x104

-19.86x107

X 10-6 10 nm 3x10 22-3x1016 19.86x107-19.86x101

Ultra 10- 100 nm 3x1016-3x1015 19.86x101-19.86x100

Visible 100 nm 1 m 3x1015-3x1014 19.86x100-19.86x10-1

Infrared 1m 1 mm 3x1014-3x1011 19.86x10-1-19.86x10-4

Microwave 1mm 10 cm 3x1011-3x1010 19.86x10-4-19.86x10-5

Radio wave 10 cm 100km 3x1010-3x106 19.86x10-5-19/86x10-9


5/13

Light Waves

(*)Wave /Particle duality for light:

Photon=elementary light particle

Mass=0

Speed(c)=3108m/sec

Energy=h

h=planck's constant=6.626210-34 J.sec

c= "dispersion relation"

=frequency (sec-1(

=wavelength (m(

And then there was a problem

However, in the early 20th century, several effects were observed which could not be

understood using the wave theory of light.

Two of the more influential observations were:

1(The Photo-Electric Effect

2(The Compton Effect

Until about 1900, the classical wave theory of light described most observed

phenomenon.

Light waves:

Characterized by:

Amplitude (A(

Frequency

((

Wavelength

((


6/13

-Photoelectric Effect (I(

If light was really a wave, it was thought that if one shined light of a fixed wavelength on

a metal surface and varied the intensity (made it brighter and hence classically, a more

energetic wave), eventually, electrons should be emitted from the surface.

Photoelectric Effect (II-(*Electrons are attracted to the (positively charged) nucleus by the

electrical force

*In metals, the outermost electrons are not tightly bound, and canbe easily liberated from the shackles of its atom.

*It just takes sufficient energy

Classically, we increase the energy of an EM wave by increasing the intensity

)e.g. brightness(

A=amplitude of the wave.

Vary wavelength, fixed amplitude

electrons

emitted?

What if we try this?

No electrons were emitteduntil the frequency of the light exceeded a critical

frequency, at which point electrons were emitted from the surface!

)Recall: small large (

No

Yes, with

low KE

Yes, with

high KE

Increase energy by

increasing amplitude

Classical Method

electrons

emitted?

No

No

No

No

Energy A2

But this doesnt work?


7/13

PhotoElectric Effect (III-( An alternate view is that light is acting like a particle

The light particle must have sufficient energy to free the

electron from the atom. Increasing the Amplitude is just simply increasing the number

of light particles, but its NOT increasing the energy of each one!

Increasing the Amplitude does diddly-squat! However, if the energy of these light particle is related to their

frequency, this would explain why higher frequency light can

knock the electrons out of their atoms, but low frequency light cannot

-Photo-Electric Effect (IV) In this quantum-mechanical picture, the energy of the

light particle (photon) must overcome the binding energy of the

electron to the nucleus.

If the energy of the photon does exceed the binding energy, the

electron is emitted with a KE = Ephoton Ebinding.

The energy of the photon is given by E=h, where the

constant h = 6.6x10-34 [J s] is Plancks constant.

Light particle

Before Collision After Collision


8/13

Photons: Quantum theory describes light as a particle

called a photon

According to quantum theory, a photon hasan energy given by

E = h = hc/ h = 6.6x10-34 [J*sec]

Plancks constant, after the scientist Max Planck.

The energy of the light is proportional to the frequency, and inversely

proportional to the wavelength! The higher the frequency (lower wavelength)

the higher the energy of the photon!

10 photons have an energy equal to ten times a single photon.

The quantum theory describes experiments to astonishing precision,

whereas the classical wave description cannot.

-de Broglies Relation

The smaller the wavelength the larger the photons momentum! The energy of a photon is simply related to the momentum by:

E = pc (or, p = E / c )

The wavelength is related to the momentum by: = h/p

The photon has momentum, and its momentum is given

by simply p = h /.

-Momentum of Photons

p = h /

If I have a photon with energy E=1 [GeV], what is its momentum?

p = E / c = (1 [GeV])/c = 1 [GeV/c] Thats it!

If I have a photon with momentum 5 GeV/c, what is its energy?

E = pc = (5 GeV/c) * c = 5 [GeV] whallah!

o, the only difference between a photons energy and momentum is:Energy [GeV]

momentum [GeV/c]

Dont forget though that the c in [GeV/c] really means 3x108

m/s].


9/13

Incident X-

ray

wavelength

i= 1.5 [nm]

ee

Electron

initially at

rest

KE=0.2 [keV]

f

BeforeBefore AfterAfter

Compute the energy of the 1.5 [nm] X-ray photon.

E = hc/ = (6.6x10-34 [J s])(3x108 [m/s]) / (1.5x10-9 [m])

= 1.3x10-16 [J]

Scattering Problem

Scattering Example (cont)


10/13

Summary of Photons Photons can be thought of as

packets of light which behave as a

particle.

To describe interactions of light with matter, one generally has to appeal to the

particle (quantum) description of light.

A single photon has an energy given by

E = hc/,

where

h = Plancks constant = 6.6x10-34 [J s] and,

c = speed of light = 3x108 [m/s]

= wavelength of the light (in [m])

Photons also carry momentum. The momentum is related to the energy by:

p = E / c = h/

Matter Waves ?One might ask:If light waves can behave like a particle, might particles act like waves?

Express this energy in [keV].1.3x10-16 [J] * (1 [eV] / 1.6 x10-19 [J]) = 825 [eV] = 0.825 [keV]

What is the magnitude of the momentum of this photon?

p = E / c= 0.825 [keV]/ c = 0.825 [keV/c[

After the collision the electrons energy was found to be 0.2 [keV]. What is the energy ofthe scattered photon?

A) 0.2 [keV] B) 0.625 [keV] C) 1.025 [keV] D) 0.825 [keV]

Since energy must be conserved, the photon must have E=0.825-0.2 = 0.625 [keV]

What would be the wavelength of the scattered photon?

HW exercise !


11/13

= h/p = h / mv

That is, the wavelength of a particle depends on its momentum,just like a photon!

The main difference is that matter particles have mass, and

photons dont !

That is, the wavelength of a particle depends on its momentum,just like a photon!

The main difference is that matter particles have mass, and

photons dont !

The short answer is YES. The explanation lies in the realm of

quantum mechanics, and is beyond the scope of this course.

However, you already have been introduced to the answer.

Particles also have a wavelength given by:

Compute the wavelength of a 1 [kg] block moving at 1000 [m/s].

= h/mv = 6.6x10-34 [J s]/(1 [kg])(1000 [m/s]) = 6.6x10-37 [m].

This is immeasureably small. So, on a large scale, we cannot

observe the wave behavior of matter

Matter Waves (cont)

Compute the wavelength of an electron (m=9.1x10

-31

[kg])moving at 1x107 [m/s].

= h/mv = 6.6x10-34 [J s]/(9.1x10-31 [kg])(1x107 [m/s])

= 7.3x10-11 [m].

This is near the wavelength of X-rays


12/13

Remarks on Particle Probes We have now asserted that high energy particles (electrons in the

case of a SEM) can provide a way to reveal the structure of matter

beyond what can be seen using an optical microscope.

The higher the momentum of the particle, the smaller the

deBroglie wavelength.

As the wavelength decreases, finer and finer details about the

structure of matter are revealed ! We will return to this very important point.

To explore matter at its smallest size, we need very high

momentum particles!

Today, this is accomplished at facilities often referred to as

atom-smashers. We prefer to call them accelerators

More on this later !


13/13

Huygens Principle In the 17th Century, Christiaan Huygens (16291695) proposed what we now know asHuygens Principle, one of the fundamental concepts of waves and wave optics.

A typical statement of the principle isevery point on a wavefront acts as a source

of a new wavefront, propagating radially outward.

Part One Optics(1)

Documents

Transcript of Part One Optics(1)