PHYS3321

FINDING NUCLEAR CHARGE DISTRIBUTIONS BY SCATTERING ELECTRONS – Part I

Scattering of photons from single slit

(Sections 3.3, 3.4 Dunlap)

Q) How would we find out the size of a slit too small to see with the eye – or even with a microscope?

A) Shine light of a fixed known wavelength on the slit and observe the diffraction pattern.

Single Slit Diffractionx.sin

x

dx

x

-b/2

+b/2

A

Light of wavelength is incident from the left – the positions of the wave fronts are shown. The wave is in phase along the line of the slit (x direction) . As seen from position A the in-phase light adds to give a maximum in the direction A (=0). However, at any other angle (such as B ) the light from different x-positions on the slit will be seen with different phases.-(some going up while others going down etc). To find the total amplitude () at angle we must sum up the contributing amplitudes from all the elements “dx” taking into account their different phases .

B

Single Slit Diffractionx.sin

x

dx

x

-b/2

+b/2

A

sin

sin2 kxx

“Phase angle” as seen at x

Small amplitude of wavelet coming from dx: dxkxAAdxd )sincos()cos(. Integrating wave amplitude at over the whole slit

2/

2/

2/

2/

sinRe)sincos()(b

b

b

b

ikx dxeAdxkxA

xixeix sincos 2

k

Single Slit Diffraction

)sinc(.)sin(

.sin

sinsin.

sin

1Re.

Re.)(

21

21

sinsin

2/

2/

sin

22

AbAbkb

kbAb

eeik

A

dxeA

bb ikik

b

b

ikx

The Amplitude at

)(sinc).0()sin(

)0()()( 2

22

III

The intensity at

Where: sin2

1kb

The (Sinc)2 diffraction pattern

sin2

1

)sin(

)0(

)(2

kb

I

I

You may think of this pattern as the “differential scattering” pattern for photons scattering from a slit.

dd

Single slit differential photon scattering cross-section

=0 = =2 =3

3,2,sin2

1 kb

By observing the position of the minima we can work out the width “b” of the slit

Minima occur when :

i.e. when:

321

321

sin

3

sin

2

sin

sin

6

sin

4

sin

2

kkk

b 2

k

40.55 10

10 100

cmrad

cm

The same principles operate if we are dealing with photons scattering from a two dimension aperture.

The amplitude squared is the probability of finding a photon going in a specified direction.

The picture below was taken using a =633nm He-Ne laser source of photons.

The distance of the photo from the slit is 10m. The distance between minima is 0.5cm

4

6331.26

sin 5 10

nmb mm

The pattern of the diffraction pattern also reveals that the aperture is squareFrom “Optics” Hect, Zajac

Single Slit Diffraction

)(.)(Re.

Re.)(2/

2/

sin

xAFTAdxexAA

dxeA

xik

b

b

ikx

x

The Amplitude at

Where the “Aperture function” A(x) is:

2/for 0

2/for ,1)(

bx

bxxA

And kx = k sin() is the component of wavevector k in the x direction.

An important diffraction principle is seen: The probability of a photon going at angle is proportional to the square of the FT of the aperture function.

X=+b/2X=-b/2

1

0

22)()()( xAFTP

Finding De-Broglie wavelengths

pc

c

p

22

The De-Broglie wavelength associated with a particle is always given by:

(1)

For a non-relativistic particle we have K.E.

i.e.

m

pmv

2

22

21

Tmcpc .2 2 (2)

Combining (1) and (2):

Tmc

FMeV

Tmc

c

.2

.1972

.22

22

(3)

For relativistic particles (and light)

TEpc (4)

Combining (1) and (4)

T

FMeV

T

c .19722

(5)

• For the general case

Finding De-Broglie wavelengths

T

E

mc2

pc

RelativisticNon-Relativistic

22

2224222

..2

)(

mcTTpc

mcTEcmcp

22 .22

mcTT

c