diffraction

Question 1

an extremely short duration pulse corresponds to an

Answer 1

infinitely wide range of frequency components, all having the same amplitude

Question 2

diffraction is

Answer 2

the spreading of waves around obstacles

Question 3

interference is most pronounced when

Answer 3

the wavelength of the light is comparable to the dimension of the obstacle it encounters

Question 4

Fraunhofer diffraction

Answer 4

diffraction observed in the image plane of the source - eg at infinity

normally dealing with plane waves and small angles

interested in determining the diffraction pattern at a large distance from the aperture

Question 5

Fresnel diffractions

Answer 5

diffraction observed close to the diffracting object

wavefront are significantly curved

complicated mathematical treatment

Question 6

fraunhofer - assuming the diffraction beams are

Answer 6

esentially parallel when they leave a slit since assuming they create an image at infinity

Question 7

fraunhofer - if a slit has width d then the path difference between the two beams originating from either end of the slit is

Answer 7

d sin theta

Question 8

can genealise d sin theta to

Answer 8

oath difference between any two beams, whose origin points are separated by y have path difference
y sin theta

path difference has a linear dependence on y

Question 9

for fresnel diffraction, assuming the

Answer 9

point of interest is relatively close to the slit

Question 10

fresenl - difference in lengths of the two beams

Answer 10

δr = sqrt (r0^2+y^2) - r0

can assume y/r0 is small so we can use binomial expansion

Question 11

path difference for near-field diffraction is

Answer 11

quadratic in y

Question 12

experimentally, how to fraunhoder

Answer 12

use a converging lens to focus parallel rays

can’t physically observe at infinity

Question 13

with respect to the path travelled by a Huygen’s wavelet originating at y=0, the path travelled by one originating at a distance y will be

Answer 13

-ysin(theta_y)

the relative phase will then be -yksin(theta y)

Question 14

if we want to know what is happening at a point P far from the slit we can

Answer 14

add up all the contributions from across the slit that reach that point, which are characterised by the angle subtended at the middle of the slit

ie integrate

Question 15

if we define the field as P at Ep then we can say that

Answer 15

Ep prop to integral between -a/2 and a/2 of
E0exp(-ikysin(thetay)) dy

Question 16

E0 represents

Answer 16

the slit itself, which we assume is constant across the width of the slit

Question 17

the sinc function represents the

Answer 17

diffracted amplitude which cannot be measured directly

diffracted irradiance can be measured

Question 18

irradiance is proportional to the

Answer 18

square of the sinc function

I=I0[sin^2 B/B^2]

Question 19

With some suitable maths – which I wouldn’t expect you to reproduce – it can be shown
that ψP, the value of ψ at a point P inside some closed surface S can be expressed as

Answer 19

an integral over that surface giving us the (Helmholtz) Kirchhoff Integral Theorem (KIT):

Question 20

KIT - S is a

Answer 20

closed surface surrounding the point P where we want to find the field

Question 21

KIT - in any given problem, we will choose a

Answer 21

sensible surface with an eye on making the integral easy to evaluate

Question 22

we can use the KIT to

Answer 22

calculate the diffraction produced when plane waves illuminate an aperture Q

Question 23

KIT - we choose our large, closed surface to include the diffracting aperture Q which we assume is

Answer 23

much larger than the wavelength of the illuminating light

Question 24

KIT - P’ is

Answer 24

the source of our light, sufficiently far away that we can consider the waves from it to be plane-parallel waves when they reach Q

Question 25

KIT - dΣ is

Answer 25

the area element of the surface is a vecotr that points normal to the element and inwards

Question 26

KIT - the gradient of the term in the square bracket is

Answer 26

resolved - by the scalar product - giving the component along the line joining the area element to the point P

Question 27

KIT - the scalar product gives the

Answer 27

normal component of the spatial derivative of the field ie the component of the field derivative along the line joining the area element to the point P

Question 28

KIT - we will let Ψ be

Answer 28

the (scalar) E-field of the light (there is a more complicated vector treatment to account for polarisation but beyond scope of this course)

Question 29

KIT - r is the

Answer 29

distance from an area element on the closed surface to the point P where we want to calculate the field

Question 30

KIT - we need to know

Answer 30

Ψ and ∇Ψ the magnitude of the filed and its gradient over the closed surface of integration

Question 31

we will denote the field emerging from the aperture by

Answer 31

ΨQ this will be the field incident on the aperture, modified by any aperture function

Question 32

aperture function

Answer 32

x1 if its a hole more complicated if the aperture was filled with something that affects the amplitude and/or phase of the transmitted light

Question 33

KIT - to make the calculation simpler, we choose

Answer 33

the big surface enclosing P such that it is a large sphere centred on P also require that the aperture is part of this surface so there will be a bit of the surface linking sphere to aperture

Question 34

first term of the KIT

Answer 34

∇[e^ikr /r] = [e^ikr /r][ik-1/r]n

Question 35

first KIT term - we can say that kr>>1 because

Answer 35

we are calculating Fraunhofer diffraction so the distance from the aperture to any point P is much larger than the wavelength

Question 36

first KIT term - n is

Answer 36

the unit vector along a line from the aperture to P need to think about its direction

Question 37

the gradient of a function points in

Answer 37

the direction in which the function increases

Question 38

our function e^ikr/r decreases as a distance from the aperture towards P increases so its gradient will

Answer 38

point back towards the aperture, along the line joining the aperture to P

Question 39

kr>>1 so k>>1/r so ∇[e^ikr/r]=

Answer 39

ik e^ikr/r (-n)

Question 40

the Fresnel-Kirchhoff diffraction formula captures the elements of the story that

Answer 40

simple Huygens wavelet picture missed 1. wavelength dependence 2. phase offset indicated by -i 3. the obliquity factor

Question 41

obliquity factor

Answer 41

(n+n'/2) . dΣ ensures that waves from the aperture propagate in a forward direction

Question 42

applying KIT to fraunhofer diffraction - for simplicity assume that

Answer 42

incident plane-wave illumination is on-axis ie k aligned with the normal to the aperture so optic axis is aligned with direction n

Question 43

Fraunhofer diffraction involves large distances and small diffraction angles. So for any point P in the image plane we can say that the corresponding unit vector n is

Answer 43

close to being aligned with the optic axis n and n' can be considered very close to parallel so (n+n'/2) . dΣ = dΣ

Question 44

We are interested in the separation because we want to know the

Answer 44

relative phase of a beam arriving at P from our area element with respect to a beam arriving there rom the origin of the optic axis that ahs travelled distance p p=sqrt(X^2+Y^2+L^2)

Question 45

p

Answer 45

this term is independent of the position of the area element within the aperture we are considering

Question 46

[xX/p +yY/p]

Answer 46

this term is the one that matters for Fraunhofer diffraction

Question 47

[x^2/2p+y^2/2p]+...

Answer 47

these terms are negligibly small in the fraunhofer region since we are far from the aperture

Question 48

In the Fraunhofer region, then, we have the path travelled by the contribution arriving at P due to our area element in the aperture as

Answer 48

r=p-[xX/p +yY/p] =p-[xθx +yθy] since θx=X/p and θy=Y/p

Question 49

The diffraction pattern results from the

Answer 49

differing phases of the components arriving at P from all the area elements in the aperture. ie a constant additive phase (p term) is irrelevant so r=-[xθx +yθy]

Question 50

If we assume plane parallel illumination that is constant over the aperture, we can replace ψQ by

Answer 50

a function A(x,y) this describes the form of the aperture in the x-y plane

Question 51

at its simplest, A(x,y)=

Answer 51

1 where the aperture is 0 everywhere else

Question 52

final Fruanhofer diffraction result is powerful because

Answer 52

**the integral is the 2D fourier transform of the aperture function**

Question 53

It is conventional to express the 2D Fourier Transform of the aperture function A as a

Answer 53

function of angular spatial frequencies

Question 54

In the diffraction pattern, at a distance ρ from the aperture, the “X” coordinate in the diffracted field (corresponding to a spatial frequency k X will be located at

Answer 54

X=pthetax = pkx/k and equivalently for the y-coord

Question 55

Small spatial features correspond to

Answer 55

large spatial frequencies and so a big diffraction pattern

Question 56

large spatial features correspond to

Answer 56

small spatial frequencies and negligible diffraction