Fraunhofer Diffraction

Question 1

Why does Fraunhofer diffraction use lenses

Answer 1

so that the source and fringe pattern can both be at convinient distances from the aperture

Question 2

What are the requirements for fraunhofer diffraction to be observed

Answer 2

the incoming and outgoing waves approach being planar over the diffracting apertures or obstacles

Question 3

What is crucial for determining the resultant field for Fraunhofer diffraction

Answer 3

the phase of each contribution at the screen due to differences in path traversed

Question 4

how can path difference be described for planar wavefronts passing through an aperture

Answer 4

linear function of the 2 aperture variables

Question 5

what is the definitive mathematical criterion for fraunhofer diffraction

Answer 5

linearity in the aperture variables

Question 6

what is the formula for the aperture size

Answer 6

D^2 = x0^2+y0^2

Question 7

what is the formula for fraunhofer diffraction if we neglect the phase factors and explicitly write the aperture function

Answer 7

E (𝑥1, 𝑦1) ∝ double integral −∞ to ∞ exp {− 𝑖𝑘/𝑧(𝑥0𝑥1 + 𝑦0𝑦1)} 𝐴 (𝑥0, 𝑦0) 𝐸(𝑥0, 𝑦0)𝑑𝑥0𝑑𝑦0

Question 8

What is notable about the fraunhofer diffraction formula

Answer 8

it is a fourier transform of the aperture field from one position x0 to another x1

Question 9

How do you write the Fraunhofer diffraction in terms of the off-axis k vectors which are kx=kx1/z and ky=ky1/z

Answer 9

𝐸 (𝑘𝑥, 𝑘𝑦) ∝ double integral from −∞ to ∞ exp{−𝑖 𝑘𝑥𝑥 + 𝑘𝑦𝑦} 𝐴(𝑥, 𝑦) 𝐸(𝑥, 𝑦)𝑑𝑥𝑑𝑦

Question 10

How can we treat an aperture (single slit) of width b

Answer 10

divide it into N coherent point sources each of extent δy=b/N

Question 11

What can we say about point sources along an aperture which is uniformly illuminated by a plane wave

Answer 11

the amplitude of each point source is proportional to δy, they are a source of spherical waves which would all be in phase

Question 12

How do you evaluate the intensity of received wave at a point on a screen

Answer 12

the sum of all the waves arriving from all point sources in the aperture

Question 13

In a direction θ, the wave emitted from a point source at y is out of phase with the wave emitted from y=0 by how much

Answer 13

kysinθ

Question 14

What is the field of a wave emitted from a point source at y on an aperture

Answer 14

𝛿𝐸 = 𝛿𝑦 exp(−𝑖𝑘𝑦𝑠𝑖𝑛𝜃)

Question 15

What is the intensity I(𝜃) on the screen for a single slit aperture of width b

Answer 15

𝐼 (𝜃) = 𝐼0 [sin 𝛽/𝛽]^2 = 𝐼0 sinc^2𝛽

Question 16

what is the formula for sinc𝛽

Answer 16

sin𝛽/𝛽

Question 17

What is the formula for 𝛽 where the wave approaches the slit at an angle i

Answer 17

𝛽=1/2 kb sin𝜃 = 𝜋/𝜆 𝑏 (sin i + sin 𝜃)

Question 18

(in words) what is 𝛽

Answer 18

half the phase shift at angle theta across the aperture b

Question 19

In single slit diffraction what is happening at the first zero?

Answer 19

the ray from the centre of the slit exactly cancels the edge ray due to the λ/2 path difference and so on across the half slit

Question 20

What are the conditions for minima from single slit diffraction

Answer 20

I(𝜃) = 0 when sin𝛽=0
𝛽 = +-mpi
sin𝜃 = mλ/b

Question 21

what are the conditions for maxima from single slit diffraction

Answer 21

𝛽=0 or when we minimise sinc^2𝛽
sin𝛽=0 or tan𝛽=𝛽

Question 22

Describe the fourier approach to fraunhofer diffraction from a slit

Answer 22

it is the fourier transform of a rect function which is the sinc function, hence the irradiance is sinc^2

Question 23

what is the fourier transform pair of a delta function

Answer 23

1

Question 24

what is the fourier transform pair of exp(iw0t)

Answer 24

2pi delta(w-w0)

Question 25

what is the fourier transform pair of cos(w0t)

Answer 25

pi [delta(w-w0)+delta(w-w0)]

Question 26

what is the fourier transform pair of sin(w0t)

Answer 26

pi/i[-delta(w-w0)+delta(w-w0)]

Question 27

what is the fourier transform pair of rect(t/T) top hat

Answer 27

Tsinc(Tw/2)

Question 28

what is the fourier transform pair of exp(-t^2/2T^2) gaussian

Answer 28

Tsqrt(2pi) exp(-T^2w^2/2)

Question 29

What happens for Fraunhofer diffraction from 2 slits

Answer 29

the rapidly varying double-slit interference pattern is modulated by the single slit diffraction pattern which then functions as an envelope to the whole pattern

Question 30

How can we think of two slit diffraction using the fourier method

Answer 30

the two slits are two rect functions

Question 31

what is 𝛽 for multi slit diffraction

Answer 31

𝛽 = 𝜋𝑏 sin 𝜃/𝜆

Question 32

When the slit separation is a, what is 𝛼 for multislit diffraction

Answer 32

𝛼 = 𝜋𝑎 sin 𝜃/𝜆

Question 33

For 2 slit diffraction, with slit width b and separation a, what happens when a=b

Answer 33

the 2 slits become a single slit of width 2b

Question 34

In diffraction patterns when does a missing order appear?

Answer 34

When the minimum of the single slit diffraction pattern overlaps a maximum of the multi slit interference

Question 35

in terms of a and 𝛼, what are the conditions for missing orders

Answer 35

𝑎 = (𝑝/𝑚) 𝑏 or 𝛼 = (𝑝/𝑚) 𝛽

Question 36

Considering N long narrow parallel slits of width d and centre to centre separation a, how do we find the diffraction pattern

Answer 36

integrate over one slit N times

Question 37

What is the N slit diffraction pattern

Answer 37

I(𝜃) = 𝐼0(sin^2 𝛽/𝛽^2)^2 (sin 𝑁𝛼/sin 𝛼)^2
where I0 is the flux density in the normal direction from any slit

Question 38

What is I(0) for multislit diffraction

Answer 38

N^2I(0)

Question 39

For multislit diffraction, when do principle maxima occur

Answer 39

alpha = 0, ±pi, ±pi, … or, equivalently, when 𝑎 sin 𝜃𝑚 = 𝑚𝜆, 𝑚 = 0, ±1, ±2

Question 40

What is a diffraction grating

Answer 40

an N slit device where N is very large and a/d is small

Question 40

How many subsidary maxima are seen in the N slit diffraction pattern

Answer 40

N-2

Question 40

How can we think of an N slit as a convolution

Answer 40

a convolution of a single slit with N delta functions

Question 41

Describe diffraction from a rectangular aperture

Answer 41

The diffracted field is a sinc in x and y because the fourier transform of a rect function is a sinc

Question 42

Describe diffraction from a circular aperture

Answer 42

it yields a diffracted airy pattern involving the bessel function Jn(u)

Question 43

What is the radius of the airy disc

Answer 43

1.22R𝜆/2a
R is dist to screen
a is aperture radius