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Modern Optics, Prof. Ruiz, UNCA doctorphys.com Chapter S. Fraunhofer Diffraction I. S0. Wave Properties 1. Reflection 2. Refraction Plastic Block. Courtesy Wikipedia. Author ajizai. Released into the Public Domain.

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3. Diffraction

Courtesy The Physics Teacher, Copyright (c) Sabina Zigman High School Student Photo, Beach in Tel Aviv, Israel

4. Interference

Courtesy Paul Doherty

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S1. Spherical Waves. The wave equation in three dimension, which we have derived earlier, is

2

2

2 2

1

c t

.

The Laplacian in spherical coordinates, which we have derived earlier, is

2

2 2

2 2 2 2 2

1 1 1( ) (sin )

sin sinr

r r r r r

We consider a spherically symmetric situation so that

( , ) ( , )r t r t

, i.e., ( , , , ) ( , )r t r t .

The Laplacian then reduces to the derivatives in r since

( , ) ( , )0

r t r t

.

2 2

2

1( )r

r r r

2

2

2 2

1

c t

=>

2

2

2 2 2

1 1( )r

r r r c t

Try

( )( , )

i tR rr t e

r

since 2

1( , )* ( , )r t r t

r .

Substituting

( )( , )

i tR rr t e

r

into

2

2

2 2 2

1 1( )r

r r r c t

gives

2

2

2 2 2

1 ( ) 1 ( )( ) ( )

i t i tR r R rr e e

r r r r c t r

2

2

2 2 2

1 ( ) 1 ( ) ( )( )

i t

i t d d R r R r d ee r

r dr dr r c r dt

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2

2 2

2 2 2

1 ( ) 1 ( ) ( )( ) ( )

i t i t i td d R r R r R re r i e e

r dr dr r c r c r

Remember ck

. Therefore

2

2

2 2

1 ( ) i tR rk e

c t r

.

2 2

2

1( )

d d R Rr k

r dr dr r r

2 2

2 2

1 '( ) 0

d R R Rr k

r dr r r r

2

2

1' 0

d RrR R k

r dr r

2

2

1( ' " ') 0

RR rR R k

r r

2

2

1( ") 0

RrR k

r r

2"0

R Rk

r r

2

" 0R k R

2

2

2

( )( )

d R rk R r

dr

( )ikr

R r Ae

=>

( )( , )

i tR rr t e

r

=> ( , )

ikr i te

r t Cr

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The wave moving outward award from some source of light is

( , )

ikr i te

r t Cr

=>

( )

( , )

i kr te

r t Cr

S2. Huygens-Fresnel Principle.

Christiaan Huygens (1629-1695) Painting by Bernard Vaillant Museum Hofwijck, Voorburg, The Netherlands Huygens-Fresnel Principle says that you can replace a wave front with "baby waves" to get the wave of the future. The baby wave:

( )

( , )

i kr te

r t Cr

Augustin-Jean Fresnel (1788-1827). From Frontispiece of his collected works (1866) Image Courtesy Wikipedia We will be analyzing diffraction patterns on a screen and we will be interested in the brightness averaged over time. So we can take the baby wave as follows.

Baby Wave (time independent):

ikre

Cr

Integrate over the source:

ikre

C dSr

Baby Waves – Integrate over the Opening Area

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ikre

C dAr

The distance a baby-wave source to a screen far away is equal to r.

Wikipedia: Arne Nordmann (norro). Creative Commons

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S3. Fraunhofer Diffraction.

Joseph von Fraunhofer (1787-1826) Physicist, Astronomer, Inventor, Lens Manufacturer Invented the spectroscope. Absorption spectra of stars called Fraunhofer lines in his honor.

The Solar Spectrum (Fraunhofer Lines)

Courtesy Wikipedia: Gebruiker:MaureenV. Released to

the Public Domain. Far-Field diffraction, i.e., the diffraction seen on a screen far away from a source, is named after Fraunhofer. Shine light at a small opening. The spatial part of the wave using the Huygens-Fresnel principle is

( )

ikr

o

er E

r .

But we need to integrate over the opening, i.e., aperture area to get the wave amplitude pE at

a point on the far away screen.

ikr

p o

A

eE E dA

r

Let the thin slit have dimension b for the width and the length (or height) be h.

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S4. Single Slit.

The integral is then

/ 2

/2

bikr ikr

p o o

A b

e eE E dA E hdu

r r

.

For large r we take

1 1

or r

but can’t in ikr

e since we have trig functions that max out at 1.

Therefore

/ 2

( sin )

/2

o

b

ik r uo

p

o b

E hE e du

r

. Redefine o

o

EC

r

.

/ 2

( sin )

/2

o

b

ik r u

p

b

E Ch e du

=>

/ 2

sin

/2

o

b

ikr iku

p

b

E Ch e e du

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/ 2

sin

/2

o

b

ikr iku

p

b

E Che e du

/ 2

( sin )

/2

o

b

ikr ik u

p

b

E Che e du

/2

( sin )

/2sin

o

bik u

ikr

p

b

eE Che

ik

( sin ) /2 ( sin ) /2

sin

oikr

ik b ik b

p

CheE e e

ik

1 1sin sin

2 2

sin

oikri kb i kb

p

CheE e e

ik

It is customary to define

1sin

2kb . Then

sin

oikr

i i

p

CheE e e

ik

=> ( sin / 2) 2

oikr i i

p

Che e eE

k i

sin( sin / 2)

oikr

p

CheE

k

=> sin

( sin / 2)

oikr

p

ChbeE

kb

sinoikr

pE Chbe

The area of the slit is hb A .

Irradiance

*

*1 1 sin sin

2 2

o oikr ikr

p pI E E CAe CAe

.

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2 2 21 sin sin 1 sin( )

2 2

o oikr ikrI CAe CAe C A

2 2sin( ) sinc

o oI I I

But the max occurs when

1sin 0

2kb for the sinc function.

0 => 0

Therefore we have the two following equivalent expressions, where ( sin ) / 2kb .

2

( ) (0)sincI I

2

2

sin( ) (0)I I

Single-Slit Diffraction

Wikipedia: jkrieger. Creative Commons In the figure you see bright and dark regions, places of constructive and destructive interference respectively. Where do the extrema occur.

2

2

sin( ) (0)I I

1sin

2kb

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(a) The minima. The angles for these are easy since we want ( ) 0I . These places of

destructive inference occur when the sine function hits 0, except for the 0° case since the sinc function is 1 there. But all the other usual places of zero for the sine function apply.

, 2 , 3 ,...

(b) The maxima. Now we do the max-min problem on

sin

.

2

sin 1 1( ) sin cos 0

d

d

=> 2

1 1sin cos

=>

sin

cos

=> tan A transcendental equation!

The above graph is a sketch. Here is how to find the values. Set your calculator to radians and use the graph as a guide as follows.

Let the case where the graphs join at zero be x0 = 0. If you hit tan for 0 you get 0.

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Call the next case 1: x1 = 1.4 from the graph since 3/2 = 1.5 Try 1.4 , 1.42 , etc. and

you will find x1 = 1.43 is very close. Similarly, try x2 = 2.4 since 5/2 = 2.5 You can

jump around a little and find that x2 = 2.46 = 7.73 is very close. Trick exploration gives you an

intuitive feel of the transcendental equation. An efficient way to find the roots is to resort to online WolframAlpha and enter tan x = x. The first few solutions are

00x

14.493...x

27.725...x

Can you find this below in the plot of sinc2(x)?

Plot of sinc2(x) from desmos.com

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S5. Double Slit.

Our previous integral

/ 2

( sin )

/2

o

b

ik r u

p

b

E Ch e du

can be adapted easily.

Since the screen is so far away, we can slip the zero reference down to the bottom of the lower slit so that

( sin ) ( sin )

0

o o

b a b

p ik r u ik r u

a

Ee du e du

Ch

.

sin sin

0

o

b a b

p ikr iku iku

a

Ee e du e du

Ch

sin sin

0sin sin

o

b a biku iku

p ikr

a

E e ee

Ch ik ik

sin ( )sin sinsin 1o

p ikr ikb ik a b ikaE

e ik e e eCh

sin sin sin sinsin 1o

p ikr ikb ika ikb ikaE

e ik e e e eCh

sin sin sinsin 1 ( 1)o

p ikr ikb ika ikbE

e ik e e eCh

sin sinsin ( 1)(1 )o

p ikr ikb ikaE

e ik e eCh

sin /2 sin /2 sin /2 sinsin ( )(1 )o

p ikr ikb ikb ikb ikaE

e ik e e e eCh

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sin /2 sinsinsin 2 sin (1 )

2

op ikr ikb ika

E kbe ik e i e

Ch

sin /2 sin /2 sin /2 sin /2sinsin 2 sin ( )

2

op ikr ikb ika ika ika

E kbe ik e i e e e

Ch

sin /2 sin /2sin sinsin 2 sin 2cos

2 2

op ikr ikb ika

E kb kae ik e i e

Ch

Recall our ( sin ) / 2kb . Now add ( sin ) / 2ka .

sin 2 sin 2cosop ikr i i

Ee ik e i e

Ch

4 cos sinsin

oikr i i

p

ChE e e e i

ik

4cos sin

sin

oikr i i

p

ChE e e e

k

2cos sin

sin / 2

oikr i i

p

ChbE e e e

kb

Recall the area of the slit is hb A .

sin(2 ) cosoikr i i

pE CA e e e

Irradiance *1

2p p

I E E

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*

1 sin sin(2 ) cos (2 ) cos

2

o oikr ikri i i iCA e e e CA e e e

21 sin sin(2 ) cos cos

2

o oikr ikri i i iCA e e e e e e

2

2 2

2

1 sin(2 ) cos

2

o oikr ikr i i i iCA e e e e e e

2

2 2

2

1 sin(2 ) 1 1 1 cos

2CA

2

2

max 2

sin( ) cosI I

The max occurs when 0 .

Then,

sin0

2

kb

and

sin0

2

ka .

With

sin1

and cos 1 . Finally, max

(0 ) 1 1I I .

2

2

2

sin( ) (0) cosI I

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Interference Due to Two Slits cos2(faint red), Diffraction sinc2(dashed), and solid red I().

Courtesy GeoGebra. Author Dave Nero. Creative Commons

Same as Above with Missing Orders Labeled

Courtesy LibreTexts, UC Davis. OpenStax University Physics. Creative Commons

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Single Slit:

2

2

sin( ) (0)I I

1sin

2kb

Wikipedia: Jordgette. Creative Commons

Double Slit:

2

2

2

sin( ) (0) cosI I

sin

2

ka

1sin

2kb