chapter 11 fraunhofer diffraction - erbionerbion.com/index_files/modern_optics/ch11.pdf · chapter...

$: Chapter 11 Fraunhofer Diffraction - Erbionerbion.com/index_files/Modern_Optics/Ch11.pdf · Chapter 11 Fraunhofer Diffraction Lecture Notes for Modern Optics based on Pedrotti & Pedrotti$
Chapter 11Fraunhofer Diffraction

Lecture Notes for Modern Optics based on Pedrotti & Pedrotti & Pedrotti

Instructor: Nayer EradatInstructor: Nayer EradatSpring 2009

4/30/2009 1Fraunhofer Diffraction

DiffractionDiffraction• Diffraction is any deviation from geometrical optics that results from

obstruction of the wavefront of light.

• Obstruction causes local variations in the amplitude or phase of the wave

• Diffraction can cause image blurriness. So the aberrations.

• An optical component that is free of aberrations is called diffraction‐p plimited optics and is still subject to blurriness due to diffraction.

• Huygens‐Fresnel principle: every point on a wavefront can be considered as a source of secondary spherical wavelets(Huygens). The field beyond a y p ( yg ) ywavefront is result of the superposition of these wavelets taking into account their amplitudes and phases (Fresnel).

4/30/2009 Fraunhofer Diffraction 2

Diffraction vs. interference:

Diffraction phenomena is calculated from interference of the wavesfrom interference of the waves originating from different points of a continuous source.

Interference phenomena is calculates interference of the beams originationinterference of the beams origination from discrete number of sources.

4/30/2009 3Fraunhofer Diffraction

Mathematical treatment of DiffractionMathematical treatment of Diffraction

• Far‐field or Fraunhofer Diffraction or : when both the source and observation screen are far from the diffraction causing aperture, so that the waves arriving at the aperture and screen can be approximated by plane waves.

• Near‐field or Fresnel diffraction: when the curvature of the wavefronts an not be ignoredignored

• We will only investigate the far‐field diffraction in this course using the Huygens‐Fresnel principle with one approximation.

• When EM waves hit the edges of the aperture there is oscillations of the electronsWhen EM waves hit the edges of the aperture there is oscillations of the electrons in the matte that cause a secondary field (edge effect).

• In HF approximation we ignore edge effect. So beyond the aperture there is no field. This holds only if the observation point is far from the aperture.


Diffraction from a single slitWe simulate the geometrical arrangement for the Frounhofer diffraction by placing a point source at the g g y p g pfocal point of a lens and a screen at the focal point of a lens after the amerture. The light reaching point

on the screen is from interference of the parallel rays from different points on the aperture. We consider each interval of as a source and calculate contribution of the all of theseP

ds sources.

0

: optical path length from the to the strength of the electric field contribution from each point.

for the field from the center of the slit. C ib i f h i l

L

r ds PEr r

d=

h f i i h i l l fP dEContribution of each interval to ds

( )-

Spherical wave

the wavefront at point is a spherical wavelet of :

P

i kr tLP

P dEE dsdE e

rω⎛ ⎞= ⎜ ⎟

⎝ ⎠

( )( )0

pamplitude

-

0

i k r tLP

E dsdE er

E d

ω+Δ⎛ ⎞= ⎜ ⎟+ Δ⎝ ⎠

⎛ ⎞

0

Path difference affets the amplitude

i kLP

E dsdE er

⎛ ⎞= ⎜ ⎟+ Δ⎝ ⎠

( )0

Path difference affets the phase

We ignore in the denuminator

r t ikeω− Δ

Δ


0

We ignore in the denuminatorsince r

ΔΔ <<

Diffraction from a single slit( )0i kr t ikL

PE dsdE e eω− Δ⎛ ⎞

= ⎜ ⎟

( )0

0

b/2 sin

/ 2

The total electric field at the point

P

i kr t iksLP P b

rP

EE dE e e dsr

ω θ−

−

⎜ ⎟⎝ ⎠

= =∫ ∫

( ) ( )( ) ( )

0 0

0

/ 2 sin / 2 sin / 2sin

0 0/ 2sin sin

1

slitb ikb ikbiks

i kr t i kr tL LP

b

r

E Ee e eE e er ik r ik

θ θθω ω

θ θ

−− −

−

⎛ ⎞ −= =⎜ ⎟

⎝ ⎠1With =β

( )0

0 0

sin and using Euler's formula2

sinsin

i i h d h i i h h di f h

i kr tL LP

kb

E b E b cE er r

ω

θ

βββ

β θ

−= =

varies with and that varies with the distance from the screen.

b 1 can be interpreted as phase difference k =k sin sin2 2

sh

k kb

β θ

β θ β θ

β

Δ → = Δ =

ows the magnitude of the phase difference at point P between the points from the center and shβ

20

ows the magnitude of the phase difference at point P between the points from the center and either endppoint of the slit. The irradiance at is proportional to the :I P E

E b⎛ ⎞⎛ ⎞2 22 2i iE bβ β⎛ ⎞


20 00

02 2Lc c E bI Er

ε ε ⎛ ⎞⎛ ⎞= = ⎜⎜ ⎟⎝ ⎠ ⎝

2 220

0 0 02 20

sin sin with I we have sin2

Lc E b I I I cr

εβ β ββ β

⎛ ⎞= = =⎟ ⎜ ⎟

⎠ ⎝ ⎠

Diffraction from a single slit2

20 02

sinThe diffraction pattern of the light from a silt by width of b at a distanc screen sin where I I I cβ ββ

= =

( )sin /2 is the phase difference betwwen the light from the center of the slit the ek kbβ

β θ= Δ =

( )0 0

ndpoints.

sinLet's plot the I. For the central maximum: limsin lim 1cβ β

βββ→ →

⎛ ⎞= =⎜ ⎟

⎝ ⎠

( )

For zeros of the sinc function: 1sin 0 sin sin where 1, 2,... 2

0 d t l d t b

kb m b m m

β

β β θ π θ λ

⎝ ⎠

= → = = → = = ± ±

f th i l t f th i f ti0 does not lead to a zero be cam = use of the special property of the sinc function. if is the distance of the slit from the screen, then the location of the minima on the screen can be found using small-angle approximation:

f

m y m fλ λsin tanθ ≅

2

m/ b

d sin cos sinFor the other maxima: 0d

mm

y m fy f yf b

λ λθ

β β β ββ β β

= → = → =

⎛ ⎞ −= =⎜ ⎟

⎝ ⎠Cental lobe: image of the slit with roundes edges. Side lobes: what causes blurriness in the image. Angular width of the central lobe: sinb θ λ= →


1 12, , The central lobe will sprad as the slit-size gets smaller.

b b bλ λ λθ θ θ−≅ ≅ → Δ =

Maxima of the sinc functionThe maxima of the sinc function are solutions of this equation:

iβ β β2

cos sin 0 tan

We can solve this equation by parametrically. A l l i i i i

β β β β ββ−

= → =

An angle equal to its tangant is intersection

of the tan

yy

ββ

=⎧⎨ =⎩

We see that the maxima are not exactly half way between the minima. They occur slightly erlier and as increases the maxima shift βtowards the center.Ratio of the irradiances at the central peak

maximum to the first of the secondary maxima ?maximum t

( )( ) ( ) ( )

2 2

0 02 2

1.43 1.43 1.43

o the first of the secondary maxima ?

sin / 1 1.43 20.18sin 1.43 0.952sin / sin /

II

β β

β π β π β π

β β ππβ β β β

= =

= = =

⎛ ⎞= = = =⎜ ⎟⎜ ⎟

⎝ ⎠


1.43 0 00.047 or only 4.7% of the II Iβ β

β π β= ==

Beam spreadingThe angular spread of the central maximm =2 /b is independent of the distance between the slit

and screen. So as screen moves away from the slit the nature of the diffraction pattern does not change

θ λΔ

. a d sc ee . So as sc ee oves away o t e s t t e atu e o t e d act o patte does ot c a ge.W is the width of the central maximum,

2W=L LbλθΔ =

All of the beams spread according to diffraction as they propagate due to the finite size of the source .

Even if we make them parallel with a lens still they will spread because of the diffraction. Example:

9

Example:

What is the width of a parallel beam of 546 and width of b=0.5mm after propagation of 10 meter?

2 2 10 546 10

nm

L

λ

λ −

=

9

-3

2 2 10 546 100.5 10

21.8This tre

LWb

W mm

λ × × ×= =

×=

atment of beam spreading isThis tre2

atment of beam spreading is

correct for far-field where

area of apertureor more generall

bL

L

λ>>

>>


por more generally Lλ

>>

Rectangular apertureFor a slit of length a and width b we calculate the diffraction pattern. We assumed a>>b in previous section. When a and b are comparable and both small we have large contriibutions to the diffraction pattern fromWhen a and b are comparable and both small we have large contriibutions to the diffractio

22

0 0

n pattern from

sin sinboth dimensions: sin and sin2 2

Z f th tt d t d 1 2

k kI I where a I I where b

m f n f

α βα θ β θα β

λ λ

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

± ±m nZeros of the pattern due occur at: y and x m,n= 1, 2,...

Using a bit more a

f fb a

= = ± ±

( )0 0 0/ 2 b/2/ /

/ 2 / 20

nalysis we can write

h di f h b i i h d h di f h f

ai kr t ikXx r ikYy rLP P a b

slit

EE dE e e dx e dyr

X Y

ω−

− −= =∫ ∫ ∫

, are the coordinates of the observation point on the screen and x,y are the coordinates of the surface element on thX Y

( )( )2 20

e aperture. The total irradiance turns out to be the product of the irradiance functions on each

dimension: sin sinI I c cβ α=


Circular aperture

0

For finding the diffraction pattern caused by a circular aperture we assume the incremental electric field amplitude at point P due to the surface element (on the aperture) is / .AdA dxdy E dA r= 0field amplitude at point P due to the surface element (on the aperture) is / . Then we

AdA dxdy E dA r

( )( ) ( )0 00- sinsin

0 0

integrate the incremental field over the entire aperture area. The resulting E at point P is

i k r t i kr tr iskA AP Ps

A

E dA EdE e E e e dAr r

ω ω θθ

+Δ −Δ<<Δ=

⎛ ⎞= ⎯⎯⎯⎯→ =⎜ ⎟+ Δ⎝ ⎠

∫∫0 0

By choosing the elemental areArear r+ Δ⎝ ⎠

2

a the rectangular area of we can reduce the double integral to a single one.

x

dA xds=

⎛ ⎞

( ) 2 20

2 2 2 2

sinP

0

x 22

2ERi kr t isk R sAR

s R x R s

E e e dsr

ω θ+− −

−

⎛ ⎞ + = → = −⎜ ⎟⎝ ⎠

= ∫

( ) 20

0

1P

0

Substituting v / and sin2E i kr t i v vA

s R kE R e e dr

ω γ

γ θ

− −

= =

=1

1v

+

−∫( )

( )

21 11

1

1

From the intergral table:

where J is the first order Bessel function of the first kind.

i v v Je dvγ π γ

γγ

+ −

−=∫


( ) ( ) ( ) ( )3 51

1 2 2 2 0

/ 2 / 2 1J and lim2 1 .2 1 .2 .3 2

Jγ

γ γ γγγγ→

= − + − =

Field from an arbitrary shape aperture on a distant screen For finding the diffraction pattern caused by an aperture of arbitrary shape we assume the incremental electric field amplitude at point P due to the surface element (on the aperture) is /dA dxdy E dA= relectric field amplitude at point P due to the surface element (on the aperture) is /AdA dxdy E dA=

( )

0

0

. Then we integrate the incremental field over the entire aperture area. The resulting E at point P is

where i t krAp

Aperture

r

EE e dA dA dxdyr

ω −= =∫∫

( ) ( )

0

1/ 2 1/ 22 2 2 2 2 20 with

Aperture

r X x Y y Z r X Y Z⎡ ⎤ ⎡ ⎤= − + − + = + +⎣ ⎦⎣ ⎦

( ) ( )1/ 22 2 2x y Xx Yy⎡ ⎤+ +

⎢ ⎥

y

x

Y

XPrΔwe ha

( ) ( )

( )

0 2 20 0

1/ 2

2 2 20 0 2

2ve: 1

21

y Xx Yyr r

r r

Xx Yyr x y r r

+⎢ ⎥= + −⎢ ⎥⎣ ⎦

⎡ ⎤+>> + → −⎢ ⎥

r0θdy

dx P0

rΔ

( )0

0 0 20

00

only first two terms.

yr

Xx Yyr x y r r

r

⎢ ⎥⎣ ⎦

+>> + → − ScreenAperture

( )( ) ( )0

0 0 0/

0 0

Xx Yyi t k r

r i t kr ik Xx Yy rA LP P

Aperture Apertu

E EE e dA E e er r

ωω

⎛ ⎞⎛ + ⎞− −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ − +⎝ ⎠⎝ ⎠= → =∫∫

any shape

For specific shape of aperture the integral limits and relationship between x and yre

dxdy∫∫


For specific shape of aperture the integral limits and relationship between x and y will change. We will look at two examples of rectangular and circular apertures.

Rectangular aperture

( ) ( )0 0/i t kr ik Xx Yy rAP

EE e e dxdyr

ω − += ∫∫

( ) ( ) ( )0 0 0

0any shape

/ 2 / 2/ /

/ 2 / 2

Aperture

b ai t kr ik Yy r ik Xx rAP b a

r

EE e e dy e dxr

ω −

− −= ∫ ∫

( )0

0

0

sin sini t krAp

r

EE e b ar

ω β αβ α

− ⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

y

x

y

xP

r( )0

0

2

sin sini t krAp

AEE er

ω β αβ α

− ⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

r0θdy

dx P0

rΔ

( )0

2

00

i t krAAEI er

ω −⎛ ⎞= ⎜ ⎟⎝ ⎠

2

ScreenAperture

( ) ( )2 2sin sin , 0 I X Y I β α

β α⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

ScreenAperture

4/30/2009 Fraunhofer Diffraction 130 0

k kWhere = asin and = bsin 2 2 2 2x yk X k Ya b

r rα θ β θ= =

Circular aperture IFor circular aperture we introduce the spherical coordinates because of the symmetry of the problem. On plane of the aperture: cos , sinx yρ φ ρ φ= = On plane of the screen: X cos ,q= Φ

( ) ( ) ( ) ( ) ( )0 0 00

2/ cos/

Y sin Differential surface element:

Ri t kr i t kr i k q rik Xx Yy rA L

qdA d d

E EE dA d dπ

ω ω ρ φ

ρ ρ φ

φ− − −Φ+

= Φ=

∫∫ ∫ ∫( ) ( ) ( ) ( ) ( )

( )

0 0 00 / cos/

0 0 0 0any shape

Wher we used: sin sin cos cos

i t kr i t kr i k q rik Xx Yy rA LP

Aperture

E e e dA e e d dr r

Yy Xx q

ω ω ρ φ

ρ φ

ρ ρ φ

ρ φ φ

Φ+

= =

= =

+ = Φ + Φ

∫∫ ∫ ∫

( )cosqρ φ= −Φ

Because of the axial symmetry of the problem, the solution must be independent of and

since point P is an arbitrary point on the screen we choose to have 0 to simplify things.

E iAE e ω

Φ

Φ =

= ( ) ( ) ( )0 0

2/ cos

at kr i k q rd e d

πρ φρ ρ φ− ∫ ∫ y yP

0

E er

=

( ) ( )

0 0

A tabulated integral. Solution is a Bessel function of the first kind

2 cosl f i ki dm

i mv u v

d e d

i d

ρ φ

π

ρ ρ φ= =

−+

∫ ∫

∫

y

x

y

xP

rdx r φ( ) ( )

( )

cos

0

2 cos0 0

Bessel function 1st kind: 2

1For m=0: and with 2

i mv u vm

iu v

iJ u e dv

J u e dvπ

π

π

+=

=

∫

∫ u k qρ=

r0θdy

dx

P0

Δρ


( ) ( )0P 0 0

0 0

E /R

i t krAE e J k q r dr

ω

ρ

ρ ρ ρ−

=

= ∫

Bessel functions of the first kind (MATLAB)Bessel functions of the first kind (MATLAB)u = (0:0.1:15)BJ0=besselj(0,u);BJ1=besselj(1,u);BJ2=besselj(2 u);

1Bessel functions of first kind

J0

BJ2=besselj(2,u);BJ3=besselj(3,u)plot(u,BJ0,u,BJ1,u,BJ2,u,BJ3);legend('J0','J1','J2','J3')title('Bessel functions of first kind');

0.5

J0J1J2J3

title( Bessel functions of first kind );xlabel('u'),ylabel('J')grid on

0

J

0 5

0

15

0 5 10 15-0.5

u

Circular aperture II

( ) ( )0P 0 0E 2 /

Ri t krAE e J k q r dω π ρ ρ ρ−= ∫ ( )

( )

P 0 00 0

E 2 /

Using the one of many properties of the

bessel functions: ' ' ' we getu

e J k q r dr

u J u du uJ u

ρ

π ρ ρ ρ=∫

∫ ( )

( ) ( ) ( ) ( ) ( )0

0 10

kRq/r20 0 0 0 0 1 0' 0

0

bessel functions: ' ' ' we get

/ / ' ' ' / /R

u

u J u du uJ u

J k q r d r kq J u u du kRq r J kRq rρ

ρ ρ ρ=

=

=

= =

∫

∫ ∫

( )0

0

2 0P 1

0

E 2i t krA rE ke R Jr kRq

ρ

ω π−= ( )0 0P 1

0 0 0

2E i t krA rAERq kRqe Jr r kRq r

ω −⎛ ⎞ ⎛ ⎞→ =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

( ) ( )

0

22 22 1*

we use / sin and calculate the irradiance at:

21Re( A

kRq r kR

JE AI E E E I

γ θ

γ

= =

⎡ ⎤= = → = ⎢ ⎥( )

( ) ( ) ( )

20

3 5

1 2 2 2

Re(2

/ 2 / 2Series expansion of the lim

2 1 2 1 2 3

P P PI E E E Ir

J

γ

γ γγγ

= = → = ⎢ ⎥⎣ ⎦

= − + − →( )1 1

2J γ

=


( )1 2 2 2p2 1 .2 1 .2 .3

γ

( )0

1 1

2 behaves like a damping sin function and J / behaves like the sinc function.J

γ γγ γ

→

Diffraction pattern of the circular aperture with Bessel functions in MATLABwith Bessel functions in MATLAB

.Fraunhofer diffraction pattern is the Fourier transform of the aperture functionWe can show this by the following plots using MATLAB. This topic will be covered in PHYS 258.

x=(-15.1:0.5:14.9);y=(-15.1:0.5:14.9);A=y.'*x;i_index=0;

2 2

2

1( , )

0

x y ag x y

x

+ ≤=

+

( )

2

2 10

1( )

0

( )( ) 2

R

r ag r

r ay a

J k aG k F k ak a

αα α π

⎧ ≤⎧⎪ → =⎨ ⎨ >⎩>⎪⎩⎡ ⎤

= = ⎢ ⎥⎣ ⎦

Diffraction pattern:

for i=-15.1:0.5:14.9j_index=0;i_index=i_index+1;for j=-15.1:0.5:14.9

j_index=j_index+1;r=sqrt(i^2+j^2); k aα⎣ ⎦

( j )if r <=5

A(i_index,j_index)=1;else A(i_index,j_index)=0;end

endende dsubplot(2,1,1);mesh(x,y,A);title('Circular Aperture')axis([-15.1 14.9 -15.1 14.9 0 1]);a=1;kx=(-15 1:0 5:14 9);kx ( 15.1:0.5:14.9);ky=(-15.1:0.5:14.9);[kax,kay]=meshgrid(kx,ky);

ka=sqrt(kax.^2+kay.^2);Gka=2*pi*a^2.*besselj(1,ka)./(ka*a);subplot(2 1 2);

17

subplot(2,1,2);mesh(kx,ky,Gka);xlabel('kx'); ylabel('ky');axis([-15.1 14.9 -15.1 14.9 -1 4]);title('Fourier Bessel of Circular Aperture')

Diffraction pattern of the circular aperture with B l f i i MATLABBessel functions in MATLAB

18

Diffraction pattern of the circular aperture with Fast Fourier Transformation(FFT) in MATLAB( )

%PHYS 258 spring 07, Nayer Eradat%A program to plot a circular aperture function %and its Fourier transform using fft and shift fft function x=(-2:0.05:2);y=(-2:0 05:2);y=(-2:0.05:2);A=y.'*x;i_index=0;for i=-2:0.05:2

j_index=0;i_index=i_index+1;_ _for j=-2:0.05:2

j_index=j_index+1;r=sqrt(i^2+j^2);if r <=0.2

A(i_index,j_index)=1;l A(i i d j i d ) 0else A(i_index,j_index)=0;

endend

endsubplot(2,1,1);mesh(x y A); %3D plotmesh(x,y,A); %3D plotxlabel('x'); ylabel('y'); zlabel('E');title('Circular aperture');fft_v=abs(fft2(A));fft_val=fftshift(fft_v); %shift zero-frequency component to center of spectrum

19

subplot(2,1,2);mesh(x,y,fft_val);xlabel('fx'); ylabel('fy'); zlabel('E');title('fft of Circular aperture');

Circular aperture diffraction patternCircular aperture better than slit or rectangular aperture for imaging First zero of the circular aperture:aperture for imaging.First zero of the circular aperture:

=3.832 Thus for an aperture of diameter D we have

1

γ

1= sin 3.832 sin 1/ 222

Disc: the

kD D

Airy

γ θ θ λ= → =

diffracted image of a circular aperture

or the central lobe of the diffraction pattern.

1/2

or the central lobe of the diffraction pattern. For far-field sin and the angular half-width of

1.22the Airy disc is: D

θ θ

λθΔ =

Example:

The beam spread for a

D

beam ( =564nm) from an

D 0 5 L 10

λ

-31/ 2

amperure D=0.5 mm at L=10m. The diameter of the Airy Disc is:

1.22 / 1.33 10 and then D radθ λΔ = = ×


1/ 2 13

That is why we pay high $$ for the large lenses. They dr L mmθ= Δ =

ca provide better image resolution.

Resolution and Rayleigh criterionRayleigh's criterion for just-resolvableRayleigh s criterion for just resolvable

images requires that the angular separation of the centers of the image

tt b l th th l

min

pattern be no less than the angular radius of the airy disc.

1.22λθΔ =min

Maximum of oneD

pattern falls directly under minimum of the next pattern.


Diffraction by the eyeDiffraction by the eye Pupil size limits the resolution of image of the objects subtended by Δθmin


Double‐slit diffraction IDiffraction pattern of a plane wavefront that is obstructed everywhere but at the two slits shown in fig.We follow our analysis for the single slit. The diffraction pattern for two slits is going to be superposition

( ) ( )0 0

(1/ 2)( ) (1/ 2)( )sin sin

1 20 01 2 (1/ 2)( ) (1/ 2)( )

of the patterns by each slit. Therefore we write:a b a b

i kr t i kr tiks iksL LP P P

slit slit a b a b

E EE dE dE e e ds e e dsr r

ω ωθ θ− − +

− −

− + −

= + = +∫ ∫ ∫ ∫( )0

0

1sin

i kr tLP

EE er ik

ω−= ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

1/ 2 sin 1/ 2 sin 1/ 2 sin 1/ 2 sin

With 1/ 2 sin and 1/ 2 sin

ik a b ik a b ik a b ik a be e e e

kb kaE b

θ θ θ θ

θ

β θ α θ

− + − − + −⎡ ⎤− + +⎣ ⎦

= =

( ) ( ) ( )( ) ( ) ( )

0

0

0

0

2

2

i kr t i i i i i iLP

i kr t i i i iLP

E bE e e e e e e er iE bE e e e e er i

ω α β β α β β

ω α α β β

β

β

− −

− −

⎡ ⎤= − + −⎣ ⎦

⎡ ⎤= + −⎣ ⎦0

Using Euler's equation:β

( ) ( )( ) ( )0 0P

0 0

2 sinE = 2cos 2 sin cos2

i kr t i kr tL LE Eb be i er i r

ω ω βα β αβ β

− −⎡ ⎤ =⎣ ⎦

( )00 0

0

2 2 2

2 sin where cos

The irradiance at point P is:

i kr t LP

E bE E e Er

ω β αβ

−= =

2


2 2 22 20 00 0

0

2 sin sincos 42 2

Lc c bEI E I Ir

ε ε β βαβ β

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= = → =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

22 0

00

2cos where 2

Lc bEIr

εα⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

Double‐slit diffraction II

2

With the modified interference pattern for double-slit due to diffraction of each slit we have:

i β⎛ ⎞2

2bE⎛ ⎞⎛ ⎞( )20

Interference pattern of the double-slit Diffraction pattern

of a single slit

sin4 cosI I βαβ

⎛ ⎞= ⎜ ⎟

⎝ ⎠0

00

2 where 2

Lc bEIr

ε ⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

max 0

sin sinwhere and sin2

Now 4 which is expected for the coherent sources.

I fi b l 6b d 6 th th i h idl t

ka a kb

I I

θ π θα β θλ

β

= = =

=2h i β In figure below a=6b and =6 thus the cos varies much more rapidly tα β α 2han sinc

We say the interference pattern of the doublle slit is modulated by the single slit diffraction pattern. β


Diffraction from many slits IDiffraction pattern of a plane wavefront that is obstructed everywhere but att the two slits shown in fig. We follow our analysis for the single slit. The diffraction pattern for two slits is going to

( )

( )

( )( )

( )

( )0 0

2 1 / 2 2 1 / 2/ 2sin sin

1 0 0( ) 2 1 / 2 2 1 / 2

be superposition of the patterns by each slit. Therefore we write:j a b j a bN

i kr t i kr tiks iksL LP Pi

j slit j j a b j a b

E EE dE e e ds e e dsr r

ω ωθ θ⎡− − + ⎤ ⎡ − + ⎤⎣ ⎦ ⎣ ⎦

− −

= ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎧ ⎫⎪ ⎪= = +⎨ ⎬⎪⎩ ⎭

∑ ∫ ∫ ∫N/2

1j= ⎪∑

( ) ( )1 0 0( ) 2 1 / 2 2 1 / 2j slit j j a b j a b= ⎡− − − ⎤ ⎡ − − ⎤⎣ ⎦ ⎣ ⎦⎪⎩ ⎭

( )

1

Here we are considering the pairs of slits that are symmetrical with respect to the center of the grating. 1, double slit, 2, 4 slits, etc. With 1/ 2 sin and 1

j

K

J j kbβ θ α

= ⎪

= = = = ( )/ 2 sinka θ( ) ( )( ) ( ) ( )

( ) ( )( ) ( )

0

0

2 1

2sin2 sin 2cos 2 1 2 Re

i kr t i i i i i iL

i j

E bK e e e e e e er ibK i j b e

ω α β β α β β

α

βββ α

− −

−

⎡ ⎤= − + −⎣ ⎦

⎡ ⎤= − = ⎣ ⎦( ) ( )( )

( ) ( )N/2

2 1 13 5

1Geometric series

2 sin 2cos 2 1 2 Re2

sin sin2 Re 2 Re ... i j i Ni i i

j

K i j b ei

S b e b e e e eα αα α α

β αβ β

β ββ β

− −

=

⎡ ⎤= − = ⎣ ⎦

⎡ ⎤ ⎡ ⎤= = + + + +⎣ ⎦ ⎣ ⎦∑Geometric series

For a ge

( )

2

/ 22

1ometric series .....1

1

nn

Ni

ra ar ar ar ar

α

⎛ ⎞−+ + + + = ⎜ ⎟−⎝ ⎠

⎡ ⎤⎛ ⎞


( )22 2

2

1sinfirst tem and ratio then 2 Re1

ii i i

i

ea e r e S b e

e

αα α α

α

ββ

⎡ ⎤⎛ ⎞−⎢ ⎥⎜ ⎟= = =⎢ ⎥⎜ ⎟−⎢ ⎥⎝ ⎠⎣ ⎦

Diffraction from many slits II

( ) / 22

Continued fromlast page:

1 1Ni iNe e

α α⎛ ⎞− ⎛ ⎞⎜ ⎟( )

( )

22

1 11

sin cos 1cos sin 12 i 2 i

ii i i

e eee e e

i NN i Ni

αα α α

α αα α

−

⎛ ⎞−⎜ ⎟ = =⎜ ⎟⎜ ⎟− −⎝ ⎠⎝ ⎠− Ν + −+ −

= Secondary

( )2 sin 2sin

sin cos 1sin sin sin2 Re2sin sin

ii N

S b S b

α αα αβ β α

β α β α

−⎡ ⎤− Ν + − Ν

= → =⎢ ⎥−⎣ ⎦2 2⎧ ⎫ ⎛ ⎞ ⎛ ⎞

ymaxima

( )0

0

sin sinsin

i kr tLP

EE e br

ω β αβ

− Ν=

2 2

0

Interference Diffraction by a between N slitssingle slit

sin sin and sin

I I β αα β α

⎧ ⎫ ⎛ ⎞ Ν⎛ ⎞=⎨ ⎬ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎩ ⎭ ⎝ ⎠ Principal maxima

( ) ( )

i d i

With 1/ 2 sin and 1/ 2 sin

sin coslim lim thsin cosa m a m

kb ka

N Nπ π

β θ α θ

α αα α→ →

= =

Ν Ν= = ± is resembles a series of

Limiting diffraction envelope

indeterminate

2principal maxima

sharp maxima that we call principal maxima.

I cantered at values 0, , 2 , 3

B t i k th 2 d k

N

N

α π π π∝ = ± ± ±


Between successive peaks there are 2 secondary peaks.

The full irradiance is

N −

product of the diffraction pattern and interference pattern.

Formation of secondary maxima


chapter 11 fraunhofer diffraction - erbionerbion.com/index_files/modern_optics/ch11.pdf · chapter...

Documents