Chapter 11 Fraunhofer

Lecture Notes for Modern Optics based on Pedrotti & Pedrotti & Pedrotti Instructor: Nayer Eradat Spring 2009

4/30/2009 Fraunhofer Diffraction 1 Diffraction • 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‐ limited 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((yg)Huygens). The field beyond a wavefront 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 waves originating from different points of a continuous source. Interference phenomena is calculates interference of the beams origination from discrete number of sources.

4/30/2009 Fraunhofer Diffraction 3 Mathematical 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 ignored • 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 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.

4/30/2009 Fraunhofer Diffraction 4 Diffraction from a single slit We simulate the gggeometrical arrangement for the Frounhofer diffraction byyp placinggp a point source at the focal point of a lens and a screen at the focal point of a lens after the amerture. The light reaching point P on the screen is from interference of the parallel rays from different points on the aperture. We consider each interval of ds as a source and calculate contribution of the all of these sources. rdsP: optical path length from the to the

EL strength of the electric field contribution from each point. rr= 0 for the field from the center of the slit.

CibifhilContribution of each interval ds to thfhe wavefront at po iihilint PdE is a spherical wavelflet of P:

⎛⎞EdsL ikr()-ω t dEP = ⎜⎟ e ⎝⎠ r Spherical wave amplitude ⎛⎞ EdsL ikr()()0 +Δ -ω t dEP = ⎜⎟ e ⎝⎠r0 +Δ ⎛⎞ EdL s ik()rt0 −ω ikΔ dEP = ⎜⎟ e e ⎝⎠r0 +Δ Path difference  affets the phase Path difference affets the amplitude We ignore Δ in the denuminator since Δ<

4/30/2009 Fraunhofer Diffraction 5 Diffraction from a single slit ⎛⎞ EdsL ikr()0 −ω t ikΔ dEP = ⎜⎟ e e ⎝⎠r0

The total electricωω field at the point P θ E b/2 EdEe==L ikr()0 −ω t edsikssin PP∫∫−b /2 slit r0 θ b /2 ⎛⎞ikssin ()ikbsinθθ / 2− ikb sin () / 2 EELLikr()00−− teee ikr() t − EeP ==⎜⎟ e rikr00⎝⎠sinθθ−b /2 ik sin 1 β With β = kbsinθ and using Euler's formula 2 β

EbLLsin ikr()0 −ω t Ebsin cβ EeP == rr00 βθ variihdhiihhdies with and that varies with the difhistance from the screen. b1 βθβθ can be interpreted as phase difference kΔ→=Δ= =k sinkkb sin ββ22 β shows the magnitude of the phase difference at point P between the points from the center and either endppoint of the slit. 2 The irradiance IP at is proportional to the E0 :

22 22 ⎛⎞ε00cc2 ε ⎛⎞EbL siiinsββε0c ⎛⎞EbL in 2 IE==⎜⎟0 ⎜ ⎟⎜⎟22 with I000=== we have II I sin cβ ⎝⎠22⎝ r0 ⎠⎝⎠2 r0 4/30/2009 Fraunhofer Diffraction 6 Diffraction from a single slit sin2 β The diffraction pattern of the light from a silt by width of b at a distanc screen II== I sin c2 where 00β 2 β βθ=Δ=kkb()sin /2 is the phase difference betwwen the light from the center of the slit the endpoints. β ⎛⎞sin β Let's plot the I. For the central maximum: limsinc()== lim⎜⎟ 1 ββ→→00⎝⎠β For zeros of the : 1 sinββ=→ 0 =()kb θπ sin = m → bθλ sin = m where m =±± 1, 2,... 2 m = 0d0 does not tl lead dt to a zero b e cause ofthf the speci al propert y of fth the si nc ftifunction. if f 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: mλλy mf sinθ ≅ tanθ =→=→=y /fy m b fbm d⎛⎞ sinβββ cos− sin β For the other maxima: ββ⎜⎟ β==2 0 d ⎝⎠ Cental lobe: image of the slit with roundes edges. Side lobes: what causes blurriness in the image. Angular width of the central lobe: b sinθλ=→ θθ θ λ λλ2 ≅≅→Δ=, , The central lobe will sprad as the slit-size gets smaller. 11bb− b 4/30/2009 Fraunhofer Diffraction 7 Maxima of the sinc function The maximaβ of the sinc function are solutions of this equation: ββcos− si βn 2 =→0tanββ =

We can solve this equation by parametrically. AlAn angle equal li to its tangant i iis intersection ⎧y = β of the ⎨ y = tan ⎩ β We see that the maxima are not exactly half way between the minima. They occur slightly erlier and as β increases the maxima shift towards the center. Ratio of the irradiances at the central peak maximum to the first of the secondary maxima ?

2 2 I ()sinββ / 1⎛⎞ 1.43 20.18 ββ==00==== ()22⎜⎟ I βπ1.43 sin / sin / sin() 1.43 0.952 = ββ βββββ==1.43π ()β 1.43π ⎝⎠

IIβπ==1.43= β0.047 0 or only 4.7% of the I 0 4/30/2009 Fraunhofer Diffractionπ 8 π Beam spreading The angular spread of the central maximm Δθ =2λ /b is independent of the distance between the slit and dscee.Soasscee screen. So as screen moves ovesaway away fro m ttestteatueotedhe slit the nature of the diffract actoion pattern does not c ha nge . W is the width of the central maximum, θ 2Lλ W=LΔ= b 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: What is the width of a parallel beam of λ = 546nm and width of b=0.5mm after propagation of 10 meter? 2Lλ 2×× 10 546 × 10−9 W == b 0.5× 10-3 Wmm= 21.8 This tre atment of beam spreading is b2 correct for far-field where L >> λ area of aperture or more generally generall L >> λ

4/30/2009 Fraunhofer Diffraction 9 Rectangular aperture For 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 d iffraction pattern from

2 2 ⎛⎞sinαβ ⎛⎞kk⎛⎞ sin ⎛⎞ both dimensions: II====00⎜⎟ where ⎜⎟ a sin and II⎜⎟ where ⎜⎟ b sin ⎝⎠αβ ⎝⎠22⎝⎠ ⎝⎠ αθ βθ mλ fffnλ f Zeros of th e patt ern d ue occur at : y= and x= m,n=±±121, 2,... mnba

a /2 b/2 EL ikr()0 −ω t ikXx// r00 ikYy r Using a bit more analysis we can write EdEePP== edxedy ∫∫∫−−ab/2 /2 slit r0 XY, are thdifhbihe coordinates of the observation poihint on the screen and x,y are th e coordinates of th e surf ace element on the aperture. The total irradiance turns out to be the product of the irradiance functions on each

22 dimension: II= 0 ()()sin cβα sin c

4/30/2009 Fraunhofer Diffraction 10 Circular aperture 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 dA= dxdy (on the aperture) is EA dA/. / r0 . Then we integrate the incremental field over the entire aperture area. The resulting E at point P is

⎛⎞EdAikr()()+Δ -ω t E ikr− t dE=⎯⎯⎯⎯AA e0 Δ<

2E +R 22 E = A eedsikr()0 −ω t isksin R− s P ∫−R r0 θ Substituting v==sR / and γθ k sin

2ER +1 2 E = A eedikr()0 −ω t iv1− v v P ∫−1 γ r0 γ +1 2 πγJ ( ) From the intergral table: edvivγ 1− v = 1 ∫−1 γ where J1 () is the first order of the first kind. γ γ 35 ()γγ/2() /2 γ J1 ()1 J1 ()=− + −" and lim = 21.21.2.3222 γ →0 2 4/30/2009 Fraunhofer Diffraction 11

γ 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 dA= dxdy (on the aperture )is) is EA dA/ r0. Then we integrate the incremental field over the entire aperture area. The resulting E at point P is E E==A eitkr()ω − dA where dA dxdy p ∫∫ r0 Aperture y Y 221/2 1/2 rXxYyZ=−+−+⎡⎤()()2222 with rXYZ =++⎡⎤ ⎣⎦0 ⎣⎦ x X 1/2 P 22 ⎡⎤(x + y ) 2(Xx+ Yy) r ⎢⎥Δ r we have: rr=+0 1 22 − 0 ⎢⎥rr00 dy θ ⎣⎦ P 1/2 dx 0 222 ⎡⎤2()Xx+ Yy rx00>> +y →rr ⎢⎥1 − 2 ⎣⎦r0 ()Xx+ Yy rxyrr>> + → − only first two terms. Aperture Screen 0 0 r0

⎛⎞⎛+⎞()Xx Yy itkr⎜⎟ω −−⎜⎟ EE⎜⎟⎜⎟0 r Ee=→=AL⎝⎠⎝⎠0 dAEeeitkr()− 0 ik() Xx+ Yy/ r0 dxdy PP∫∫ ∫∫ rr00Aperture Aperture any shape For specific shape of aperture the integ ral limits and relationship between x an dyd y will change. We will look at two examples of rectangularω and circular apertures.

4/30/2009 Fraunhofer Diffraction 12 Rectangular aperture E Ee= A itkr()ω − 0 eik() Xx+ Yy/ r0 dxdy P ∫∫ r0 Aperture any shape

ba/2 /2 EA itkr()ω − 0 ik() Yy// r00 ik Xx r ( ) EeP = edyedx r ∫∫−−ba/2 /2 0 y y

E itkr()ω − ⎛⎞sinβα⎛⎞ sin Eeba= A 0 x p ⎜⎟⎜⎟ x P r0 ⎝⎠βα⎝⎠ r Δ r0 AE itkr()ω − ⎛⎞sinβα⎛⎞ sin Ee= A 0 dy θ p ⎜⎟⎜⎟ P r0 ⎝⎠βα⎝⎠ dx 0 2 ⎛⎞ AEA itkr()ω − 0 Ie0 = ⎜⎟ ⎝⎠r0

2 2 Aperture Screen ⎛⎞sinβα⎛⎞ sin IXY()(),0= I ⎜⎟⎜⎟ ⎝⎠βα⎝⎠ kXkk kY Where α ==ab= asinθβ and θ= bsin 22rrx 22y 4/30/200900 Fraunhofer Diffraction 13 Circular aperture I For circular aperture we introduce the spherical coordinates because of the symmetry of the problem. On plane of the aperture: xy==ρφ cos , ρφ sin On plane of the screen: X=Φq cos , Ysin=Φq Differential surface element: dA= ρρφ d d R 2π EEitkr itkr ikqr/cos EdAddE ==ALee(ωωρφ−−−000) ik( Xx+ Yy)/ r0 dA ee( ) ( ) ( Φ)ρddρ φ P ∫∫ ∫ ∫ rr00Aperture ρφ==00 any shape Wher we used: Yy+= Xxρφ q () sin sin Φ+ φ cos cos Φ=−Φρφq cos() Because of the axial symmetry of the problem, the solution must be independent of Φ and since point P is an arbitrary point on the screen we choose to have Φ= 0 to simplify things. E a 2π E = A ei()ωtkr− 00ρρ φde ikqr/cos (() d ) y y P r ∫∫ρφ 0 ρφ==00 A tabulated integral. Solution is a x x Bessel function of the first kind P −m 2π r φ i imvu( + cos v) dx Besselfl functi on 1st ki kidnd: J (ue)= dv Δ r0 m ∫0 2π dy θ π ρ 1 2π P0 For m=0: Ju()= eiucos v dv and with ukq= ρ 0 2 ∫0 E R E/= A eJkqrditkr()ω − 0 ρρρ P00∫ () r0 =0 4/30/2009 Fraunhofer Diffraction 14 ρ Bessel functions of the first kind (MATLAB) u = (0:0.1:15) BJ0=besselj(0,u); BJ1=besselj(1,u); BJ2=besselj(2, u); BJ3=besselj(3,u) plot(u,BJ0,u,BJ1,u,BJ2,u,BJ3); Bessel functions of first kind legend('J0','J1','J2','J3') 1 title('Bessel functions of first kind' ); J0 xlabel('u'),ylabel('J') J1 J2 grid on J3

0.5 J

0

-050.5 0 5 10 15 u

15 Circular aperture II E R E2/= A eJkqrditkr()ω − 0 πρρρ P00∫ ( ) r0 ρ =0 Using the one of many properties of the u bessel functions: uJ''' '( u ') du '= uJu we get ∫0 01 R 2 kRq/r0 J()()()()() kqrρρρ// d== r kq J u '''// udu kRqr J kRqr ∫∫00 0u'0= 0 010 ρ =0 ⎛⎞ ⎛⎞ EA itkr()ω − 0 2 r0 kRq2AEA itkr()ω − 0 r0 kRq E2P1= eRJπ ⎜⎟→=EP1eJ ⎜⎟ rkRq0 ⎝⎠rrkRqr00 ⎝⎠ 0

we use γθ==kRq / r0 kR sin and calculate the irradiance at:

2 22 2 1 * 2EAA ⎡⎤J1 () IEEEI==→=(Re( PPP) 2 ⎢⎥ 2 r0 ⎣⎦γ γ γγ35 ()/2() /2 J1 ()γ 1 Series expansion of the J1 (γ )= −+−→" lim = 21212321222.21γ.2 .3 γ →0 γ 2

J11 behaves like a damping sin function and J()γγ / behaves like the sinc function. 4/30/2009 Fraunhofer Diffraction 16 Diffraction pattern of the circular aperture with Bessel functions in MATLAB x=(-15.1:0.5:14.9); Fraunhofer diffraction pattern is the of the aperture function. y=(-15.1:0.5:14.9); We can show this by the following plots using MATLAB. This topic will be A=y.'*x; i_index=0; covered in PHYS 258. for i=-15.1:0.5:14.9 ⎧1 xya22+≤ j_index=0; ⎪ ⎧1 ra≤ gxy(, )= ⎨⎨→=grR () i_index=i_index+1; ⎪0 x2 + ya2 > αα⎩0 ra> for j=-15.1:0.5:14.9 ⎩

j_index=j_index+1; 2 ⎡⎤Jka1()α Diffraction pattern: Gk0 ()== Fk() 2π a⎢⎥ r=sqrt(i^2+j^2); ka if r <=5 ⎣⎦α A(i_index,j_index)=1; else A(i_index,j_index)=0; end end eednd subplot(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); 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); 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') 17 Diffraction pattern of the circular aperture with BlBessel ffiunctions 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); 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; elA(lse A(iidi_index,j jid_index)0)=0; end end end subplot(2,1,1); mesh(xyAx,y,A); % 3D plot xlabel('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 subplot(2,1,2); mesh(x,y,fft_val); xlabel('fx'); ylabel('fy'); zlabel('E'); title('fft of Circular aperture'); 19 Circular aperture diffraction pattern Circular aperture better than slit or rectangular aperture for imaging .First zero of the c ircular aperture: γ =3.832 Thus for an aperture of diameter D we have 1 γ =kD sinθθλ=→ 3.832 D sin = 1/ 22 2 Airy Disc: the diffracted image of a circular aperture or the central lobe of the diffraction pattern. For far-field sinθθ and the angular half-width of 1.22 the Airy disc is: Δ= 1/2θ D λ Example: The beam spread for a beam (λ =564nm) from an amperure D05DJJJJJJJJJJJM=0.5 mm at L10L=10m. The diameter of the Airy Disc is: -3 Δ=θλ1/2 1.22 /Drad = 1.33 × 10 and then rLd =Δθ1/2 =13 mm That is why we pay high $$ for the large lenses. They ca provide better image resolution. 4/30/2009 Fraunhofer Diffraction 20 Resolution and Rayleigh criterion Rayleigh's criterion for just-resolvable images requires that the angular separation of the centers of the image pattern b e no l ess th an th e angul ar radius of the airy disc. 1.22λ Δ=θ min D Maximum of one pattern falls directly under minimum of the next pattern.

4/30/2009 Fraunhofer Diffraction 21 Diffraction by the eye

Pupil size limits the resolution of image of the objects subtended by Δθmin

4/30/2009 Fraunhofer Diffraction 22 Double‐slit diffraction I Diffraction 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 of the patterns by each slit. Therefore we write: EE−−(1/ 2)(ab ) (1/ 2)( ab + ) E=+ dEdEe =LLikr()00−−ωω t edseikssin + ikr() t eds iks sin PP∫∫12 P ∫ ∫ slit12 slitrr00−+ (1/2)() a b (1/2)() a − b θθ E 1 Ee= L ikr()0 −ω t ⎡⎤eeee()(1/2ik−+ a b sinθθθθ )−++ 1/2 ik −−()( a b sin 1/2 ) ik a + b sin ()() 1/2 ik a − b sin ()() P βθαθ⎣⎦ rik0 sinθ With ==()1/ 2kb sin and 1/ 2 ka () sin E b Ee=−+−L ikr()0 −ω t ⎡⎤ eeeeeeii i− ii i P ⎣⎦()( ) ri0 2 β αβ β αβ β

EL ikr()0 − t bββiiii− Ee=+−ω ⎡⎤ eeeeαβ α P ⎣⎦()() ri0 2 β ααββ Using Euler's equation:

EELLikr()0 −−ωω tbb ikr()0 t 2sinβ E=P eie⎣⎦⎡⎤()() 2cos 2sin= cos ri002 r E 2sinb β ikr()0 −ω t L β EEeP ==00 where E cosα r0 The irradiance at point P is: α 2 22 2 ⎛⎞εε00cc22 ⎛⎞⎛⎞2bEL ⎛⎞sin ⎛⎞ sin 2 ⎛⎞ε0c ⎛⎞2bEL IE==⎜⎟00 ⎜⎟⎜⎟⎜⎟cos →= II 4 ⎜⎟cos where I0 = ⎜⎟⎜⎟ ⎝⎠22 ⎝⎠⎝⎠r0 ⎝⎠ ⎝⎠ ⎝⎠2 ⎝⎠r0 4/30/2009 Fraunhofer Diffraction 23 ββ ββ α Double‐slit diffraction II With the modified interference pattern for double-slit due to diffraction of each slit we have: 2 2 ⎛⎞sin β ⎛⎞ε c ⎛⎞2bE II= 4cos2 ()α where I = 0 L  0 ⎜⎟ 0 ⎜⎟⎜⎟ ⎝⎠ ⎝⎠2 r0 Interference pattern  ⎝⎠ of the double-slit Diffractionβ pattern αβθof a single slit kasinθ πθ a sin where == and =kbsin 2 λ Now IImax= 4 0 which is expected for the coherent sources. IfiIn figure b blelow a=6b 6bd and αβ=6 β th ththus the cosα varihies much more rapidl idlty thihan sinc2β We say the interference pattern of the doublle slit is modulated by the single slit diffraction pattern.

4/30/2009 Fraunhofer Diffraction 24 Diffraction from many slits I Diffraction 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 be superposition of the patterns by each slit. Therefore we write:

N /2 N/2 ⎧⎫⎣⎦⎡−−+⎤()21jab /2 ⎣⎦⎡() 21 jab −+⎤ /2 EELLikr()0 −−ωω t⎪⎪ikssin ikr()0 t iks sin EPPi== dEe⎨⎬ edse + eds ∑ ∫∫rr∑ ∫ j=1 slit() j 00j=1 ⎪⎡⎤⎡−−−(21jab) ⎤⎡ /2 ⎡⎤( 21 jab−−) ⎤ /2 ⎪ ⎩⎭⎣⎦ ⎣⎦ θθK Here we are considering the pairs of slits that are symmetrical with respect to the center of the grating. Jj==1, double slit, 2, 4 slits, etc. With βθα =( 1/ 2) kb sin and =( 1/2)ka sinθ E b L ikr()0 −ω t ⎡⎤ii i− ii i Ke=−+−⎣⎦ eeeeee()() ri0 2 β αβ β αβ β

b sin ij21− Ki=−=(2sinβα)( 2cos2( j 1) ) 2 b Re⎡⎤ e() 2iββ⎣⎦ N/2 sin⎡⎤ij()21−−αα sin ⎡ii35 i iN() 1 ⎤ Sb==++++∑ 2ββ Re⎣⎦ e 2 b Re ⎣ eeeeβ ... ⎦ j=1  ββGeometric series ααα n α 2 n ⎛⎞r −1 For a geometric series aarar++ + ..... + ara =⎜⎟ ⎝⎠r −1 N /2 ⎡⎤⎛⎞2iα sin β (e ) −1 first tem ae===iiαα and ratio re22 then S 2 b α Re ⎢⎥ e i⎜⎟ β ⎢⎥⎜⎟e2i −1 ⎣⎦⎢⎥⎝⎠ 4/30/2009 Fraunhofer Diffraction 25

α Diffraction from many slits II Continued fromlast page:

N /2 ⎛⎞(e2iα ) −1 iN 2iα ⎜⎟αα⎛⎞e −1 e 2iiiααα==⎜⎟− ⎜⎟eee−−1 ⎝⎠ ⎝⎠α ββα cosNiN+− sin 1 −Ν+sinααiN() cos − 1 βαβα= αα Secondary 2i2isinα −2 2isinα maxima sin⎡⎤−Ν+siniN() cos − 1 sin sin Ν Sb=→=2Re⎢⎥αβα Sb ⎣⎦−2sin sin 2 E ⎧⎫⎛⎞sinβ sin Να sinβ sin Να 2 L ikr()0 −ω t β ⎛⎞ EebP = ⎨⎬ and II= 0 ⎜⎟⎜⎟ r0 ⎩⎭⎝⎠sin ⎝⎠sin   Principal maxima Diffraction by a Interference single slit between N slits

With βθαθ==()1/ 2kb sin and (1/ 2) ka sin Limiting diffraction envelope sinΝΝN cos limαα==± limN this resembles a series of am →→ππsin am cos idindetermi nate αα sharp maxima that we call principal maxima.

2 Iprincipal maxima ∝=±±±N cantered at values απππ 0, , 2 , 3 BtBetween successi ve peak s th ere are N − 2 secondkdary peaks. The full irradiance is product of the diffraction pattern and interference pattern. 4/30/2009 Fraunhofer Diffraction 26 Formation of secondary maxima

4/30/2009 Fraunhofer Diffraction 27 4/30/2009 Fraunhofer Diffraction 28 4/30/2009 Fraunhofer Diffraction 29 4/30/2009 Fraunhofer Diffraction 30 4/30/2009 Fraunhofer Diffraction 31 4/30/2009 Fraunhofer Diffraction 32