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Chapter 11 Fraunhofer Diffraction

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Chapter 11 Fraunhofer Diffraction Chapter 11 Fraunhofer Diffraction 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<<r0 4/30/2009 Fraunhofer Diffraction 5 Diffraction from a single slit ⎛⎞ EdsL ikr()0 −ω t ikΔ dEP = ⎜⎟ e e ⎝⎠r0 The total electric field at the point P E b/2 EdEe==L ikr()0 −ω t edsikssinθ PP∫∫−b /2 slit r0 b /2 ⎛⎞ikssinθ ()ikbsinθθ / 2− () ikb sin / 2 EELLikr()00−−ωω teee ikr() t − EeP ==⎜⎟ e rikr00⎝⎠sinθθ−b /2 ik sin 1 With β = kbsinθ and using Euler's formula 2 EbLLsin β ikr()0 −ω t Ebsin cβ EeP == rr00β 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 22 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 Isin cβ ⎝⎠22⎝ r0 ⎠⎝⎠ββ2 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== Isin c2β where 00β 2 βθ=Δ=kkb()sin /2 is the phase difference betwwen the light from the center of the slit the endpoints. ⎛⎞sin β Let's plot the I. For the central maximum: limsinc()β == lim⎜⎟ 1 ββ→→00⎝⎠β For zeros of the sinc function: 1 sinββ=→ 0 =()kb sin θπθλ = m → bsin = m where m =±±1, 2,... 2 m = 0d0 does no tlt lead dt to a zero b e cause ofthf the spec ia l 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: ⎜⎟==2 0 dββ⎝⎠ β Cental lobe: image of the slit with roundes edges. Side lobes: what causes blurriness in the image. Angular width of the central lobe: bsinθλ=→ λ λλ2 θθ≅≅→Δ=, , θ 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 maxima of the sinc function are solutions of this equation: ββcos− sin β =→0tanββ = β 2 We can solve this equation by parametrically. AlAn angle equa lil 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⎛⎞ 1.43π 20.18 ββ==00==== 22⎜⎟ I 1.43 sinββ / sin ββ / sin() 1.43π 0.952 βπ= ()βββ==1.43π ()β 1.43π ⎝⎠ IIβπ==1.43= 0.047 β 0 or only 4.7% of the I 0 4/30/2009 Fraunhofer Diffraction 8 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, 2Lλ W=LΔ=θ 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 treatment 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 αθ ⎜⎟ asin and II⎜⎟ where βθ ⎜⎟ bsin ⎝⎠αβ ⎝⎠22⎝⎠ ⎝⎠ mλ fffnλ f Zeros o f the 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 an d x,y are t he coordinates o f t he sur face 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−ω t dE=⎯⎯⎯⎯AA e()()0 Δ<<r0 →= E e()0 eisk sinθ dA PP⎜⎟ Δ=ssinθ ∫∫ ⎝⎠rr00+Δ Area By choosing the elemental area the rectangular area of dA= xds we can reduce the double integral to a single one. 2 ⎛⎞x 22 22 ⎜⎟+=sR →= x2 Rs − ⎝⎠2 2E +R 22 E = A eedsikr()0 −ω t isksinθ R− s P ∫−R r0 Substituting v==sR / and γθ ksin 2ER +1 2 E = A eedikr()0 −ω t ivγ 1− v v P ∫−1 r0 +1 2 πγJ ( ) From the intergral table: edvivγ 1− v = 1 ∫−1 γ where J1 ()γ is the first order Bessel function of the first kind. 35 γ ()γγ/2() /2 J1 () γ1 J1 ()γ =− + −" and lim = 21.21.2.3222 γ →0 γ 2 4/30/2009 Fraunhofer Diffraction 11 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.
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