Fraunhofer Diffraction Tuesday, 12/5/2006 Physics 158 Peter Beyersdorf Document info 20. 1 Class Outline Introduction Diffraction from a set of discrete sources Diffraction from a line source Diffraction of a slit Diffraction from a circular aperture Babinet’s Principle 20. 2 Introduction from Latin “diffraction” means to break apart Francesco Maria Grimaldi (1613 - 1663) Isaac Newton (1643-1727) Discovered the diffraction of light from Repeated and improved on Grimaldi’s hard edges and gave it the name experiments. Newton noticed that the bands diffraction. He noticed that the shadow of of red lights were largest while the blue small objects like pencil tips is wider than were least with the green of middle size. the computed geometrical shadow. Newton used the term "inflection" but Sometimes the shadow was encircled by Grimaldi's word "diffraction" survived. rainbow-like colored bands. Newton was in favor of the picture of light as a stream of ballistic particles. He was one of the earliest physicists to suggest that light was wavelike in nature. He formulated a geometrical basis for a wave theory of light in his “Physico- mathesis de lumine” (1666). 20. 3 Introduction Thomas Young (1773-1829) Augustin Jean Fresnel (1788-1827) In 1800 he published his Experiments on He was able to calculate formulae which Sound and Light in the Philosophical gave the position of the bright and dark Transactions of the Royal Society lines based on where the vibrations were in phase and where they were out of phase. To explain the double refraction Using Fresnel’s formulas Poisson predicted a phenomenon observed in certain crystals he bright spot in the center of the shadow postulated that light must be a transverse behind a dark disk, a prediction that Arago wave. found hard to believe before it was shown experimentally. He is best known for his double-slit experiment that was strongly in favor of Joseph von Fraunhofer (1787 - 1826) the wave theory of light, allowing him to He re-discovered the dark lines in the explain the phenomenon of interference of spectra, that he labeled from A-Z (today light. He also explained the color bands in known as the Franhofer lines). soap films an the Newton rings between two glass plates. He investigated diffraction in the far-field, In the limit where the distance of the He also showed that light exhibits a phase source is at infinity: a special case of the flip upon reflection from a denser Fresnel diffraction formula, equivalent to medium. having a “low Fresnel number” (a2/λR) 4 20. Diffraction The intensity pattern of light 1.4000000209 1.2000000179 transmitting through an 1.0000000149 0.8000000119 z = 1 x λ 0.6000000089 0.4000000060 0.2000000030 aperture (or the shadow of a 0 50 µm 2.0000000298 1.8000000268 1.6000000238 1.4000000209 mask) can not be explained 1.2000000179 1.0000000149 z = 40 x λ 0.8000000119 0.6000000089 0.4000000060 0.2000000030 by geometrical optics. The 0 50 µm 3.5 3.0 further one gets from the 2.5 2.0 z = 140 x λ 1.5 1.0 0.5 aperture, the more the 0 50 µm 3.0 2.5 intensity pattern deviates 2.0 z = 200 x λ 1.5 1.0 0.5 0 from that predicted by 100 µm 0.18 0.16 0.14 geometrical optics. In the 0.12 0.10 λ 0.08 z = 1000 x 0.06 0.04 0.02 0 far field limit this phenomena 500 µm is called Fraunhofer The intensity profile from a uniformly Diffraction illumiaetd square aperture of size x by x at various distances from the aperture. 20. 5 Diffraction effects from a discrete set of sources Interference from coherent oscillators double slit gratings The field at point p is i(kr1 ωt) i(kr1 ωt) i(krn ωt) Ep = E0e − + E0e − + . + E0e − i(kr1 ωt) ik(r2 r1) ik(rn r1) Ep = E0e − 1 + e − + . + e − ! " Assumptions: If the sources are equally spaced along the size of the oscillators is small compared to the wavelength a line such that the phase difference they act like point sources of equal amplitude from source n to point p (relative to The observer at point P is far from the array of sources such that all the rays from the source 1) is nδ=nkdsinθ point sources appear almost parallel. i(kr1 ωt) iδ i(n 1)δ Ep = E0e − 1 + e + . + e − ! " which is a geometric series 20. 6 Diffraction effects from a discrete set of sources Geometric series 2 n 1 S = 1 + a + a + . + a − multiply S by a and subtract S to get aS S = an 1 − − Allowing us to solve for S Assumptions: an 1 the size of the oscillators is small compared S = − a 1 to the wavelength − they act like point sources of equal amplitude The observer at point P is far from the array so of sources such that all the rays from the point sources appear almost parallel. i(kr1 ωt) iδ inδ Ep = E0e − 1 + e + . + e inδ i(kr1 ωt) e 1 Ep = E0e − ! − " eiδ 1 20. 7 ! − " Diffraction effects from a discrete set of sources einδ/2 einδ/2 e inδ/2 i(kr1 ωt) − Ep = E0e − − eiδ/2 eiδ/2 e iδ/2 ! " − − #$ " # i(kr ωt) iδ(n 1)/2 sin (nδ/2) E = E e 1− e − p 0 sin (δ/2) giving an intensity at point P of Assumptions: sin2 (nδ/2) the size of the oscillators is small compared Ip = I0 to the wavelength sin2 (δ/2) they act like point sources of equal amplitude or The observer at point P is far from the array sin2 ((nkd/2) sin θ) of sources such that all the rays from the Ip = I0 point sources appear almost parallel. sin2 ((kd/2) sin θ) inδ rapid oscillation i(kr1 ωt) e 1 Ep = E0e − − slow envelope 20. 8 eiδ 1 ! − " Gratings Maxima of interference pattern occur where denominator is zero (kd/2)sinθ=mπ Using k=2π/λ we get sinθ=m(λ/d) If we generalize this to allow a phase shift between oscillators of ε=kdsinθi such that δ→δ+ε, the maxima occur at 2 sin ((nkd/2) sinsinθθ)i-sinθ=m(λ/d) Ip = I0 2 Which is sinthe ((grkadt/in2)g sinequaθ)tion 20. 9 Grating Diffraction Pattern ty ensi t d in e iz l ma r o n -5 -2.5 0 2.5 5 k d sinθ 20.10 Diffraction from a Slit Consider the case where the point sources are really just secondary sources of Huygen’s wavelets along an illuminated slit Since there are infinitely many sources over the slit of width D we have n→∞ and d→D/n # then "3! "2! "! ! 2! 3! 2 2 sin ((nkd/2) sin θ) sin ((kD/2) sin θ) 2 Ip = I0 I0 = I0sinc ((kD/2) sin θ) sin2 ((kd/2) sin θ) → ((kd/2) sin θ)2 20.11 Diffraction from a Slit • P D/2 x 2 θ Ip = I0sinc ((kD/2) sin θ) θ L At a point of minimum intensity ΔR=λ modulo λ i.e. xm=mλL/D Δ θ At a point of maximum intensity -D/2 R≈ D ΔR=λ/2 modulo λ width of central peak is Δθ=2λ/D 20.12 Diffraction from a Circular aperture For a circular aperture the far field intensity at a point P is 0 i(ωt "k "r) E = E e − · dS r !app S(x’,y’)• R r •P(x,y) !k !r k z2 + (x x )2 + (y y )2 · ≈ − ! − ! 2 2 2 2 2 = k!x + y + z + 2xx! + 2yy! + x! + y! 2 2 2 2 2 = k!R (1 + (2xx! + 2yy!)/R + (x! + y! )/R ) kR + kxx!/R + kyy!/R ≈ ! Where x’ and y’ are coordinates on the aperture surface S and x and y are coordinates in the far field 20.13 Diffraction from a Circular aperture In spherical coordinates this reduces to a 2π 0 i(ωt kR) i(kρq/R) cos φ Ep = E e − e ρdρdφ r 0 0 ! ! S(x’,y’)• R q which evaluates to a r • 2J (ka sin θ) 2 I (θ) = I(0) 1 P(x,y) p ka sin θ ! " where J1(r) is called the first order Bessel function and is defined by i m 2π J (u) = − ei(mv+u cos v)dv m 2π !0 λ θ = 1.22 1 a 20.14 Diffraction from a Circular aperture The intensity distribution 0 1 2 3 4 5 6 has a first minima at 2 2J (ka sin θ) I (θ) = I(0) 1 p ka sin θ 2 ! " (J1(πr)/πr) The region inside of this is called the “Airy disk”. The size of the airy disk is the minimum size a circular lens can focus light to. If this is larger than any other aberrations the lens or optical systemλ is said to be θ = 1.22 “diffraction limited” 1 a 20.15 Example A spy satellite has a 1m diameter circular mirror and is in low Earth orbit at an elevation of 145 km. What is its resolution of objects on the surface of the Earth? A point source 145 km away would be imaged to an airy disk with an angular radius 1.22λ/a. Assuming λ=500nm, i.e. it is in the middle of the visible spectrum and with a=5m gives θ1=1.22 μrad, thus two points that are to be clearly resolved by the satellite must be separated by more than this so the resolution δx 6 δθ = 1.22 10− 145 103 m ≥ × so δxmin=18 cm × 20.16 Example Is your eye diffraction limited? take a≈4mm, thus Δθmin=1.22 500 nm/4mm=150 μrad so at a distance of 10m you should be able to resolve objects that are δx 6 δθ = 150 10− 10 m ≥ × δxmin=1.5 mm apart.