29. Diffraction of waves, Schemes and Mind Maps of Quantum Physics

Download 29. Diffraction of waves and more Schemes and Mind Maps Quantum Physics in PDF only on Docsity! 29. Diffraction of waves Light bends! Diffraction assumptions The Kirchhoff diffraction integral Fresnel Diffraction diffraction from a slit Diffraction Light does not always travel in a straight line. It tends to bend around objects. This tendency is called "diffraction.“ Shadow of a razor blade illuminated by a laser This is why radio communications does not necessarily require a line of sight. C. J. Davisson and L. H. Germer, 1927 Diffraction of particles This experiment was completed only a few years after Louis de Broglie had described the wavelength of a particle in terms of its momentum, λ = h/p. (h = Planck’s constant) Electrons do this too. The observation of this fact was one of the first important confirmations of quantum physics. What is E(x0,y0) at a distance z from the plane of the aperture? This region is assumed to be much smaller than this one. We wish to find the light electric field after a screen with a hole in it. This is a very general problem with far-reaching implications. The Diffraction Problem (x1, y1) incident plane wave plane of the aperture z observation plane ( ) ( )2 22 01 0 1 0 1= + − + −r z x x y y (x0, y0) Diffraction Assumptions This set of assumptions actually over-determines the problem. But even so, this is a useful starting point. A more accurate treatment is very complicated! The first thorough treatment of this problem was due to Kirchhoff. He made a few assumptions: Gustav R. Kirchhoff (1824 - 1887) 1) Maxwell's equations 2) Inside the aperture at z = 0, the field and its spatial derivative are the same as if the screen were not present. 3) Outside the aperture (in the shadow of the screen) at z = 0, the field and its spatial derivative are zero. Incident wave ( , )E r t
( , )E r t∇

( , ) 0E r t =
( , ) 0E r t∇ =

( , ) 0E r t =
( , ) 0E r t∇ =

z 0 The Solution: Kirchhoff Diffraction Integral The field in the observation plane, E(x0,y0), at a distance z from the aperture plane is given by a convolution: 1 1 0 0 0 1 0 1 1 1 1 1 Aperture( , ) ( , ) ( , ) ( , )= − −∫∫ x y E x y h x x y y E x y dx dy 01 0 1 0 1 01 exp( )1 ( , ) jkr h x x y y j rλ − − =where : A very complicated result! In order to use this, we must make some approximations… In the denominator, we can approximate r01 by z. But not in the exponent, because k is large so kr01 cannot be neglected. Paraxial approximation In the spirit of the paraxial approximation, we will assume that the aperture is small compared to the distance z, so that z >> x0 − x1 and y0 − y1. 2 2 0 1 0 1 01 1 − −   = + +        x x y y r z z z First, we note that we can factor z out of the square root in the expresson for r01: ( ) ( )2 22 2 0 1 0 10 1 0 1 01 1 1 1 small corrections 2 2 2 2   − −− −   ≈ + + = + + = +           x x y yx x y y r z z z z z z z Make use of the Taylor expansion: 1 1 1 2 + ≈ +ε ε ( ) ( ) ( )2 2 0 1 0 1 0 1 0 1 exp exp exp 2 21 ( , ) x x y y jkz jk jk z z h x x y y j zλ    − −            − − ≈ Replace r01 in the exponent of h(x0−x1, y0−y1): ( ) ( ) 1 1 2 2 0 1 0 1 0 0 1 1 1 1 ( , ) 1 ( , ) exp ( , ) 2 2   − − ≈ + +       ∫∫ Aperture x y x x y y E x y jk z E x y dx dy j z z zλ ( ) 1 1 2 2 2 2 0 0 1 1 0 0 1 1 0 0 1 1 1 1 ( , ) ( 2 ) ( 2 ) , exp ( , ) 2 2   − + − + = +      ∫∫ jkz Aperture x y x x x x y y y ye E x y jk E x y dx dy j z z zλ Multiplying out the squares, and factoring out the constants: Thus, we have: Paraxial approximation: Fresnel diffraction If the incident wave is a plane wave, as is typically assumed, then: 1 1( , ) constant=E x y (constant with respect to x1 and y1) At a certain angle T, the path difference between two waves (from the top of the slit and the mid-point) equals half of a wavelength. This leads to destructive interference, and therefore a dip in the intensity at that angle. Diffraction causes fringes More generally, we can imagine dividing the slit into an even number of zones. At certain angles, the light from each zone can destructively interfere with the light from the neighboring zone, leading to dark regions in the diffraction pattern. These alternating light and dark regions are known as “fringes”. Write the Fresnel integral for this one-dimensional problem: Next step: define new variables and1 1 = x b ξ 0 0 = x b ξ ( ) ( ) ( ) 1 2 2 0 0 0 1 1 1 exp j b E x E d z πξ ξ ξ ξ λ −   → ∝ −   ∫Then: 2=k π λ since ( ) ( ) 2 2 0 0 1 1 0 1 1 2 exp 2 x x x x E x jk Aperture x dx z   − + ∝       ∫ Fresnel integral for a slit The aperture function is given by: ( ) 1 1 1 0 otherwise b x b Aperture x − < < =   ( ) ( )2 0 1 0 1exp 2 −   − =        ∫ b b x x E x jk dx z Thus the integral becomes: Fresnel diffraction example: a slit ( ) ( ) 1 2 2 0 0 1 1 1 exp j b E d z πξ ξ ξ ξ λ −   ∝ −   ∫ As a shorthand, we define a dimensionless quantity known as the “Fresnel number”: 2 = b N zλ ( ) ( ){ } 1 2 0 0 1 1 1 expE j N dξ π ξ ξ ξ − ∝ −∫ This is not an integral that can be solved in closed form. It must be computed numerically. and, of course, we are really interested in the intensity ( ) ( ) 2 0 0I Eξ ξ∝ Fresnel Diffraction through a slit: far field ( ) ( ){ } 1 2 0 0 1 1 1 expE j N dξ π ξ ξ ξ − ∝ −∫ In the limit that N << 1 (very far from the aperture), the integral can be performed analytically. The math is a bit tedious, so we just quote the result here: ( ) 0 0 0 2 sin 2      ∝ Nx b E x Nx b π π Our old friend the sinc function! In this regime (the “far field”), the diffraction pattern no longer changes shape as z increases, but merely expands in size uniformly. Fresnel number 0.03 0.02 0.01 -100b 100b Fuzzy shadow edges make it hard to see diffraction. An incident plane wave or spherical wave (which, like a plane wave, has a flat and well-behaved wave front) is required to see diffraction. Here, rays from a point source yield, in principle, a perfect shadow of the hole, allowing diffraction ripples to be seen. Screen with slit In general, a large source (like the sun or a light bulb) casts blurry shadows, masking the diffraction ripples. Rays from other regions of the source blur the shadow. A large source masks diffraction effects. Fresnel Diffraction through a slit: isn’t there an awesome java applet that illustrates all this? http://www.falstad.com/ripple/ More than one, actually http://www.falstad.com/diffraction/