29. Diffraction of waves

Transcription

1 29. Diffraction of waves Light bends! Diffraction assumptions The Kirchhoff diffraction integral Fresnel Diffraction diffraction from a slit

2 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.

3 Diffraction of Ocean Water Waves Ocean waves passing through, and bending around the edges of, slits (regions between wave brakes) in Tel Aviv, Israel: Diffraction occurs for all waves, whatever the phenomenon.

4 Diffraction of a wave by a slit Passing light through a small slit yields a diffraction pattern that depends on the size of the slit and the wavelength of the wave. Large slit This phenomenon is general, and can be observed using waves of any kind. Smaller slit Very small slit

5 Diffraction of particles Electrons do this too. The observation of this fact was one of the first important confirmations of quantum physics. C. J. Davisson and L. H. Germer, 927 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)

6 The Diffraction Problem 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. plane of the aperture observation plane (x, y ) (x, y ) ( ) ( ) = + + r z x x y y z incident plane wave This region is assumed to be much smaller than this one. What is E(x,y ) at a distance z from the plane of the aperture?

7 Diffraction Assumptions The first thorough treatment of this problem was due to Kirchhoff. He made a few assumptions: ) Maxwell's equations 2) Inside the aperture at z =, 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 =, the field and its spatial derivative are zero. 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! Incident wave Gustav R. Kirchhoff ( ) E( r, t ) = E( E( r, t) r, t) = E( r, t) E( r, t ) = E( r, t) = z

8 Huygens Principle Huygens Principle says that every point along a wave-front emits a spherical wave that interferes with all others. Christiaan Huygens Our solution for diffraction uses this idea.

9 A solution based on Huygen s principle Huygen s wavelet (x, y ) Incident field (x, y ) r r = z + ( x x ) + ( y y ) z Each point (x, y ) in the aperture emits a spherical wave, with amplitude determined by the wave incident on the aperture. The net field at the point (x, y ) is therefore given by a superposition: (, y ) ( spherical wave propagating a distance r ) ( incident field at x,y ) E x = all points (x,y ) in the aperture Of course, the sum becomes an integral

10 The Solution: Kirchhoff Diffraction Integral The field in the observation plane, E(x,y ), at a distance z from the aperture plane is given by a convolution: E( x, y ) = h( x x, y y ) E( x, y ) dx dy where : x y Aperture(, ) h( x x, y y ) = exp( jkr ) jλ r A very complicated result! In order to use this, we must make some approximations In the denominator, we can approximate r by z. But not in the exponent, because k is large so kr cannot be neglected.

11 Paraxial approximation First, we note that we can factor z out of the square root in the expresson for r : 2 2 x x y y r = z + + z z In the spirit of the paraxial approximation, we will assume that the aperture is small compared to the distance z, so that z >> x x and y y. Make use of the Taylor expansion: + ε + ε 2 ( x x ) ( y y ) x x y y r z + + = z + + = z + small corrections 2 z 2 z 2z 2z Replace r in the exponent of h(x x, y y ): h( x x, y y ) ( ) ( x x ) ( y y ) 2 2 exp jkz exp jk exp jk 2z 2z jλ z

12 Paraxial approximation: Fresnel diffraction Thus, we have: ( x x ) ( y y ) 2 2 E( x, y ) exp jk z + + E( x, y ) dx dy jλz 2z 2z Aperture ( x, y ) Multiplying out the squares, and factoring out the constants: E x y jk E x y dx dy jkz e ( 2 + ) ( 2 + ), = x x x x y y y y exp + (, ) jλz 2z 2z Aperture ( x, y ) ( ) If the incident wave is a plane wave, as is typically assumed, then: E( x, y ) = constant (constant with respect to x and y )

13 The Fresnel Diffraction Integral Usually, instead of writing an integral over an aperture, we will explicitly write the aperture function in the integral: E x + y ( 2x x 2 y y ) ( x + y ) jλz z 2z 2z in (, ) = exp + exp + (, ) E x y jk z jk Aperture x y dx dy And we ll usually neglect the factors in front of the integral, to obtain: E x y jk Aperture y dx dy 2 2 ( 2 2 ) ( + ), x x y y x y exp + (x, ) 2 2 z z ( ) This is the Fresnel Diffraction integral. Even with all the approximations we ve made, it is usually difficult to evaluate.

14 Fresnel diffraction example: a slit Consider a uniform plane wave incident on a metal screen with a slit of width 2b in the x -direction. A one-dimensional problem, this may be the simplest of all possible diffraction problems. It s still not easy. 2b Before solving it, let s first try to anticipate what we might expect the answer to look like. Destructive interference when the path length difference is λ/2, 3λ/2, 5λ/2, etc. observation screen

15 Diffraction causes fringes 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. 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.

16 Fresnel integral for a slit Write the Fresnel integral for this one-dimensional problem: E x jk Aperture x dx 2 2 x 2x x + x 2z ( ) exp ( ) The aperture function is given by: Aperture( x ) b < x < b = otherwise Thus the integral becomes: b ( x ) 2 x = exp 2z ( ) E x jk dx b Next step: define new variables and ξ Then: = x b 2 j π b 2 λz ( ) ( ) exp ( ) E x E ξ ξ ξ dξ ξ = x b since k = 2π λ

17 Fresnel diffraction example: a slit E 2 j π b 2 d λz ( ) exp ( ) ξ ξ ξ ξ As a shorthand, we define a dimensionless quantity known as the Fresnel number : 2 N = b λz { 2 } ( ) exp ( ) E ξ jπ N ξ ξ dξ and, of course, we are really interested in the intensity I ( ξ ) E ( ξ ) 2 This is not an integral that can be solved in closed form. It must be computed numerically.

18 Fresnel Diffraction from a Slit (cont d) The irradiance vs. position for different distances from the slit: Close to the slit z Far from the slit Slit I(x,z) Incident plane wave

19 Fresnel Diffraction through a slit: numerical results for I(ξ ) Fresnel number: 2 N = b λz Example: green light (λ =.5 µm) Far from the slit: Fresnel number N.5-2b 2b a) slit width b = millimeter = 2λ N = at a distance of 2 meters b) slit width b = microns = 2λ N = at a distance of 2 µm Recall: this Fresnel calculation is only valid for z >> b, which is the paraxial approximation. Closer to the slit: # of ripples ~ Fresnel number!

20 Fresnel Diffraction through a slit: far field { 2 } ( ) exp ( ) E ξ jπ N ξ ξ dξ In the limit that N << (very far from the aperture), the integral can be performed analytically. The math is a bit tedious, so we just quote the result here: ( ) E x 2π Nx sin b 2π Nx b Our old friend the sinc function! Fresnel number.3 -b b In this regime (the far field ), the diffraction pattern no longer changes shape as z increases, but merely expands in size uniformly..2.

21 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. A large source masks diffraction effects. Screen with slit Here, rays from a point source yield, in principle, a perfect shadow of the hole, allowing diffraction ripples to be seen. Rays from other regions of the source blur the shadow. In general, a large source (like the sun or a light bulb) casts blurry shadows, masking the diffraction ripples.

22 Fresnel Diffraction through a slit: isn t there an awesome java applet that illustrates all this? More than one, actually