Diffraction Light bends! Diffraction assumptions Solution to - PowerPoint PPT Presentation

Diffraction Light bends! Diffraction assumptions Solution to Maxwell's Equations The far-field Fraunhofer Diffraction Some examples

Diffraction Shadow of a Light does not hand illuminated always travel in a by a straight line. Helium- Neon laser It tends to bend around objects. This tendency is called diffraction . Any wave will do Shadow of this, including a zinc oxide matter waves and crystal acoustic waves. illuminated by a electrons

Why it’s hard to see diffraction Diffraction tends to cause ripples at edges. But poor source temporal or spatial coherence masks them. Example: a large spatially incoherent source (like the sun) casts blurry shadows, masking the diffraction ripples. Screen Untilted rays with hole yield a perfect shadow of the hole, but off-axis rays blur the shadow. A point source is required.

Diffraction of a wave by a slit λ  slit size Whether waves in water or electromagnetic radiation in air, passage through a slit yields a diffraction pattern that will appear more dramatic as the size of the slit approaches the λ < slit size wavelength of the wave. λ ≈ slit size

Diffraction of ocean water waves Ocean waves passing through slits in Tel Aviv, Israel Diffraction occurs for all waves, whatever the phenomenon.

Diffraction Transmission by an Edge x Even without a small slit, diffraction Light can be passing strong. by edge Simple propagation past an edge yields an Electrons unintuitive passing by irradiance an edge pattern. (Mg0 crystal)

Radio waves diffract around mountains. When the wavelength is km long, a mountain peak is a very sharp edge! Another effect that occurs is scattering, so diffraction’s role is not obvious.

Diffraction Geometry 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 applications. y 1 y 0 A ( x 0 , y 0 ) x 0 x 1 P 1 0 Incident wave This region is assumed to be much smaller than this one. What is E ( x 1 ,y 1 ) at a distance z from the plane of the aperture?

Diffraction Solution The field in the observation plane, E ( x 1 ,y 1 ) , at a distance z from the aperture plane is given by: ∫∫ = − − E x y z ( , , ) h x ( x , y y , ) ( z E x , y ) dx dy 1 1 1 0 1 0 0 0 0 0 A x ( , y ) 0 0 1 exp( ikr ) − − = 01 ( h x x , y y , ) z where : λ 1 0 1 0 i r 01 Spherical ( ) ( ) = + − 2 + − 2 2 r z x x y y wave and : 01 0 1 0 1 A very complicated result! And we cannot approximate r 01 in the exp by z because it gets multiplied by k , which is big, so relatively small changes in r 01 can make a big difference!

Fraunhofer Diffraction: The Far Field We can approximate r 01 in the denominator by z , and if D is the size of the aperture, D 2 ≥ x 0 2 + y 0 2 , so when k D 2 / 2 z << 1 , the quadratic terms << 1, so we can neglect them: ( ) ( ) ( ) ( )   = + − 2 + − 2 ≈ + − 2 + − 2 2 2 2 r z x x y y z 1 x x / 2 z y y / 2 z   01 0 1 0 1 0 1 0 1 ( ) ( ) ≈ + − + + − + 2 2 2 2 kr kz k x 2 x x x / 2 z k y 2 y y y / 2 z 01 0 0 1 1 0 0 1 1 Independent of x 0 and Small, so neglect these terms. y 0 , so factor these out.  ∫∫   +   2 2 exp( ikz ) x y ik ( ) ( ) ( ) = − +   1 1   E x y , exp ik exp x x y y E x y , dx dy λ 1 1  0 1 0 1  0 0 0 0  i z 2 z z A x ( , y ) 0 0 This condition means going a distance away: z >> kD 2 /2 = π D 2 / λ If D = 1 mm and λ = 1 micron, then z >> 3 m.

Fraunhofer Diffraction We’ll neglect the phase factors, and we’ll explicitly write the aperture function in the integral: ∞ ∞ ∫∫   ik ( ) ( ) ∝ − +   E x y , exp x x y y A x y ( , ) ( E x y , ) dx dy 1 1 0 1 0 1 0 0 0 0 0 0   z −∞ −∞ E ( x 0 , y 0 ) = constant if a plane wave This is just a Fourier Transform! Interestingly, it’s a Fourier Transform from position, x 0 , to another position variable, x 1 (in another plane). Usually, the Fourier “conjugate variables” have reciprocal units (e.g., t & ω , or x & k ). The conjugate variables here are really x 0 and k x = kx 1 /z , which have reciprocal units. So the far-field light field is the Fourier Transform of the apertured field!

The Fraunhofer Diffraction formula We can write this result in terms of the off-axis k-vector components: ∞ ∞ E ( x , y ) = const if a plane wave ∫∫ ( ) ( )   ∝ − + E k k , exp i k x k y A x y E x y dxdy ( , ) ( , )   x y x y Aperture function −∞ −∞ ( ) { } where we’ve dropped ∝ F E k k , A x y E x y ( , ) ( , ) the subscripts, 0 and 1, x y k x k x = kx 1 /z and k y = ky 1 /z and: k z θ x = k x /k = x 1 /z and θ y = k y /k = y 1 /z k y or:

The Uncertainty Principle in Diffraction! ( ) { } ∝ F k x = kx 1 /z E k k , A x y E x y ( , ) ( , ) x y Because the diffraction pattern is the Fourier transform of the slit, there’s an uncertainty principle between the slit width and diffraction pattern width! If the input field is a plane wave and ∆ x = ∆ x 0 is the slit width, ∆ ∆ > x k 1 x Or: ∆ ∆ > x x z k / 0 1 The smaller the slit, the larger the diffraction angle and the bigger the diffraction pattern!

Fraunhofer Diffraction from a slit Fraunhofer Diffraction from a slit is simply the Fourier Transform of a rect function, which is a sinc function. The irradiance is then sinc 2 .

Fraunhofer Diffraction from a Square Aperture The diffracted field Diffracted is a sinc function in irradiance both x 1 and y 1 because the Fourier transform of a rect function is sinc. Diffracted field

Diffraction from a A circular aperture yields a diffracted Circular Aperture "Airy Pattern," which involves a Bessel function. Diffracted Irradiance Diffracted field

Diffraction from small and large circular Far-field apertures intensity pattern from a small aperture Recall the Scale Theorem! This is the Uncertainty Principle for diffraction. Far-field intensity pattern from a large aperture

Fraunhofer diffraction from two slits w w x 0 -a 0 a A ( x 0 ) = rect[( x 0 +a )/ w ] + rect[( x 0 -a )/ w ] ∝ F E x ( ) { ( A x )} 1 0 ∝ + + sinc[ ( w kx / ) / 2]exp[ z ia kx ( / )] z 1 1 − sinc[ ( w kx / ) / 2]exp[ z ia kx ( / )] z 1 1 ∝ kx 1 /z E x ( ) sinc( wkx / 2 ) cos( z akx / ) z 1 1 1

Diffraction from one- and two-slit screens Fraunhofer diffraction patterns One slit Two slits

Diffraction Gratings •Scattering ideas explain what happens a when light impinges on a periodic array of Scatterer grooves. Constructive interference occurs if D the delay between adjacent beamlets is an θ m C integral number, m , of wavelengths. θ m θ i a Incident B wave-front Path difference: AB – CD = m λ Potential diffracted A θ i wave-front [ ] θ − θ = λ a sin( ) sin( ) m AB = a sin( θ m ) m i Scatterer CD = a sin( θ i ) where m is any integer. A grating has solutions of zero, one, or many values of m , or orders . Remember that m and θ m can be negative, too.

Because the diffraction angle depends on λ , Diffraction orders different wavelengths are separated in the nonzero orders. Diffraction angle, θ m ( λ ) First order No wavelength Incidence Zeroth order dependence angle, θ i occurs in zero Minus first order. order The longer the wavelength, the larger its deflection in each nonzero order.

Real diffraction gratings m = m = m m = -1 0 =1 Diffracted 2 white light White light diffracted by a real grating. The dots on a Diffractio CD are equally n spaced gratings (although some

World’s largest diffraction grating Lawrence Livermore National Lab