Wave Phenomena Physics 15c Lecture 19 Diffraction
What We Did Last Time Studied interference > waves overlap Amplitudes add up Intensity = (amplitude) does not add up Thin-film interference Reflectivity of thin film depends on thickness, plus hard/soft-ness of the two boundaries Newton s rings 1/4-wavelength anti-reflective coatings Two-body interference πd sinθ I = I Intensity depends on angle 0 cos λ Young s experiment demonstrated wave-ness of light
Goals for Today Extend -body n-body interference Radiation becomes highly directional Diffraction gratings Diffraction Light through a finite-size hole on the wall Light does not follow the exact shape of the hole Fundamental limit of the resolution of optical system Crystallography As an example of technology that uses interference
N-Body Interference It s easy to extend -body interference to n bodies Result is interesting There are practical applications Difference between paths from neighboring sources is d sinθ r m = r 1 + (m 1)d sinθ Add up E from all sources n E = E 0 e i(kr m ωt ) = E 0 e i(kr 1 ωt ) e m=1 n m=1 d ik(m 1)d sinθ = E 0 e i(kr 1 ωt ) e iknd sinθ 1 e ikd sinθ 1 r 1 d sinθ r n n α (m 1) = α n 1 m=1 α 1
N-Body Interference E = E 0 e i(kr 1 ωt ) e iknd sinθ 1 e ikd sinθ 1 = E 0 ei(kr 1 ωt ) e i k(n 1)d sinθ e i We are interested in E phase doesn t matter E = E 0 knd sinθ sin( ) kd sinθ sin( ) = E 0 x is half the phase difference between neighbors How does this look like? k(n 1)d sinθ = E 0 e i(kr 1 ωt ) i e sinnx sin x It s a periodic function (period π) of x where x What happens at x = 0, π, π, where sinx = 0? knd sinθ knd sinθ i e kd sinθ kd sinθ i i e e knd sinθ sin( ) sin( kd sinθ ) kd sinθ
N-Body Interference Wherever sinx = 0, sinnx is also 0 sinnx Find the limit lim x 0 sin x = lim nx + x 0 x + = n n = 5 n = 3 n = n = 4
N-Body Interference Express the intensity relative to θ = 0 E = E 0 knd sinθ sin( ) kd sinθ sin( ) = E 0 sinnx sin x I(θ) I(0) = sinnx n sin x x = Peaks near x = 0, π, π, gets narrower as n increases With large enough n, waves vanish except for very sharp peaks at kd sinθ i.e., when the difference between the neighbors is a multiple of the wavelength First peak at = mπ d sinθ = mλ sinθ = λ d I I 0 0 π π π Large n π 3π 3π x
Diffraction Grating Consider a plate with fine stripes e.g., a transparent film with regular scratches Gaps between scratches transmit light Like Young s experiment, with many many many slits Light is diffracted at angle θ only if Trivial solution: θ = 0 for any wavelength First non-trivial solution: m = 1 sinθ = λ d Light goes to different θ depending on λ light d sinθ = mλ Can observe light intensity as a function of wavelength
Wide Slit Now we make a wide slit on the wall We know what happens: a shadow shaped just like the slit We also know that we get spreading waves from a narrow slit Observer behind the wall sees light with a narrow slit, but not with a wide slit Does this make sense? What happens if the width of the slit is in between? Light Light
Wide Slit Huygens principle tells us what to do Imagine a lot of wave sources in the slit Waves spread from each source Add up the amplitude This is n-body interference with big n We know, at large distance away from the wall, I(θ) I(0) = sinnx kd sinθ n sin x x d a Make n while keeping nd = a = width of the slit constant
Wide Slit Given that nd = a, we can re-express x kd sinθ Take the limit of n infinity lim n sin n sin ka sinθ ka sinθ n = = lim n ka sinθ n sin n ka sinθ ka sinθ n = sinnx n sin x = sin ka sinθ n sin sin ka sinθ ka sinθ ka sinθ n a Let s define α ka sinθ Then the above limit becomes I(θ) I(0) = sin (α) α = πa sinθ λ sinα α = sinc α = sinc(α) a sinθ
Fourier and Sinc We saw sinc x = sin x in Lecture #10 x Fourier transform of a square pulse was F(ω) = 1 πωt sin ωt = 1 π sinc ωt Is there a connection? In fact there is
Fourier Integral Represent the slit by the transmittivity as a function of position y It s a square function Waves from y is i {k(r y sinθ ) ωt } E 0 f (y)e Integrate this to get the total amplitude E(θ) = + f (y) a 1 E 0 f (y)e i {k(r y sinθ ) ωt } dy = E 0 e i(kr ωt ) + f (y)e iky sinθ dy = πe 0 e i(kr ωt ) F(k sinθ) y a y r y sinθ Amplitude as a function of angle θ follows the Fourier transform of the transmittivity as a function of position y r
Interpretation Shape of the slit and the amplitude of the waves are related by Fourier integrals f (y) Width of f(y) and F(k sinθ) are inversely proportional to each other Remember the discussion of speed and bandwidth? Wide slit light continues straight F(k sinθ) The wider the slit, the narrower the spread of the direction of the waves Narrow slit light spreads all directions If you remember the δ -function, you know that a perfectly straight beam of light must be infinitely wide
Fraunhofer Diffraction Let s look at I(θ) I(0) = sin α α α πa sinθ λ a λ = 0.1 a λ = 1 a λ = a λ = 5 a λ = 10 narrow - - - - - - wide First zero is at sinθ = λ a
Fresnel Diffraction So far we considered diffraction at large distance Fraunhofer diffraction as a function of θ At short distance, things get much more complicated Angle from each source differs Path difference not exactly dsinθ Solution is known as the Fresnel diffraction Principle is unchanged, but the result cannot be expressed in a simple form Must be calculated numerically d r 1 d sinθ r n
Optical Devices Optical devices use lenses to collect light from object Telescopes, microscopes, cameras, human eyes Lenses have finite aperture It s like passing light through a hole Diffraction Light spreads out according to This blurs the image sinθ = λ a
Optical Resolution Consider two points of the object θ θ d Light from these points end up in two points But they are spread out due to diffraction by If the angle θ between two points is θ < θ d, the points cannot be distinguished (or resolved) in the image We can consider θ d as the ultimate resolution It s fundamental due to the wave nature of light It s determined by the aperture of the device sinθ d = λ a
Airy Disc Fraunhofer diffraction was calculated for a slit Lenses are usually round We need the diffraction formula for a round hole Calculation tedious, but not fundamentally different Solution known as Airy Disc light a sinθ d = 1. λ a θ d
Airy Disc Actual shape of light pattern looks like a = λ a = λ a = 5λ There should be fringes, but impossible to see
Diffraction Limit Resolution of an optical device with aperture a is This is known as the diffraction limit or Rayleigh criterion It can get worse, of course Example: 10-inch telescope Eyes are good for 10-3 rad θ sinθ > 1. λ a If magnification > 400, you start to see the blur 1. 550 10 9 m 0.54m =.6 10 6 rad No point in making >500 power 10-inch telescope Rule of thumb for telescope designers: 50x per inch
Large Telescopes The larger, the better It s that simple Better resolution + better light collection Keck + KeckII (10 m) on Mauna Kea, Hawaii Location chosen to minimize the disturbance due to air density fluctuation Index refraction of air limits all telescopes on Earth
Crystallography Crystal = regular array of molecules Each molecule can scatter light Crystal makes 3-D version of n-body interference Consider it as light-reflecting planes separated by a fixed distance d d What is the condition for constructive interference?
Bragg Condition Constructive interference is achieved if d sinθ = n λ d d Waves reflected by adjacent layers have the same phase Bragg condition: d sinθ = nλ This is the basis of all crystallography d sinθ d sinθ
Crystallography d sinθ = nλ There are many different sets of planes in a crystal Different angle and different plane spacing Each set reflect light when Bragg condition is met Experiment can control Angle: rotate the crystal Wavelength: use white light and analyze wavelength of the reflection using a diffraction grating (spectrometer) Crystalline structure can be reconstructed from measured reflection
X-Ray Diffraction Wavelength must be comparable to d X-rays To make X-rays, accelerate electrons and hit a wall When the electrons stop, deceleration creates radiation λ = 0.6975 Å on cubic SiC crystal -D diffraction pattern of lysozyme
Summary Two-body n-body interference Waves become more focused as n gets larger Diffraction gratings analyze light spectrum Diffraction Wide slits Fraunhofer diffraction Slit and diffraction connected by Fourier The wider the slit, the narrower the angular distribution Diffraction limits resolution of optical devices Airy disc for round aperture 1. Crystallography Bragg condition Next : Optics I(θ) I(0) = sin α α I(θ) I(0) = sinnx n sin x kd sinθ x α πa sinθ λ θ > 1. λ a