Lecture notes for Biophysics IIHana DobrovolnyDepartment of Physics, York University, Toronto, ONSeptember 12, 2011Contents1 Images 51.1 Imaging theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.1.1 One-dimensional images . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.1.2 Two-dimensional images . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2 Image reconstruction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.2.1 Projections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.2.2 Fourier transform technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.2.3 Back projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.3 Image ltering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.4 Microscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312 Optical tweezers 362.1 Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.1.1 Radiation force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.1.2 Gradient force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.2 Sphere in a focused beam of light . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.3 Determining strength of the trap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.3.1 Constant velocity calibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.3.2 Brownian motion calibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423 Photon interactions with matter 433.1 Types of interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433.1.1 Photoelectric eect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433.1.2 Compton scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.1.3 Coherent scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.1.4 Pair production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503.2 Photon attenuation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523.3 Energy transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543.3.1 De-excitation of the atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543.3.2 Mass energy transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.4 Charged particle interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563.4.1 Stopping power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563.4.2 Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613.4.3 Measures of energy transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 633.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6614 Crystallography 694.1 Lattice structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 694.1.1 Bravais lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714.1.2 Primitive cells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 724.1.3 Volume of a cell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 754.1.4 Index system for crystal planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 764.2 Bragg diraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 764.2.1 Fourier analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 774.2.2 Reciprocal lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 784.2.3 Diraction in Fourier space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 794.2.4 Structure Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 814.2.5 The phase problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 834.3 Protein Crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 854.3.1 Making protein crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 854.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 885 Spectroscopy 925.1 Determining energy levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 925.1.1 Particle in a box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 925.1.2 Two-dimensional particle in a box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 955.1.3 Circular well . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 975.1.4 Rotational spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 975.1.5 Vibrational spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 995.1.6 Franck-Condon principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1035.2 Absorption and emission rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1035.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1066 Nuclear Magnetic Resonance 1096.1 Magnetic dipoles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1096.2 Magnetization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1106.2.1 Average magnetic moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1106.3 Precession . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1116.3.1 Dephasing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1126.4 Relaxation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1136.4.1 Rotating coordinate system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1156.5 Oscillating magnetic eld . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1166.5.1 Pulses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1176.5.2 Fluctuations in the magnetic eld . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1186.5.3 Autocorrelation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1196.5.4 Measuring the magnetic resonance signal . . . . . . . . . . . . . . . . . . . . . . . . . 1196.5.5 Pulse sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1216.6 Imaging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1216.6.1 Slice selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1226.6.2 Image reconstruction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1226.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1272List of Figures1.1 Impulse response of a temporal signal. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2 Point spread function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3 Spatial frequencies. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.4 Projections. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.5 Sinogram. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.6 Back projection. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.7 Examples of image lters. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.8 The microscope. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.9 Microscopy techniques. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.10 Staining in bright eld microscopy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.11 Oblique illumination microscopy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.12 Dark eld microscopy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261.13 Dispersion staining microscopy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.14 Phase contrast microscopy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.15 Dierential interference contrast microscopy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.16 Interference reection contrast microscopy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291.17 Fluorescence microscopy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301.18 Confocal microscopy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.1 Gradient force in a laser beam. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.2 Forces on a sphere in a focused beam. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.3 Eect of numerical aperture on a laser trap. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402.4 Power spectrum of a trapped particle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.1 Photon interactions with matter. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443.2 Photoelectric eect. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.3 Applications of the photoelectric eect. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.4 Compton scattering. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.5 Angular dependence of energy in Compton scattering. . . . . . . . . . . . . . . . . . . . . . . 483.6 Compton scattering cross section. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.7 Coherent scattering. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513.8 Attenuation coecient. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.9 De-excitation of the atom. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.10 Tracks of charged particles in matter. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.11 Mass stopping power. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 583.12 Energy transfer at high energies. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603.13 Range. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 623.14 Energy transferred and energy imparted. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 643.15 Dose and kerma. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 644.1 Crystals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7034.2 A two-dimensional lattice. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 704.3 Two-dimensional Bravais lattices. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 724.4 Three-dimensional Bravais lattices. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 734.5 Primitive vectors of BCC and FCC lattices. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744.6 The Wigner-Seitz primitive cell. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 754.7 Crystal planes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 764.8 Bragg diraction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 774.9 DNA diraction pattern. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 804.10 Laue cones. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 814.11 Brillouin zones. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 824.12 The phase problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 844.13 Isomorphous replacement. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 854.14 Extended X-ray absorption ne structure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 864.15 Protein crystallization. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 864.16 Vapor diusion methods of protein crystallization. . . . . . . . . . . . . . . . . . . . . . . . . 875.1 Wavefunctions and probabilities for a particle in a box. . . . . . . . . . . . . . . . . . . . . . . 945.2 Degeneracy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 975.3 The rigid rotor model of a diatomic molecule. . . . . . . . . . . . . . . . . . . . . . . . . . . . 985.4 Simple harmonic oscillator spectrum. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1015.5 Simple harmonic oscillators. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1025.6 Franck-Condon principle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1046.1 Larmor precession. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1126.2 Dephasing. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1136.3 Relaxation of magnetization in a constant magnetic eld. . . . . . . . . . . . . . . . . . . . . 1156.4 Magnetic eld of a quadrupole magnet. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1236.5 Conguration of magnetic elds in an MRI. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1246.6 Phase encoding. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1256.7 Image reconstruction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1264Chapter 1ImagesImages, whether formed by the eye, a camera or other detection device are important in medical and biologicalphysics. We have already seen many examples of imaging in medical physics such as x-rays, optical coherencetomography (OCT) or positron emission tomography (PET). Imaging plays a crucial role in observing andunderstanding biological processes, so we need to understand how images are formed and how to correctlyinterpret them.1.1 Imaging theoryImages are map some quantity as a function of position, typically in two dimensions. Before we attemptto describe two-dimensional images, we will introduce the basic concepts of images with a one-dimensionalsystem. Since spatial one-dimensional images are dicult to conceptualize, we will use a temporal system,namely sound, to develop the mathematics needed to understand images.1.1.1 One-dimensional imagesImagine that you are listening to music coming from your radio. Suppose that the original soundtrackis described by some function, f(t). Ideally, the music perceived by your ear will also be f(t). In reality,however, the electronics of the radio and travel through the air to your ear distorts the original input (see Fig.1.1). The majority of these distortions essentially add some delay to parts of the original signal. For example,while some of the sound wave will travel directly to your ear, other parts will bounce o nearby structuresbefore being detected by your ear. The signal perceived by your ear, g(t), then will be a combination of theoriginal input and the input at earlier times. If we assume the system is linear, we can simply add all thecontributions to determine g(t):g(t) =_ f(t

)h(t, t

) dt

(1.1)where h(t, t

) is a weighting function that describes how the signal is delayed.Suppose f is a function at time t

0 you can imagine it to be some loud, sudden, short-lived noise likea thunderclap. Then the noise you will hear is given byg(t) =_ (t

t

0)h(t, t

) dt

= h(t, t

0) (1.2)This means that h(t, t

) describes how our system responds to an impulse at time t

, so its called the impulseresponse of the system. If we know the impulse response of the system, then its possible to calculate theoutput for any arbitrary input which means that if you want to calculate the output signal for any inputsignal, you will rst need to determine the impulse response for your system by applying some sudden loudnoise and measuring the response.5Figure 1.1: Impulse response of a temporal signal. An input signal is transformed by the impulseresponse function h(t) to the signal that is actually detected.If the system response to an impulse is the same regardless of when it occurs, the system is called astationary system. For example, if you are listening to music in a room where nothing (including you)is moving, this is a stationary system the impulse response will be the same no matter when you testit. If, however, you move to another point in the room, the system is no longer stationary there willbe one particular impulse response before you move and a dierent one after you move. For a stationarysystem, h(t, t

) depends only on the time dierence between the input and the detection of the signal:h(t, t

) = h(t t

). Then the superposition integral takes the formg(t) =_ f(t

)h(t t

) dt

. (1.3)This is called the convolution integral and is denoted g(t) = f(t) h(t).Example: ConvolutionFind the convolution of f(t) = etand h(t) = e2tfor x 0.Solution: Note that the integral goes from 0 to rather than from to because f(t) = 0when x < 0.g(t) = f(t) h(t)=_ 0et

e2(tt

)dt

= e2t_ 0et

dt

= e2tConvolution theoremRecall that any function can be approximated by a Fourier integral, which converts a function to an inniteseries of sine and cosine functions:f(t) = 12_ [Cf() cos t +Sf() sin t] d, (1.4)6whereCf() =_ f(t) cos t dt (1.5)Sf() =_ f(t) sin t dtare the Fourier transform pairs. We will use the Fourier transform to help characterize the convolution.We can write the input signal, f(t), and the impulse response, h(t t

), in their Fourier integral formf(t

) = 12_ [Cf() cos t

+Sf() sin t

] d (1.6)h(t t

) = 12_ [Ch() cos (t t

) +Sh() sin (t t

)] d .This means that the output function isg(t) =_ f(t

)h(t t

) dt

(1.7)=_ 12_2_ dt

_ [Cf() cos t

+Sf() sin t

] d_ [Ch() cos (t t

) +Sh() sin (t t

)] d.We can use the trigonometric relations sin(tt

) = sin(t) cos(t

)sin(t

) cos(t) and cos(tt

) = cos(t) cos(t

)+sin(t) sin(t

) to expand the h(t t

) integralh(t t

) =_ [Ch()(cos(t) cos(t

) + sin(t) sin(t

)) (1.8)+ Sh()(sin(t) cos(t

) sin(t

) cos(t))] d.We can then integrate g(t) over t

to getg(t) = 12_ [Cf()Ch() Sf()Sh()] cos(t) d (1.9)+ 12_ [Cf()Sh() +Sf()Ch()] sin(t) d.Comparing this to the Fourier integral for g(t), we see that we must haveCg() = Cf()Ch() Sf()Sh() (1.10)Sg() = Cf()Sh() +Sf()Ch().This is known as the convolution theorem and it tells us that the convolution can be determined from theFourier transforms of the original functions. We now have two possible methods for determining the outputfunction for an arbitrary input function. We can directly perform the convolution integral or we can calculatethe Fourier transforms for f(t) and h(t t

), use the convolution theorem to nd the Fourier coecientsfor g(t) and calculate the inverse transform to nd g(t). Note that this can be done more compactly usingcomplex exponentials, in which case the Fourier transform isF() =_ f(t)eitdt, (1.11)and the convolution theorem becomesG() = F()H(). (1.12)7Example: Convolution theoremFind the convolution of f(t) = etand h(t) = e2tfor x 0 using the convolution theorem.Solution: We need to nd the Fourier transform coecients for both f(t) and h(t).Cf() =_ 0etcos t dtCf() = 11 +2Sf() =_ 0etsin t dtSf() = 1 +2.andCh() =_ 0e2tcos t dtCh() = 22 +2Sh() =_ 0e2tsin t dtSh() = 2 +2.This allows us to nd the Fourier coecients of gCg() = Cf()Ch() Sf()Sh()Cg() = 2(1 +2)(2 +2) 2(1 +2)(2 +2)Cg() = 2 2(1 +2)(2 +2)Sg() = Cf()Sh() +Sf()Ch()Sg() = (1 +2)(2 +2) + 2(1 +2)(2 +2)Sg() = 3(1 +2)(2 +2).Now we can take the inverse transform to nd g(t)g(t) = 12_ [Cg cos t +Sg sin t] dg(t) = 12_ _ (2 2) cos t(1 +2)(2 +2) + 3 sin t(1 +2)(2 +2)_ dg(t) = e2t81.1.2 Two-dimensional imagesIn medical and biological physics, we most often use two-dimensional images, so we want to extend the math-ematics of images from one dimension to two dimensions. In two dimensions, an object can be representedby a function f(x

, y

) in the object plane. We have developed many ways to look at the object. If weare using direct visual observation (looking with our eyes), we detect light rays that have bounced o theobject and are processed by our brains. There are also other ways of observing the object x-rays measurethe attenuation coecient as a function of position or MRIs measure the relaxation time as a function ofposition. The image we see, g(x, y) in the image plane, is given byg(x, y) =_ _ f(x

, y

)h(x, x

; y, y

) dx

dy

(1.13)where h(x, x

; y, y

) describes the measurement process.We can dene a space invariant image, similar to a stationary one-dimensional signal, as one in whichthe object is not moving in space and where the measurement process does not change in time. In this case,the image produced at (x, y) depends only on relative distances x x

and y y

and the image can bedetermined from the two-dimensional convolutiong(x, y) =_ _ f(x

, y

)h(x x

, y y

) dx

dy

(1.14)g(x, y) = f(x, y) h(x, y).Suppose we have an object given by f(x

, y

) = L(x

x

0)(y

y

0), which describes a single brightspot, of amplitude L, in a plane, then the image is given by:g(x, y) =_ _ L(x

x

0)(y

y

0)h(x x

, y y

) dx

dy

(1.15)= Lh(x, y; x

0, y

0)So h(x, y; x

0, y

0) describes the distortion of the imaging system, analogous to the impulse response in aone-dimensional system. h(x, y; x

0, y

0) is known as the point spread function (PSF) of the system (see Fig.1.2). Note that, in reality, the PSF may vary for each point in the object plane, however, for simplicity weassume that there is a single PSF for the entire object.Example: MagnicationThere are some specic distortions of the object that may be benecial when trying to observethe object. One such distortion is the magnication of an object. Verify that magnication isdescribed by the following point spread function:h(x, y; x

, y

) = (x mx

)(y my

) (1.16)Solution: Substituting into the equation for the image:g(x, y) =_ _ f(x

, y

)(x mx

)(y my

) dx

dy

= f_ xm, ym_,which tells us that the point ( xm, ym) maps to the point (x, y) or that the object is magnied by afactor of m in both the x and y directions.9Figure 1.2: Point spread function. An object is transformed by the point spread function h(x, y), whichdescribes the measurement or observation process, to the image that is actually detected.10Convolution theorem in two dimensionsWe can write a Fourier transform in two dimensionsf(x, y) =_ 12_2_ dkx_ dky[Cf(kx, ky) cos(kxx +kyy) (1.17)+ Sf(kx, ky) sin(kxx +kyy)]whereCf(kx, ky) =_ dx_ f(x, y) cos(kxx +kyy) dy (1.18)Sf(kx, ky) =_ dx_ f(x, y) sin(kxx +kyy) dyare the Fourier transform pairs. We can write a similar expression for the Fourier transform of h(x x

, y y

) and substitute them into the expression for g(x, y). Using the trigonometric relationships to expandh(x x

, y y

) and integrating over x

and y

, we can derive the convolution theorem in two dimensionsCg(kx, ky) = Cf(kx, ky)Ch(kx, ky) Sf(kx, ky)Sh(kx, ky) (1.19)Sg(kx, ky) = Cf(kx, ky)Sh(kx, ky) +Sf(kx, ky)Ch(kx, ky).Transfer functionsThe optical transfer function (OTF) is dened as the Fourier transform of the point spread function:Ch(kx, ky) =_ dx_ h(x x

, y y

) cos(kx(x x

) +ky(y y

)) dySh(kx, ky) =_ dx_ h(x x

, y y

) sin(kx(x x

) +ky(y y

)) dy,or if we use complex numbers we can write this more compactlyH(kx, ky) =_ 12_2_ dx_ h(x x

, y y

)ei(kx(xx

)+ky(yy

))dy (1.20)The OTF is useful because it breaks up the point spread function into two components: the modulationtransfer function (MTF) which describes how the modulation is altered, and the phase transfer function(PTF) which describes how the phase is altered. The three transfer functions are related throughOTF = MTFeiPTF(1.21)Suppose we have an object of the form L(x, y) = a + b cos(kxx + kyy). The modulation of the object isdened as:M = LmaxLminLmax +Lmin(1.22)= (a +b) (a b)(a +b) + (a b) = ba.which is the ratio of the amplitude of the signal to the oset of the signal. The modulation transfer functionis the amplitude of the optical transfer function:MTF(kx, ky) = |H(kx, ky)| =_C2h(kx, ky) +S2h(kx, ky) (1.23)11Figure 1.3: Spatial frequencies. The left image is the original. The center image has the high frequenciesremoved and the right image has the low frequencies removed.and it gives the ratio of image modulation to object modulation MTF = Mimage/Mobject.The phase transfer function is the phase of the optical transfer function:PTF(kx, ky) = tan1_Sh(kx, ky)Ch(kx, ky)_ (1.24)and it describes phase shifts from the object to the image. Note that for each angle, there may be a dierentphase shift.Spatial frequenciesJust as sounds are a superposition of waves in time, images can be thought of as a superposition of waves inspace. The lowest spatial frequency determines the eld of view. Low frequencies determine shape, contrastand brightness. The highest frequency determines the resolution of the image. High frequencies determineedges and sharp detail. The role of various frequencies can be seen in Fig. 1.3. The center image shows theoriginal with the high frequencies removed the image appears blurry with all sharp edges and small detailsremoved. The right image shows the original with the low frequencies removed the image has retained allthe edges, but other than those edges the image is a uniform gray.1.2 Image reconstructionMost often the measurements we make somehow convert a three-dimensional object to a two-dimensionalimage. Many times we are interested in re-creating the full three-dimensional object by making a series oftwo-dimensional images and mathematically reconstructing the three-dimensional object. In this section,we will study the mathematical techniques used in image reconstruction. For simplicity, however, instead ofconsidering reconstruction of a three-dimensional object using two-dimensional images, we will consider thesimpler case of reconstructing a two-dimensional surface from one-dimensional projections. The techniquesdeveloped here for two dimensions can easily be extended to three dimensions.1.2.1 ProjectionsFirst, well state the problem more formally. Suppose we have a two-dimensional function f(x, y) thatdescribes the surface of an object. We measure the function along projections essentially integratingf(x, y) along various lines F(, x

) =_f(x, y) dy

, getting several one-dimensional measurements, asshown in Fig. 1.4. If we do this along several angles, can we reconstruct the original surface?12Figure 1.4: Projections. A projection is essentially an integration of the surface along a particular line.Each arrow indicates a particular projection and results in a single numerical measurement.Example: ProjectionSuppose an image is given byI =__5 2 93 0 36 2 7__.What are the projections along the horizontal (x) and vertical (y) directions?Solution: For the projection along the x-axis, we sum all the columnsF(x) =_ 5 + 3 + 6 2 + 0 + 2 9 + 3 + 7F(x) =_ 14 4 19,and for the projection along the y-axis, we sum all the rowsF(y) =_ 5 + 2 + 9 3 + 0 + 3 6 + 2 + 7F(y) =_ 16 6 15.The set of all projections can be plotted in a sinogram. The sinogram plots the projections with x

on thex-axis, on the y-axis and F(, x

) as an intensity (Fig. 1.5), essentially stacking the projections we measureas the detector rotates around the object.13Figure 1.5: Sinogram. An object and its sinogram consisting of a stack of projections at dierent valuesof .1.2.2 Fourier transform techniqueIt turns out that there are several possible methods for reconstructing images, one of which uses Fouriertransforms. We can nd the Fourier transform of the surface f(x, y) (Eq. (1.17)):f(x, y) =_ 12_2_ dkx_ dky[Cf(kx, ky) cos(kxx +kyy)+ Sf(kx, ky) sin(kxx +kyy)]where the Fourier transform coecients are given by Eqs. (1.18).First consider the slice ky = 0; along this slice, the Fourier cosine coecient isCf(kx, 0) =_ cos(kxx) dx_ f(x, y) dy (1.25)=_ cos(kxx)F(0, x) dxwhere F(x) is simply the projection along the y-axis. Similarly, the other Fourier coecient can be writtenas a function of the projection along the y-axis,Sf(kx, 0) =_ sin(kxx)F(0, x) dx. (1.26)This tells us the projections along the y-axis determine the Fourier transform pairs along the ky = 0 axis.This is a specic example of the Fourier slice theorem. You can write equations of this form along any line inFourier space, correlating slices in Fourier space to slices in real space. In general, the Fourier slice theoremstates that the following two calculations are equal:1. Take a two-dimensional function f(x, y), project it onto a line, and do a Fourier transform of thatprojection.14Figure 1.6: Back projection. (Left) Every point on the object surface contributes in some way to eachprojection and every point in every projection contributes in some way to the reconstructed image. (Right)The location of a point on the object can be described in either the coordinate system of the object (x, y)or in the coordinate system of the projection (x

, y

).2. Take that same function, do a two-dimensional Fourier transform rst, and then slice it through itsorigin, parallel to the projection line.This means that the projections can be used to reconstruct the original image through Fourier transforms inthe following way: Take the Fourier transforms of the projections each projection gives one slice in Fourierspace and put the slices together in Fourier space then take the inverse Fourier transform. Unfortunately,this is not always an easy technique to implement; the object must be fairly densely sampled and the inversetransform calculation can be time-consuming.1.2.3 Back projectionBack projection is based on the idea that every point on the surface contributes in some way to eachprojection and every point in every projection contributes in some way to the reconstructed image (Fig.1.6 (left)). The general strategy for back-projection is to take the contributions from each projection toreconstruct the original images.Let the coordinate system of the object be (x, y) and the coordinate system of the projection at angle to the object be (x

, y

), see Fig. 1.6 (right). We can nd equations to transform from the object coordinatesystem to the projection coordinate system;x

= xcos +y sin (1.27)y

= xsin +y cos and the inverse transformsx = x

cos y

sin (1.28)y = x

sin +y

cos Suppose we have an object f(x, y). We dene the projection, F, at angle as an integral along the liney

F(, x

) =_ f(x, y) dy

(1.29)=_ f(x

cos y

sin , x

sin +y

cos ) dy

.15This is known as the Radon transform.We now need to nd some way to convert the projections back into the image. We dene the backprojection as:fb(x, y) =_ 0F(, x

) d. (1.30)This sums the projections over a half circle we only need to measure projections over half the circle, sincethe other half will result in identical projections. We want to show that the back-projection is in some wayrelated to the original object. Consider the projections at the origin:F(, 0) =_ f(y

sin , y

cos ) dy

(1.31)and at the angle

= +/2, which is the angle to the y

-axis (see Fig. 1.6 (right)),F(

, 0) =_ f(y

cos

, y

sin

) dy

. (1.32)This looks like integration in polar coordinates, so re-name y

as r

,F(

, 0) =_ f(r

,

) dr

. (1.33)We substitute this into the back projection, Eq. (1.30), which becomesfb(0, 0) =_ 0_ f(r

,

) dr

d

. (1.34)We can change the limits of integrationfb(0, 0) =_ 20_ 0f(r

,

) dr

d

; (1.35)multiply and divide by r to make it look like polar integrationfb(0, 0) =_ 20_ 0f(r

,

)r

r

dr

d

; (1.36)and convert back to Cartesian coordinatesfb(0, 0) =_ 20_ 0f(x

, y

)(x2+y2)1/2 dx

dy

. (1.37)Recall that this result was only for projections at the origin; the general result (away from the origin) isfb(x, y) =_ 20_ 0f(x

, y

)((x x

)2+ (y y

)2)1/2 dx

dy

(1.38)which looks like the convolution of the object withh(x x

, y y

) = 1((x x

)2+ (y y

)2)1/2 (1.39)So the back projection is related to the object, but it is not an exact replica of the original object.16Example: Back ProjectionSuppose an image is given byI =_ 1 23 4_.Find all the projections then reconstruct the image using the back projection method.Solution: In addition to the horizontal and vertical projections, we can take projections alongthe diagonal.F(0) =_ 1 + 3 2 + 4F(0) =_ 4 6F(45) =_ 3 1 + 4 2F(45) =_ 3 5 2F(90) =_ 1 + 2 3 + 4F(90) =_ 3 7F(45) =_ 1 2 + 3 4F(45) =_ 1 5 4.The back projection method essentially puts all the data from the projections back into the originalimageI =_ 4 + 5 + 3 + 1 6 + 2 + 3 + 54 + 3 + 7 + 5 6 + 5 + 7 + 4_=_ 13 1619 22_.You can see that although all the information has been put back, the back projection methoddoes not quite reconstruct the original image.It would be more convenient if we could take the projections in some clever way so that when we calculatethe back back projection, we get the original object. That is, can we create a function G(, x

) from F(, x

)such that back projection of G will give us exactly f(x, y) instead of the convolution of Eq. (1.38). To begin,note that there is a function g(x, y) which when projected and back-projected leads to f(x, y):f(x, y) = gb(x, y) =_ 20_ 0g(x

, y

)((x x

)2+ (y y

)2)1/2 dx

dy

(1.40)17We can now use the convolution theorem. Since f is a convolution of g and h = 1((xx

)2+(yy

)2)1/2, theFourier transforms are related by the convolution theorem,Cf(kx, ky) = Cg(kx, ky)Ch(kx, ky) Sg(kx, ky)Sh(kx, ky) (1.41)Sf(kx, ky) = Cg(kx, ky)Sh(kx, ky) +Sg(kx, ky)Ch(kx, ky).Solving for Sg and Cg givesCg = ChSf ShCfC2h +S2h(1.42)Sg = ChCf +ShSfC2h +S2h.In this case, we have a specic functional form for h, so we can nd the Fourier transform,Ch(kx, ky) = 2(k2x +k2y)1/2(1.43)Sh(kx, ky) = 0,which makes Sg and CgCg = 12(k2x +k2y)1/2Cf (1.44)Sg = 12(k2x +k2y)1/2Sf.So we can nd a function g(x, y) from f(x, y) that when projected and back projected will give f(x, y)exactly. This means that we have to somehow manipulate the original object, f(x, y), to get the functiong(x, y) which will give the projections G(, x

) which when back-projected gives exactly f(x, y). Instead oftrying to manipulate the object, however, it is easier to manipulate the projections. So its more useful tond G(, x

) from F(, x

), entirely avoiding the function g(x, y) which we do not know. Consider projectionsalong the x axis. We can write G as a Fourier transform of gG(0, x) = 12_ [Cg(kx, 0) cos(kxx) +Sg(kx, 0) sin(kxx)] dkx, (1.45)where we have set ky = 0 because of the Fourier slice theorem. Eq. (1.44) gives a relationship between theFourier coecients of g and f, so G can be written as a function of the Fourier coecients of fG(0, x) =_ 12_2_ [Cf(kx, 0) cos(kxx) +Sf(kx, 0) sin(kxx)] |kx| dkx. (1.46)Since this is independent of the choice of axis, it must be true for all projections. This means that there isa function that will transform the projections of f into the projections of g. We can nd that function fromthe Fourier coecients. Since we haveCg(kx, 0) = |kx|2 Cf(kx, 0) (1.47)Sg(kx, 0) = |kx|2 Sf(kx, 0),we must haveCh = |kx|2 (1.48)Sh = 0,18which meansh(x) =_ 12_2_ |kx| cos(kxx) dkx (1.49)If we take the each projection of the object, F(, x), and convolve it with h(x) then take the back projection,we will recover exactly f(x, y). This process is called ltered back projection.1.3 Image lteringFiltering of images is the process of modifying an image to highlight certain aspects or minimize others.Filters are deliberate distortions of the images and, so can be written as convolutions. When discussinglters, it is sometimes easier to consider the image as consisting of discrete pixels. In this case, ltersare represented as matrices which are passed over the entire image by the convolution process. The lteressentially replaces the value of a pixel with a weighted sum of its nearest neighbours. Some examples oflters are discussed below.BlurringA blurring lter smooths out dierences between neighbouring pixels by replacing a pixel with the mean ormedian of the pixel and some of its neighbours. A simple 33 blurring lter (larger ones can also be made)has the formFblur =__0 0.2 00.2 0.2 0.20 0.2 0__. (1.50)Note that we have to normalize the lter i.e. the sum of all elements must be one or the resulting imagewould be brighter than the original. In this example, all pixels used in the averaging are given equal weight,but it is also possible to create lters with unequal weights and to vary the number of neighbouring pixelsused in the averaging. The eect of a blurring lter is shown in Fig. 1.7 (center left).Motion BlurringMotion blurring gives the impression that an image is moving in one direction. This is achieved by blurringin only one direction. A motion blurring lter that gives the appearance of diagonal movement has the formFmotion =__13 0 00 13 00 0 13__, (1.51)where we have again normalized the lter. Note that this example is a 33 lter, but again larger ones arepossible. An image ltered with a motion lter is shown in Fig. 1.7 (center right).Edge FinderAn edge-nding lter can be seen as a sort of discrete version of the derivative; you take the current pixeland subtract the value of the previous one from it, so you get a value that represents the dierence betweenthose two or the slope of the function. An edge-nding lter that will detect horizontal dierences isFedge =__0 0 01 1 00 0 0__. (1.52)This lter is not normalized; the sum of all the elements is 0. Since this is the case, most of the image willbe darker than the original, as can be seen in Fig. 1.7 (bottom left).19SharpeningSharpening an image is very similar to nding edges. The dierence is that we want to add the enhancededges to the original image. So we add an edge-nding lter with the identity lter (which just returns theoriginal image) and the result will be a new image where the edges are enhanced, making it look sharper.An example of this type of lter isFsharpen =__1 1 11 9 11 1 1__, (1.53)where we have again ensured that the lter is normalized since we want to retain the original image. Notethat this particular lter nds edges in all directions. The result of a sharpening lter is shown in Fig. 1.7(bottom right).Example: FiltersSuppose an image is given byI =__4 5 2 4 53 0 6 1 38 4 3 0 26 7 1 2 79 4 9 3 4__,where you can assume all pixels outside the image are 0. Use the motion blurring lter given byEq. (1.51) to lter the image.Solution: For each pixel in the image, we replace it with the weighted sum indicated by the lterInew =__0 + 43 + 0 0 + 53 + 63 0 + 23 + 13 0 + 43 + 33 0 + 53 + 00 + 33 + 4343 + 0 + 3353 + 63 + 0 23 + 13 + 2343 + 33 + 00 + 83 + 7333 + 43 + 13 0 + 33 + 2363 + 0 + 7313 + 23 + 00 + 63 + 4383 + 73 + 9343 + 13 + 3333 + 23 + 43 0 + 73 + 00 + 93 + 0 63 + 43 + 0 73 + 93 + 0 13 + 33 + 0 23 + 43 + 0__Inew =__113 323 1 213 123213 213 323 123 2135 223 123 413 1313 8 223 3 2133 313 513 113 2__.1.4 MicroscopyOne of the original imaging techniques used to study biology is microscopy. The microscope contains a seriesof lenses that produce a magnied image of the object under study (Fig. 1.8). While the original use ofthe microscope was to simply observe specimen under normal light (bright eld microscopy), the eld ofmicroscopy has advanced and now uses many dierent techniques for viewing objects. These include: Bright eld Oblique illumination Dark eld20Figure 1.7: Examples of image lters. Image lters can be used to distort or enhance images. Someexapmles include a blurring lter (center left), a motion lter (center right), an edge lter (bottom left) anda sharpening lter (bottom right).21Figure 1.8: The microscope. The objective produces a slightly magnied real image that is seen as ahighly magnied virtual image when viewed through the eyepiece. Dispersion staining Phase contrast Dierential interference contrast Interference reection microscopy Fluorescence ConfocalMany of these techniques can be used in conjunction to highlight dierent aspects of the specimen. Fig. 1.9shows the same specimen (algae lament) viewed with three dierent microscopy techniques. The image isquite faint in the bright eld and many features are indistinct. The phase contrast and dierential interferencecontrast images are clearer and sharper, but each highlights dierent features of the lament. Images takenusing a variety of microscope techniques can be seen at http://www.microscopyu.com/smallworld/gallery/and http://www.olympusmicro.com/galleries/index.html.Bright eld microscopyBright eld microscopy is the simplest microscopy technique. The sample is illuminated from below andtransmitted light is viewed through the eyepiece. The sample is seen because light is absorbed by someparts of the sample, but passes through others. This technique is cheap to implement and very easy tounderstand and interpret. Unfortunately, the contrast is typically low since biological samples tend not toabsorb much light, although this can be corrected to some extent with stains. Fig. 1.10 shows stained andunstained mites under bright eld microscopy. Without stain, the mite is faint and dicult to see; withstain, even small details such as the cilia on the legs are clearly visible.22Figure 1.9: Microscopy techniques. The images show an algae lament imaged using (A) bright eldmicroscopy (B) phase contrast microscopy and (C) dierential interference contrast microscopy. Taken fromhttp://www.microscopyu.com/articles/formulas/specimencontrast.html.Oblique illumination microscopyFor oblique illumination, the light source is moved o-axis, so that some light is reected from the sidesof the sample, producing a 3-D eect. This is another simple, easy-to-interpret technique that is cheap toimplement. It has the advantage of showing parts of the sample that are not seen in normal bright eldmicroscopy. Unfortunately, the contrast is again typically low since biological samples tend not to absorbmuch light, although again this can be ameliorated with stains. Fig. 1.11 shows a polyp under varying degreesof oblique illumination. As the angle increases, the sample develops more contrast and a 3-D appearancesince light is being reected o the edges of the specimen.Dark eld microscopyDark eld microscopy removes most of the transmitted light by physically blocking the central portion of theillumination source. Light that enters the eyepiece has been scattered or refracted by the sample (Fig. 1.12(left)). This is an extreme version of oblique illumination where none of the direct illumination is allowedto form the image. This is another simple, easy-to-interpret technique that is cheap to implement. It hasthe advantage of producing sharp images without the use of stains. Unfortunately, light levels tend to below, requiring a strong light source to see the image. The images produced by dark eld microscopy can bequite striking as the edges of the specimen appear bright on a dark background as seen in the image of amosquito wing in Fig. 1.12 (right).Dispersion stainingDispersion staining takes advantage of dispersion of materials to produce colour images without staining.Recall that dispersion is the relationship between the index of refraction of a material and wavelength.Suppose two substances in contact have the same dispersion at some wavelengths, but not at others. Thensome wavelengths will pass through the contact region undisturbed (since there is no change in the index ofrefraction) while others will be refracted or reected. We can collect the refracted and reected light (whichwill be of a certain colour) to create an image. This technique can be used to identify substances, and iscommonly used to identify asbestos (Fig. 1.13). Unfortunately, it is of limited usefulness since the sampleunder study must be immersed in a uid whose index of refraction is the same at all but a few wavelengths23Figure 1.10: Staining in bright eld microscopy. The images show two species of plant mites, Orthoty-deus kochi (top row) and Raphignathus gracilis (bottom row), under bright eld microscopy either unstained(left column) or stained with a potassium-based dye (right column). Taken from Faraji and Bakker (2008),Eur. J. Entomol. 105: 793-795.24Figure 1.11: Oblique illumination microscopy. The images show a Cnidaria (coe-lenterate) polyp Obelia in the hydroid generation stage, under (a) bright eld illumi-nation and under oblique illumination at 5 (b), 10 (c), and 15 (d). Taken fromhttp://micro.magnet.fsu.edu/optics/olympusmicd/anatomy/micdoblique.html.25Figure 1.12: Dark eld microscopy. (left) Light rays used in dark eld microscopy are reected orrefracted from the sample. (right) A dark eld image of a mosquito wing.and these few wavelengths should be in the visible spectrum. Pictures of objects imaged with dispersionstaining are available at http://www.microlabnw.com/index/gallery-ds.html.Phase contrastPhase contrast microscopy uses the phase shift introduced by the change in index of refraction of the sampleto produce an interference pattern with a reference beam. In phase shift microscopy, the light from theillumination source is split into a reference beam and a beam that goes through the sample. The beamsare later combined to produce an interference pattern, which converts dierences in phase to dierencesin amplitude, which are detectable by our eyes. This method highlights dierences in densities which is aparticular advantage for biological samples, allowing otherwise transparent structures within cells to be seen.Phase contrast is, however, susceptible to a particular aberration which causes halos around dark spots thatmay obscure small details. Fig. 1.14, shows glial cells in both the bright eld and phase contrast modes. Inthe bright eld mode, the cells appear semi-transparent with only the membrane and nucleus being somewhatvisible. In phase contrast mode, the same eld of view reveals more structural detail; cellular attachmentsare visible and the internal structure is visible in much more detail. Images taken using phase contrastmicroscopy can be viewed at http://www.olympusmicro.com/primer/techniques/phasegallery.html.Dierential interference contrastDierential interference contrast (DIC) microscopy splits a single light beam into two beams polarized at90 (Fig. 1.15 (top)). The beams sample two nearby parts of the sample and may develop dierences inphase due to dierent optical properties at the sites. When the light is recombined, an interference pattern isproduced showing regions of contrast. Similar to the phase contrast method, DIC has much better contrastand resolution than the bright eld mode. Fig. 1.15 (bottom) shows several specimens in both DIC andphase contrast. The DIC images do not show the halo eect that is seen in phase contrast images, so thatthe edges are sharper in the DIC images. However, the phase contrast method sometimes shows featuresthat DIC does not, such as the cell nuclei in murine kidney tissue (center column).26Figure 1.13: Dispersion staining microscopy. An image of asbestos using dispersion staining. Theindex of refraction of asbestos matches that of the surrounding uid at all wavelengths except blue. Bluelight is reected/refracted from the asbestos strands making them visible. Copyright: 2006 EnvironmentalMicrobiology Laboratory, Inc.Figure 1.14: Phase contrast microscopy. Human glial cells shown in(a) bright eld and (b) phase contrast microscopy. Image taken fromhttp://www.microscopyu.com/articles/phasecontrast/phasemicroscopy.html.27Figure 1.15: Dierential interference contrast microscopy. (top) Paths of the lightrays used for DIC. (bottom) Images showing human buccal mucosa epithelial (cheek) cell (leftcolumn), murine kidney tissue (center column), and Obelia polypoid annulated stem peris-arc (right column) shown in DIC (upper row) and phase contrast (lower row). Taken fromhttp://www.olympusmicro.com/primer/techniques/dic/dicphasecomparison.html.28Figure 1.16: Interference reection contrast microscopy. (left) The paths of light rays used to produceIR images. (right) A mouse embryo imaged using IR microscopy. Image taken from Verschueren (1985), J.Cell Sci. 75:279-301.Interference reection contrastThe specimen is placed under a piece of glass. A beam of light is directed through the objective towards theglass covering the specimen. When the cell membrane is close to the glass surface, there will be destructiveinterference with the original beam, resulting in a dark spot. When it is further from the glass, there willbe less interference, resulting in lighter spots (Fig. 1.16 (left)). This technique is particularly useful formeasuring cell migration as it can detect parts of the cell moving towards or away from the glass surface.The mouse embryo cell shown in Fig. 1.16 (right) shows the cell moving to the left there are dark regionson the edge of the extension indicating that this part of the cell is close to the glass.Fluorescence microscopyWhen some substances are excited with one wavelength of light, they emit light at another wavelength.Fluorescence microscopy uses light with the excitation wavelength to illuminate the specimen and collectsthe emitted light to create the image (Fig. 1.17). Since excitation and imaging are done at dierent wave-lengths, there is no interference between the illumination source and collected light. Unfortunately, theemitted light levels can be low and tend to degrade over time through photobleaching. While most bi-ological samples do not naturally uoresce, uorescent labels have been developed that can be used toimage biological samples and biological processes. There is a gallery of uorescence microscopy images athttp://www.microscopyu.com/galleries/fluorescence/.Confocal microscopyInstead of using a wide eld of illumination, confocal microscopes use a single point of light to illuminate asmall section of the sample. A full image is made by scanning the point source over the entire sample. Thisgreatly improves resolution by eliminating out-of-focus light (Fig. 1.18) and allows for 3-D reconstruction ofa specimen.29Figure 1.17: Fluorescence microscopy. (left) The paths of light rays used to produce uo-rescence images. (right) An endothelial cell stained with uorescent labels; red labels the mi-tochondria, green labels the F-actine cytoskeleton and blue labels the nucleus. Taken fromhttp://www.meade.com/dsi/gallery/molecular probes.html.Figure 1.18: Confocal microscopy. Images of human medulla (left column), rabbit mus-cle bers (center column) and sunower pollen grain (left column) imaged using standard wideeld uorescence microscopy (top row) and confocal microscopy (bottom row). Image taken fromhttp://www.olympusfluoview.com/theory/confocalintro.html.301.5 ProblemsProblem 1: (a) The equationC(x, t)dx = 14Dt__ C(, 0)e(x)2/4Dtd_dxdescribes the concentration of particles that are subject to diusion and drift, where D is thediusion coecient. What is the impulse response of this system? (b) The equationv(R) =_ a2i4o2vix21rdxdescribes the exterior potential due to current ow along an axon, where i and o are theinterior and exterior conductivities. What is the impulse response of this system?Problem 2: The identity lter is dened as the lter that takes an input to itself Id[x(t)] = x(t). Show thatthe impulse response of the identity lter is (t).Problem 3: Show that the convolution function is commutative interchanging the order of variables givesthe same results.Problem 4: Calculate the convolution, f(x) g(x), of f(x) = g(x) = e|x|.Problem 5: Find the output of a system if the input is x(t) = 2u(t 10) and the impulse response ish(t) = sin(2t)u(t).Problem 6: A linear time invariant system has the following impulse response,h(t) = 2eatu(t)Use convolution to nd the response y(t) to the following inputx(t) = u(t) u(t 4)Sketch y(t) for the case when a = 1.Problem 7: Calculate the cosine and sine transforms of ebx, where b is a positive integer.Problem 8: (a) Derive the expression for the Fourier transform of df(x,y)dx in terms of the Fourier transformof f(x, y). (b) Based on the result in (a), what is the Fourier transform of d2f(x,y)dx2 ?Problem 9: Show that the Fourier transform of(x) =_ n=cneinx, 0 < x < 20, otherwise,does not vanish outside of any nite interval.Problem 10: Dene two rectangular functions asf1(x, y) =_ 1, if 0 x X; 0 y Y0, otherwise,andf2(x, y) =_ 1, if |x| X/2; |y| Y/20, otherwise.Find the Fourier transforms of the two functions and explain any dierences between the twofunctions in the spatial and frequency domains.31Problem 11: (a) Use the convolution integral to calculate the convolution g(t) of the functionh(t t

) =_ 0, t < t

1e(tt

)/, t > t

,where is a constant, withf(t) =_ 1, 0 < t < T,0, otherwise.(b) Calculate the Fourier transform of g(t), h(t t

), and f(t) from part (a) and show thatthey obey the convolution theorem.Problem 12: Use the convolution integral to calculate g(x) fromh(x x

) = aa2+ (x x

)2and f(x) = cos(kx). This can be interpreted as a spatial frequency lter. Hint:_ cos(kx)dxx2+b2 = bekb,_ sin(kx)dxx2+b2 = 0.Problem 13: What are the two dimensional images whose Fourier transforms are shown below?32Problem 14: Calculate the two-dimensional Fourier transform of the functionf(x, y) =_ 1, a/2 < x < a/2, b/2 < y < b/2,0, otherwise.Problem 15: Calculate the two-dimensional Fourier transform of the functionf(x, y) = sech_xa_sech_yb_.You may need the following relationship_ 0sech (uz) cos(wz)dz = 2usech_w2u_.Problem 16: Calculate the two-dimensional Fourier transform of the functionf(x, y) =_ 1,_x2+y2< a,0,_x2+y2> a.Hint: convert to polar coordinates in both x-y and kx-ky and use the fact thatJ0(u) = 12_ 20cos(ucos(x))dx, J1(u) =_ uJ0(u)du,where J0 and J1 are Bessel functions of order zero and one.Problem 17: Find the Fourier transform of the point-spread function for magnication (Eq. (1.16)).Problem 18: Show that if the Fourier transform of a function f(x) is g(k), then the Fourier transform off(x a) is eika(k).Problem 19: Use the convolution theorem to show that the sum of the squares of the Fourier coecientsof the image is equal to the sum of squares of the Fourier coecients of the object times thesquare of the modulation transfer function.Problem 20: How does magnication change the spatial frequencies in going from object to image?Problem 21: This problem shows how increasing the detail in an image introduces high-frequency compo-nents. Find the Fourier transform of the two functionsf1(x) =___0, x < 0,1, 0 < x < 1,0, x > 1f2(x) =___0, x < 0,_3/2, 0 < x < 1/3,0, 1/3 < x > 2/3,_3/2, 2/3 < x < 1,0, x > 1.Plot a(kx) = [C2(kx) +S2(kx)1/2for each function and compare the features of each plot.Problem 22: Suppose that the object is a point at the origin, so that f(x, y) = (x)(y). Find the projectionF(x) and the transform functions C(kx, 0) and S(kx, 0). Use the results to reconstruct theimage using the Fourier technique.33Problem 23: Derive the coordinate transformations Eqs. (1.27) & (1.28).Problem 24: An object is described by the functionf(x, y) = e(x2+y2)/b2.(a) Find the Fourier transform C(kx, ky) and S(kx, ky). (b) Find the projection F(, x

). Thentake the one-dimensional Fourier transform of F(, x

) usingC(, k) =_ F(, x

) cos(kx

)dx

S(, k) =_ F(, x

) sin(kx

)dx

,where k = (k2x +k2y)1/2. Your answer should be the same as in (a). You will need the followingintegrals:_ eax2dx =_a_ eax2cos(bx)dx =_aeb2/4a_ eax2sin(bx)dx = 0Problem 25: A system performs the operation g(x, y) = f(x, y)f(x1, y). Derive the MTF of the system.Problem 26: Assume you have just measured the following projection functionF(, x

) =be(x

a cos )2/b2.Find f(x, y).Problem 27: Assume you have just measured the following projection functionF(, x

) = a2 ex2/a2_1 + cos2_2x2a2 1__.Find f(x, y).Problem 28: Suppose an object is a point at the origin, f(x, y) = (x)(y). The projection is also a point:F(, x

) = (x

). Calculate the back projection fb(x, y) (without ltering). You will need thefollowing property of functions(g(u)) =

i(u ui)|dg/du|u=uiwhere the ui are the points such that g(ui) = 0. Note that the back projection does not recoverthe original image.Problem 29: Let f(x, y) = (x x0)(y y0). (a) Calculate F(, x

). You will need the following propertiesof the function:_ (b z)(z a)dz = (b a)(az) = (z)|a| .(b) Use the function F(, x

) you found in part (a) to calculate the back projection fb(x, y). (c)Show that fb(x, y) is equivalent to the convolution of f(x, y) with the function 1/_(x x

)2+ (y y

)2.34Problem 30: A 22 image has the pixel values_ 7 24 0_.Compute the projections of the image at 0 and 90. Compute a reconstruction of the imageusing the simple back projection method.Problem 31: If f(x, y) is dimensionless, determine the units of F(, x

) and fb(x, y). Do the image and theback projection have the same units?Problem 32: Calculate the projections, F(, x

) for the following objects: (a) f(x, y) = 1/[(xa)2+y2+b2]and (b) f(x, y) = x/(x2+y2)2.Problem 33: Consider the object f(x, y) = a/_a2x2y2for |x| < a, and 0 otherwise. (a) Calculatethe projection F(, x

). (b) Use the projection from part (b) to calculate the back projectionfb(x, y). (c) Compare the object and back projection.Problem 34: An image is given by the 33 matrix__1 2 12 3 21 2 1__,and is 0 outside this matrix. What is the output if the image is processed with the followinglters: (a)h = 19__1 1 11 1 11 1 1__,(b)h = 15__0 1 01 1 10 1 0__,Problem 35: The 55 imagef(m, n) =__0 0 0 0 00 10 10 10 00 10 10 10 00 10 10 10 00 0 0 0 0__,is processed by two systems in cascade. The rst system produces the output g1(m, n) =f(m, n)f(m1, n). The second system produces the output g2(m, n) = g1(m, n)g1(m, n1).Compute the images g1 and g2. Does the sequence of application of the two operators aectthe result?Problem 36: What are the impulse response and transfer function of the 33 mean lter?Problem 37: The output of a lter at (m, n) is dened as the average of the four immediate neighbours of(m, n); the pixel at (m, n) is itself not used. Derive the MTF of the lter.35Chapter 2Optical tweezersOptical tweezers are an experimental technique that is used to study and manipulate small objects. Since thistechnique is non-invasive, it has found wide application in manipulating and studying dielectric spheres, livingcells, DNA, bacteria, and metallic particles. Optical tweezers are routinely applied to measure elasticity,force, torsion and position of a trapped object. The technique relies on forces caused by the electromagneticwaves in a laser beam. We begin with an examination of the forces on a particle trapeed in a laser beam.2.1 Forces2.1.1 Radiation forceWhen light hits an object, it causes a force. If we consider light as a particle, then the cause of this forceis quite straightforward the light transfers some of its momentum to the object. This radiation pressurealways acts in the direction of propagation of the light. We can write the radiation pressure as a function ofthe Poynting vector, S, which is dened asS = 10EB, (2.1)where E is the electric eld, B is the magnetic eld of the electromagnetic wave, and 0 is the permeabilityof free space (a constant). The Poynting vector gives the energy ux (energy per unit area per unit time) ofan electromagnetic wave and it can be related to radiation pressureP = nc < S >, (2.2)where n is the refractive index of the object, c is the speed oight, and < S > is the average value of thePoynting vector. The force due to radiation pressure on the surface of an object isF = nc_ _ (SinSout) dA (2.3)where we integrate the dierence in pressure due to incoming and outgoing radiation over the surface area.36Example: Force on a mirrorCalculate the force due to radiation pressure from a 60 W light bulb on a mirror.Solution: For light reecting o a mirror, Sin = Sout. The force then isF = 2nc_ _ Sin dA= ncP,where the integral represents the total power, P. Putting in the index of refraction for glass (1.5),the speed of light (3 108m/s), and the total power incident on the mirror (60 W), we ndF = 2 1.53 10860 W= 4 107N.We see that for macroscopic objects, the force due to radiation pressure is very small.2.1.2 Gradient forceLight incident on a particle creates a dielectric response, due to the polarizability of the constituent atoms orions. We can think of this force as a consequence of the refraction of light as it enters the particle. Refractionat the surface changes the direction of propagation and thus the momentum of the light. This momentumis transferred to the particle causing a force. For an atom or ion in a monochromatic, linearly polarized,continuous light eld, E, the time-averaged induced dipole moment isp = E, (2.4)where is the relative complex polarizability of the particle to the surrounding medium. The interaction ofthe induced dipole with the electric eld of the light creates an electrostatic potentialU = p E. (2.5)Thus in a light eld with a spatially varying intensity, there is a gradient forceF = U= p E (2.6)= E E= 2E2.We see that the gradient force is linearly dependent on the spatial gradient of the intensity of the electro-magnetic wave. Note that the force acts towards the point of highest intensity, which for a focused laserbeam will be the focal point.Since this is a restoring force, we can model this force as a Hookean spring with stiness, k,F = kx. (2.7)This linear regime is seen in Fig. 2.1 which shows the gradient force as a function of the distance from thecenter of the trap. Although not immediately obvious from these simple expressions, the trap stiness isgreatest when the particle to be trapped is the same size as the beam waist; as particle size decreases, therestoring force decreases rapidly, but decreases only modestly when the particle size increases.37Figure 2.1: Gradient force in a laser beam. Near the focus of the laser beam, a small particle experiencesa linear restoring force.Example: Gradient forceIn a typical optical tweezers setup, the trap stiness can be varied from 1060.1 pN/nm. Whatis the range of restoring force if a particle is displaced 0.1 m?Solution: The gradient force is given byF = kxand substituting the trap stiness and displacement, we nd F = 10410 pN.2.2 Sphere in a focused beam of lightSuppose we focus a beam of light to a very narrow point. If we place a sphere within the beam, light will exertseveral forces on it as shown in Fig. 2.2. The forces F1 and F2 represent the gradient (or refractive) forcesdue to the edges of the laser beam. In reality, there will be forces pointing in all directions between F1 andF2 due to all the light rays in the beam. It is clear, however, that the horizontal components (perpendicularto the laser) of these forces will cancel, leaving only the vertical components (parallel to the laser). Therewill also be a force due to reection from the spheres surface. Recall that this force acts in the directionof propagation of the light. If the sphere is displaced towards the light source, the radiation force helps topush the sphere back towards the focal point. If the sphere is displaced in the other direction, however, theradiation force helps to push the sphere out of the trap. The net result is to shift the center of the trapslightly away from the focal point of the laser.For ideal optical tweezers, we want to minimize the reective force and maximize the refractive force.The greatest contributions to the refractive force come from the edges of the beam. Increasing the numericalaperture of the lens will increase the force by increasing the angle of incidenceNA = nsin , (2.8)38Figure 2.2: Forces on a sphere in a focused beam. Refraction forces tend to pull the beam toward thefocus, while reection forces tend to push the bead away from the light source.while this does not increase the reective force (it is independent of angle of incidence), but it comes at thecost of creating a shallower trap (Fig. 2.3).2.3 Determining strength of the trap2.3.1 Constant velocity calibrationIn a typical experiment, a bead, cell or part of a cell sits in a viscuous uid. If we trap the bead and thenmove it through the medium at a constant velocity, we can determine the strength of the trap. At a constantvelocity, there is no net force on the object. The force exerted by the trap must equal the viscuous dragforce. The viscuous is force is given byF = 6rv (2.9)where is the viscosity, r is the radius of the sphere, and v is the velocity of the object, all of which can bemeasured. This means we can determine the force exerted by the trap.2.3.2 Brownian motion calibrationThe trap can also be calibrated by observing Brownian motion. The trapped object will experience smallrandom forces from Brownian motion. Since there is no net acceleration, all forces must balance and we canwriteFvis +Ftrap = FBrown (2.10)dxdt +kx = F(t),39Figure 2.3: Eect of numerical aperture on a laser trap. Increasing the NA increases the trap strength,but decreases the trap depth and increases the power lost.where is the coecient of viscous drag and we have assumed an ideal trap with no reective force. SinceBrownian motion is due to thermal uctuations, it has an average value of zero. This also means that thepower spectrum (Fourier transform) is constant since there is no dominant frequency in the motion.|F(f)|2= 4kBT, (2.11)where kB is the Boltzmann constant, and T is the temperature.Recall that, if the Fourier transform of g(x) is G(f), then the transform of its derivative is 2ifG(f).We can transform the entire dierential equation:2ifX(f) +kX(f) = F(f) (2.12)Squaring and solving for X(f):|X(f)|2= kBT(fc +f)2 (2.13)where fc = k/2. The power spectrum of the motion of a trapped particle is constant at low frequenciesand it bends at fc (Fig. 2.4). By measuring the power spectrum of a trapped particle, we can determine fcwhich can be used to determine k and therefore Ftrap.40Figure 2.4: Power spectrum of a trapped particle. The power spectrum of a particle trapped in a laserbeam can be used to determine the strength of the trap.412.4 ProblemsProblem 1: Estimate the radiation pressure due to a 75 W bulb at a distance of 8.0 cm from the center ofthe bulb. Estimate the force exerted on your ngertip if you place it at this point.Problem 2: Suppose that a laser pointer is rated at 3 mW. If the pointer produces a 2 mm diameter spoton a screen, what is the radiation pressure exerted on the screen?Problem 3: Laser light can be focused (at best) to a spot with a radius r equal to its wavelength . Supposethat a 1.0 W beam of green laser light ( = 5 107m) is used to form such a spot and thata cylindrical particle of about that size (let the radius and height equal r) is illuminated bythe laser. Estimate the acceleration of the particle, if its density equals that of water and itabsorbs the radiation.Problem 4: Consider the total internal reection of a plane wave with wavelength = 800 nm incident atan angle = 70 from the normal of a glass-air interface. The plane wave is incident fromthe glass-side. The normal of the interface is parallel to the gravitational axis and the air-sideis pointing to the bottom. A tiny glass particle is trapped on the air-side in the evanescenteld generated by the totally internally reected plane wave. Calculate the minimum requiredintensity I of the plane wave to prevent the glass particle from falling down ( = 1.25). Thespecic density of glass is = 2.2 103kg/m3and the particle diameter is d0 = 100 nm. Whathappens if the particle size is increased?Problem 5: A spherical glass particle in water is trapped at the focus of a laser beam with = 800 nm.The polarizability of the particle is = 30V0 w + 2wwhere V0 is the volume of the particle, and the dielectric constants of glass and water are = 2.25 and w = 1.76, respectively. (a) Show that for small transverse displacements (x)from the focus the force is proportional to x. Determine the spring constan