* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Note
Diffraction grating wikipedia , lookup
Ultrafast laser spectroscopy wikipedia , lookup
Silicon photonics wikipedia , lookup
Optical rogue waves wikipedia , lookup
Optical tweezers wikipedia , lookup
Optical flat wikipedia , lookup
Optical coherence tomography wikipedia , lookup
Magnetic circular dichroism wikipedia , lookup
Fiber-optic communication wikipedia , lookup
Atmospheric optics wikipedia , lookup
Surface plasmon resonance microscopy wikipedia , lookup
Anti-reflective coating wikipedia , lookup
Interferometry wikipedia , lookup
Photon scanning microscopy wikipedia , lookup
Ray tracing (graphics) wikipedia , lookup
Fourier optics wikipedia , lookup
Birefringence wikipedia , lookup
Thomas Young (scientist) wikipedia , lookup
Harold Hopkins (physicist) wikipedia , lookup
Diffraction wikipedia , lookup
Nonlinear optics wikipedia , lookup
Nonimaging optics wikipedia , lookup
Physics 408 Optics J Kiefer © 2011 Table of Contents TABLE OF CONTENTS ............................................................................................................. 2 I. A. GEOMETRICAL OPTICS .................................................................................................. 4 Huygens.................................................................................................................................. 4 1. Huygens’ Principle.............................................................................................................. 4 2. “Laws” of Reflection and Refraction .................................................................................. 4 B. 1. 2. Fermat .................................................................................................................................... 7 Path Length ......................................................................................................................... 7 Fermat’s Principle ............................................................................................................... 8 1. 2. Refracting Surfaces ............................................................................................................... 9 Spherical surfaces ............................................................................................................... 9 Imagery ............................................................................................................................. 12 C. D. Optical Systems ................................................................................................................... 14 1. Ray Tracing ....................................................................................................................... 14 2. Matrices............................................................................................................................. 15 E. 1. 2. Mirrors ................................................................................................................................. 16 Planar Mirrors ................................................................................................................... 16 Curved Mirrors.................................................................................................................. 17 1. 2. Prisms ................................................................................................................................... 18 Dispersing Prisms ............................................................................................................. 18 Reflecting Prisms .............................................................................................................. 20 F. G. Fiber Optics ......................................................................................................................... 21 1. Large Diameter Fiber Optics ............................................................................................ 21 2. Sending Signals on Glass Optical Fiber............................................................................ 23 II. WAVE MECHANICS ........................................................................................................ 27 A. 1. 2. 3. B. Wave Equation .................................................................................................................... 27 Wave Profile ..................................................................................................................... 27 Harmonic Waves ............................................................................................................... 27 Three Dimensional Wave Forms ...................................................................................... 28 Fourier Analysis .................................................................................................................. 30 1. Superposition of Waves .................................................................................................... 30 2. Waves Having Different Frequencies ............................................................................... 32 3. Fourier Sums ..................................................................................................................... 33 2 C. Fraunhoffer Diffraction...................................................................................................... 37 1. Far Field (Fraunhoffer) Diffraction .................................................................................. 37 2. Appertures ......................................................................................................................... 38 D. 1. 2. Fresnel Diffraction .............................................................................................................. 40 Fresnel Zones .................................................................................................................... 40 Fresnel Lens ...................................................................................................................... 42 1. 2. 3. Interference ......................................................................................................................... 43 Coherent Interference........................................................................................................ 43 Wave Front Splitting ......................................................................................................... 43 Amplitude Splitting........................................................................................................... 44 E. F. Vector Wave Theory ........................................................................................................... 44 G. Polarization .......................................................................................................................... 44 III. A. 1. 2. 3. EXTRA............................................................................................................................. 46 Absorption and Emission of Radiation ............................................................................. 46 Transition Coefficients...................................................................................................... 46 Transition Rates ................................................................................................................ 46 Population Inversion ......................................................................................................... 47 B. Fourier Optics ..................................................................................................................... 47 C. Lenses, Thick and Thin ...................................................................................................... 47 D. Aberration ........................................................................................................................... 48 3 I. Geometrical Optics A. Huygens 1. Huygens’ Principle “Every point on a primary wave front serves as the source of spherical secondary wavelets. The wavelets advance with speed and frequency equal to those of the primary wave. The primary front at some later time is the envelope of the wavelets.” a. Wave front A wave is a disturbance that propagates from one place to another. The disturbance is described by a wave function, r, t . The wave front is the surface of uniform phase, meaning that all points on the surface share the same disturbance. In the case of light waves, the “disturbance” is the electromagnetic field, so the electric field has the same magnitude and direction at every point on the wave front. Say that at t = 0 the wave front has a shape, or profile, . Then at a later time, t, the wave front shape is , possibly different from . b. Wavelets Spreading circles or spheres are easier to deal with than advancing arbitrary shapes. So, we imagine that a real wave results from the superposition of spherical waves emanating from points forming the wave front. The radii of the secondary wavelets are vt , where v is the speed of light in the medium. If the medium is inhomogeneous, then some wavelets expand faster than others, and the wave profile changes. c. Huygens’ construction The new wave front is constructed by drawing circles of radius vt , each centered on a point on the primary wave front. The speed, v, need not be the same for all the wavelets. 2. “Laws” of Reflection and Refraction We know that, in general, when light impinges on an interface between two media, a portion of the energy is reflected and a portion is transmitted across the boundary. 4 a. Reflection & transmission at normal incidence Assume: i) both media are homogeneous and isotropic and ii) both media respond linearly. Thus, i t , but because vi > vt, i t . On the other hand, r i and vr = vi, so r i . b. Snell’s “Law” 1i strikes point A at t = 0. Secondary wavelets propagate from point A to radii vi t and vt t , at which moment.. . i2 strikes point A and 1i arrives at point B. By t 2t , secondary wavelets have spread from point A to 2tvi and 2tvt , and from point B to tvi and tvt . i2 has reached point B and 3i hits at point A. Clean up the picture by removing the wavelets and look at the angles. 5 Measured from the interface, i = angle of incidence r = angle of reflection t = angle of refraction t = time required for the whole of wave front, i , to “impact” on the interface. BD AC and sin r . But BD vi t and AC vi t , so AD AD sin i sin r . This is the “Law” of Reflection. We see by inspection that sin i By a similar token, sin i BD AE and sin t . But BD vi t and AE vt t , so AD AD sin i BD vi or vt sin i vi sin t . This is the “Law” of Refraction, or Snell’s “Law”. sin t AC vt c. Index of refraction c , where c is the speed of light in vacuum. v ni sin i nt sin t Define a light ray as a line or arrow indicating the direction of energy flow. If the medium is isotropic, the light ray will be orthogonal (or normal) to the wave front. Define the index of refraction, n Define corresponding points as the intersections of a specified light ray with successive wave fronts. In terms of light rays rather than wave fronts the picture of reflection and refraction looks like this: 6 In summation, i) all the rays lie in the same plane (by symmetry), ii) i r , and iii) ni sin i nt sin t . B. Fermat 1. Path Length a. Spatial path length The spatial path length is the total distance traversed by a light wave in propagating from point S to point P. b. Optical path length The elapsed time in traversing from S to P is t s1 s2 s3 1 s1n1 s2 n2 s3n3 . The optical v1 v2 v3 c n path length (OPL) is the “effective” distance OPL si ni . If the medium is inhomogeneous, i 1 P then OPL n( s)ds ; sort of like work with a changing force. S 7 This could be, say, the atmosphere, whose density varies with altitude. 2. Fermat’s Principle a. Least action In going from point S to point P, a light wave traverses a path for which the optical path length is a minimum. b. Snell’s “Law” via Fermat’s Principle Between S and P OPL ni h 2 x 2 nt b 2 (a x) 2 . We wish to minimize this with respect to x. [Notice that within each medium, if homogeneous, the paths are straight lines—the shortest distance between two points.] OPL 0 x a x 1 0 x ni nt h2 x2 b 2 (a x) 2 ni sin i nt sin t 0 ni sin i nt sin t c. Principle of Reversibility 8 A light ray going from P to S follows the same path as a ray going from S to P. This follows from symmetry arguments. C. Refracting Surfaces The purpose of optical systems is to manipulate wave fronts, or to take rays emanating from one location and cause them to converge at another. 1. Spherical surfaces a. Huygens’ picture The edges of the plane wave front overtake the center, which has already entered the second medium and slowed. b. Fermat’s picture 9 According to Fermat’s Principle, the path SAP of the light ray is a minimum. OPL n1 o n2 i We want to rewrite this in terms of the polar coordinates of the point A, using C as the origin and measuring the polar angle clockwise from the optical axis, SP. By the law of cosines, 2o R 2 ( s o R ) 2 2 Rs o R cos 2i R 2 ( si R ) 2 2 R( si R) cos OPL n1 R 2 ( s o R) 2 2 Rs o R cos 1 2 n2 R 2 ( si R) 2 2 R( si R) cos 1 2 OPL 0 1 n1 2 R ( s o R ) 2 2 Rs o R cos 2 2 R s o R sin 2 1 n2 2 R ( si R) 2 2 R( si R) cos 2 2 R si R sin 0 2 n1 n2 R( s o R) R( si R) 0 o i n1 n n n Rs o 1 R 2 2 Rs i 2 R 2 0 o o i i n1 n2 n2 si n1 s o o i iR oR This specifies the relationship among the incident ray, R, n1, n2, and the refracted (transmitted) ray. 10 c. Gaussian optics If is small, then the incident ray is called a paraxial ray. In that case, we can make simplifying approximations by expanding the cosine and sine in power series. 2 4 6 cos 1 2! 4! 6! 3 5 7 sin 1 3! 5! 7! If is very small, the first order approximation gives cos 1 . Then o so and i si and n1 n2 n2 si n1 so 1 n2 n1 . s o si Rs i Rs o R That treatment of lenses and light rays that utilizes the first order approximation is called Gaussian Optics. It is valid for rays arriving at shallow angles to the optical axis. d. Focal points There is a focal length on either side of the refracting surface. n1 R. n2 n1 n2 The image or second focal length is obtained by setting so ; then f i si R. n2 n1 By convention, R > 0 if C is to the right of V. The object or first focal length is obtained by setting si . Then f o so e. Some terminology i) A light ray originates at the point S, traverses an optical system, arrives at the point P. S & P are called conjugate points, since their roles could be reversed. ii) In a perfect optical system, every point in object space is perfectly imaged in image space. iii) A real optical system is diffraction limited, since a real system collects only a segment of a wave front. iv) In geometrical optics, we ignore diffraction, as if 0 . 11 2. Imagery a. focal planes Off-axis light rays incident normally to the interface will be undeviated and pass through C. Light rays close to such a ray of either side will be deviated to converge with the undeviated ray. The ray bundle converges at a focal point lying on a curved focal surface, . For paraxial rays, only a small section of the focal surface is involved, centered on the optical axis. We pretend the surface is flat and call it the focal plane. It is an approximation of Gaussian Optics that parallel rays entering an optical system are focused on a plane surface. b. Imaging extended objects Each point on an extended object emits light rays in all directions. Those that enter the optical system are imaged in image space. Tracing only two rays from each point is enough to locate the image, though often three are used. Defn. Optical center of a refracting surface or optical system. Consider parallel surfaces of opposite sides of a lens. A ray passing through those parallel surfaces will be undeviated in direction, though off set. Where such a ray crosses the optical axis is called the optical center. 12 Ray tracing rules of thumb i) paraxial rays passing through the optical center are undeviated. ii) Incident rays passing through fo emerge parallel to the optical axis. iii) incident rays parallel to the optical axis emerges to pass through fi. By convention, the extent of the object (image) is (+) above the optical axis, (-) below. c. Magnification Transverse or lateral magnification M T By similar triangles, yi si f s s i . Now, because of the sign conventions, M T i . yo yo f so Longitudinal magnification M L We have yi . yo dxi . dxo y f s f . Cross multiply. i i so f yo f f 2 si f so f xi xo xi f2 xo dxi f2 2 M T2 dxo xo M L M T2 13 d. Stops and pupils Nice sharp images are formed on a flat focal plane only for paraxial rays. Rays entering the optical system far off-axis will degrade the desired flat image, as they will not focus on the focal plane, but on the curved focal surface. Thus, while in general the size of a lens determines the brightness of an image, it may be desirable to prevent some of the light rays from entering the system for the sake of image quality. See cheap astronomical telescopes. This is a relative question depending on the size of the object and its distance from the lens. Aperture stop: an opening or diaphragm that restricts the amount of light reaching the image. It may be the rim of the lens itself. Field stop: an element limiting the angular breadth of the image. It determines the field of view. Pupils are images of the aperture stop, as seen from different points. Entrance pupil: the image of the aperture stop as seen from an object point on the optical axis, looking through any elements preceding the stop. If there are none, then the entrance pupil coincides with the aperture stop. Exit pupil: the image of the aperture stop as seen from an axial image point, looking through any elements following the stop. If there are none, then the exit pupil coincides with the aperture stop. Chief ray: a ray from an off-axis point on the object that passes through the midpoint of the aperture stop. Marginal ray: a ray directed from an axial object point to the rim of the entrance pupil. Relative aperture: the ratio of the aperture diameter and the focal length of the optical system. f-number: reciprocal of the relative aperture. f /# D. Optical Systems 1. Ray Tracing 14 f . D a. Refraction equation Rewrite Snell’s “Law” in terms of vectors defining the directions of the wave propagation and of the surface normal. ni kˆi uˆ nt kˆt uˆ Next, write the angles of incidence and of refraction in terms of the optical axis rather than the interface normal. Further, we’ll consider paraxial rays. ni1 i1 nt1 t1 ni1 i1 1 nt1 t1 1 y y ni1 i1 1 nt1 t1 1 R1 R1 n ni1 nt1 t1 ni1 i1 y1 t1 R 1 This is the refraction equation for the first surface. The transfer equation takes the ray through the lens to the second surface. y 2 y1 tan t1 t1 d 21 y 2 y1 d 21 t1 y1 d 21 i 2 b. Process i) given i 1 , y1 at the first interface, ii) compute t1 ; iii) given d 21 V1V2 , solve for y2. iv) repeat surface to surface through the system. 2. Matrices a. Propagation 15 Each light ray is represented by a column matrix . y The refraction equation can be rewritten as a matrix multiplication. nt1 t1 1 D1 ni1 i1 y y 0 1 i1 t1 rt1 R1 ri1 R1 is the refraction matrix for the first surface. Similarly, the transfer equation becomes 1 0 n n ri 2 i 2 i 2 d 21 1 t1 t1 T21rt1 y t1 y i 2 nt1 where d21 is the approximate distance between the two surface vertices. b. System matrix The chain of events is strung together: 1 0 1 D n ni 2 i 2 1 D21 d 1 t1 t1 21 1 1 n 1 yt1 0 yi 2 0 t1 for two surfaces. Any number of such matrices can be multiplied together, in the correct (reverse) order, so that a many-surface optical system can be traced quickly on a computer. The system matrix, A, is just the product of all the individual matrices. For the two-surface system, A = R2T21R1 so that rt2 = Ari1. E. Mirrors 1. Planar Mirrors a. Ray tracing By similar triangles, si so . Also, the image is virtual, so we use a slightly different sign convention: (+) to the left of the mirror(that is, the side the object is on), (-) to the right(the side opposite the object). 16 b. Extended objects A planar mirror produces a virtual image which is perverted, that is, switched left-for-right.. More technically stated, the right-handed coordinate system in object space is transformed into a left-handed system in image space. This is not simply a rotation. Contrast this result with the refracting surface, which produces a real image that is inverted both left-to-right and up-to-down. This called reversion, and is equivalent to a rotation by 180o about the optical axis. The virtual image produced by a refracting surface is erect—it is not inverted. c. Magnification For a planar mirror, of course M T 1 . 2. Curved Mirrors a. Parabolic mirrors We’ll use Fermat’s Principle to figure out what shape of mirror will focus parallel incident light rays to a point. The condition is that the optical path lengths of parallel rays striking arbitrary points on the mirror (A1 and A2) and passing through the focal point F must be the same. OPL W1 A1 A1F W2 A2 A2 F Establish a base line D1D2 , parallel to the wave fronts (perpendicular to the incident rays) such that W1 A1 A1D1 W2 A2 A2 D2 Subtract to eliminate the WAs. A1F A1D1 A2 F A2 D2 We find that the curvature should be such that the quantity AF AD constant . This defines a paraboloid, with focus F and directrix D1D2 . b. Spherical mirrors Sadly, a parabolic surface is difficult to make. Spherical mirrors are easier to create. In the paraxial region, a sphere of radius R = 2f will be very close to the paraboloid surface of focal 17 length f. That is, the shape of a small region of the parabola near the axis is almost the same as spherical. This can be demonstrated geometrically. The radius CA bisects the angle SAP, so it divides SP into pieces proportional to SA and to SC CP AP . I.e., . Notice, too, that SC so R and CP R si . SA AP Now, so and si are both positive, but R < 0, so SC so R and CP R si . Near the axis, SA soand PA si , whence so R R si so si 1 1 2 1 so si R f This is the Mirror Formula—be careful of the +/- signs. The ray tracing is just like a refracting surface, although of course the rays do not go through the mirror. F. Prisms 1. Dispersing Prisms a. Angle of deviation 18 i1 t1 t 2 i 2 i1 t 2 From Snell’s “Law” t 2 sin 1 n sin i 2 t 2 sin 1 n sin t1 t 2 sin 1nsin cos t1 sin t1 cos 1 t 2 sin 1 n(sin cos t1 sin i1 cos ) n n 2 sin 2 i1 1 t 2 sin 1 n sin sin cos i1 2 n n t 2 sin 1 sin n 2 sin 2 i1 1 2 sin i1 cos i1 sin 1 sin n 2 sin 2 i1 So, if and n are fixed, then f (i1 ) . 1 2 sin i1 cos b. Angle of minimum deviation 0 , but that’s too messy. Instead we’ll go back to the beginning. We could carry out i1 i1 t 2 d d 1 t2 0 di1 di1 d t 2 1 di1 To obtain dt 2 , differentiate Snell’s “Law” at each interface. d i1 d sin i1 n sin t1 cos i1di1 n cos t1dt1 d n sin i 2 sin t 2 n cos i 2 di 2 cos t 2 dt 2 From d t1 i 2 0 , dt1 di 2 . So we can divide the two equations to obtain 19 2 2 cos i1 cos t1 cos cos t2 i2 2 2 1 sin i1 n sin 2 i1 1 sin 2 t 2 n 2 sin 2 t 2 Now, n 1, so this can be true only if i1 t 2 . Therefore, the minimum deviation occurs when i1 t 2 and t1 i 2 . These mean that the light ray is traversing the prism parallel to its base. sin m 2 . Setting m and solving for n, we obtain an expression: n sin 2 This provides a way of measuring the index of refraction, by observing m with a spectrometer. 2. Reflecting Prisms a. Non-dispersing prism Let a light ray pass through a prism in such a way as to experience one internal rerflection. The angle of deviation is independent of wavelength and index of refraction, so there is no dispersion. Proof: Evidently, 180o . On the other hand, 2 360o 180o 360o 2 180o 360o 2(90o i1 ) 2i1 20 If the triangle isn’t isosceles, i1 t 2 . b. Other Another example is an erecting prism, used in binoculars to un-invert the image. No disoerson, again. G. Fiber Optics 1. Large Diameter Fiber Optics Compared to the wavelengths of IR, visible and UV light, the diameters of optical fibers are large. Light pipes are used to reflect and reorient light rays. The communication fibers are used to transmit signals in the form of light pulses. a. Total internal reflection Consider the situation of a light ray approaching an interface between two media. Snell’s “Law” relates the incident angle to the transmitted angle. n sin i t sin t ni There is a critical angle, c , of incidence at which the transmitted angle is 90o. n n sin c t sin 90o t ni ni If nt < ni, there exists a critical incident angle for which the wave is entirely reflected; no energy is transmitted past the interface. This phenomenon is called total internal reflection. A light ray traveling within a glass cylinder, and striking the side of the cylinder will undergo total internal reflection the incident angle is greater than or equal to the critical angle. The critical angle is determined by the indices of refraction of the cylinder, nf, and the medium 21 surrounding the cylinder, na. The light ray will bounce from one side of the cylinder to the other again and again as it propagates along the cylinder. The path length traversed by the light ray between successive reflections is h D . The sin t D . If L is the total length of the cos t L cos t cylinder, then the total number of reflections undergone by the light ray is N . Finally, D the total path length for the light ray is distance along the cylinder between reflections is g sin 2 i L 1 n 2f L cos t D cos t Nh L sin i D sin t sin t nf 1 2 L n 2f sin 2 i sin i 1 2 . The elapsed time for the light ray to travel the length of the cylinder is 2 2 n f n f Ln f sin i 2 . t vt c c sin i 1 b. Acceptance angle Now the light ray was injected into the fiber through one end. We might ask, at what angle must the light ray enter the end of the cylinder in order to be totally internally reflected along the length of the cylinder? 22 sin c na sin 90o t nf cos t na nf 1 sin 1 2 t sin max 2 na nf 1 2 n f na2 na 1 2 The numerical aperture is defined to be NA na sin max n 2f na2 1 2 . If the light ray enters the end of the cylinder at an angle max , then it will hit the inside surface of the cylinder at c and be totally internally reflected. Commonly, optical fibers are clad with a material other than air so as to decrease the NA. c. Dispersion Rays entering the end of the cylinder at different angles travel different total distances along the cylinder, according to how often the ray bounces. Over a long cylinder, dispersion occurs. The Ln f minimum transit time is tmin for a light ray traveling along the axis of the cylinder. The c longest time occurs for a ray incident at the critical angle, c . n L f 2 na Ln f tmax vf vf cna Ln f n f 1 . c na A pulse of light injected into the cylinder would be comprised of several light rays, at a range of angles, so it would spread in space as it propagated along the cylinder. Input pulses must be far enough apart that they don’t begin to overlap and become indistinguishable by the time they reach the far end of the cylinder. In optical fibers, a narrow NA is desirable to reduce t . The intermodal dispersion over the length of the cylinder is t tmax tmin 2. Sending Signals on Glass Optical Fiber Optical pulses are generated by a laser and injected into the end of a glass fiber. The light waves travel through a material medium and therefore the pulses change as they travel along the fiber. a. Attenuation 23 Let z be distance along the optical fiber. The power of the light wave will be attenuated along dP P . At the end of a fiber of length L, Pout (t ) Pin (t )e L . There are several the fiber: dz loss mechanisms. Here are two of them. i) the molecules (silica) of the glass as well as impurity atoms/molecules absorb EM waves at different frequencies. There is a spectral loss profile. ii) Rayliegh scattering occurs due to microscopic fluctuations in the index of refraction. The scattering depends on the wavelength and the size of the fluctuation regions. In other words, the glass is not perfectly uniform. b. Chromatic dispersion The index of refraction is a function of wavelength. An optical pulse starts out as a (near) square wave, which is comprised of a sum of harmonic components over a range of wavelengths. Each component travels at a slightly different speed in the fiber—hence the pulse spreads out as it propagates. Also, the laser bandwidth is not infinitesimally narrow, itself. At some point separate pulses may begin to overlap. c. Bit rates The optical signals are composed of a continuous sequence or string of on/off pulses. The length of each bit is the bit period, TB. The reciprocal of the bit period is the bit rate, B. The shorter TB, the greater is B. However, as TB decreases, the frequency distribution in the pulses widens, which leads to greater dispersion as the pulse travels down the fiber. 24 Multiple channels of bit strings can be sent into the fiber, at slightly differing carrier wavelengths, e.g., 1.5, 1.55, 1.45 m . But, if the channel spacing is too close, the dispersion will cause pulses from different channels to overlap, so the channels are not completely separate anymore. Information on one channel is mixed with information on another. d. Engineering the fiber In designing and manufacturing the optical fiber, the goal is to minimize these effects, and others. Minimize attenuation: i) select carrier wavelengths to which the silica is most transparent—IR, ii) remove impurities, iii) re-amplify periodically along the fiber. Minimize dispersion: i) single mode fibers—if the fiber core radius is less than about 3 times the wavelength, then only one mode can propagate along the fiber, rather than many, ii) regenerate pulses every few kilometers to correct chromatic dispersion (expensive), iii) fiddle with the composition of the fiber to reduce the dependence of n on wavelength, iv) give initial pulses a shape that does not disperse—solitons, v) graded index fibers. The engineering aims are to maximize distance between amplifications and to maximize bit rate. e. Fiber amplifiers Here’s something interesting. Ordinary amplifiers require the reading of the optical signal, conversion to electrical, amplification, re-conversion to optical pulses. This means a relative bulking and power-hungry gadget, with several possible points of malfunction. There is a fiber amplifier that does not convert the optical pulses to/from electric signals, and can be placed in-line with the fiber. A section of fiber itself is the amplifier. 25 The incoming signal stimulates coherent emission in the erbium-doped fiber, thereby exactly duplicating the incoming signal with increased amplitude. 26 II. Wave Mechanics Light is an E-M wave, so we’ll review the mathematics of wave motion. A. Wave Equation 1. Wave Profile Define a function, ( x, t ) ( x vt) which describes the shape and motion of a wave pulse traveling along the x-axis. a. General description We will need the slope and curvature of the function, ( x, t ) . Let x x vt . x x x x x 2 2 x 2 x x x 2 on the other hand, x x vt , therefore, t 1 x x t v t 2 2 2 2 2 v v v v v v x x t x x x t 2 x 2 x 2 b. Solution The general solution to the wave equation is c1 f x vt c2 g x vt , where the f or g is the wave profile. The solution is a superposition of wave pulses traveling in the +x and –x directions. 2. Harmonic Waves a. Definitions x, t A cos( x vt ) or A sin( x vt ) Wavelength, , such that x, t x , t . Amplitude, A = maximum disturbance. Propagation number, k, such that k 2 . Period, , such that x, t x, t . 1 Frequency, . Note that the wave speed, c . Angular frequency, 2 . 1 Wave number, . Notice the distinction we are making now between k and . b. Phase velocity The argument of the cosine is called the phase. The phase is a function of x and of t. ( x, t ) kx t 27 If we select a fixed value of phase, o , then as the wave travels, this point propagates along the x x-axis with a velocity . t o o t c t k k c. Complex representation Another solution to the wave equation is x, t Aei t kx Aei . This is more convenient to use than the sine or cosine, but it’s complex. The actual, physical wave is Re Aei . 3. Three Dimensional Wave Forms a. Wave equation in three dimensions 2 2 2 1 2 2 x 2 y 2 z 2 c 2 t 2 x r x, y , z k k k x , k y , k z Specific forms of are labeled according to the shapes of the constant-phase surfaces. b. Plane Waves The surface of constant phase is a plane, perpendicular to the wave vector, k . The equation for such a plane is r ro k 0 where ro is a point in the plane. Write it out 28 r ro k x xo k x y yo k y z z o k z 0 k r xkx yk y zk z xo k x y o k y z o k z a constant The phase of a plane wave, then, is r k t c. Spherical Waves The surfaces of constant phase are spherical shells, propagating outward from or inward toward a central point. How do we write the profile of such a wave? We require that r r . We need to write the wave equation in terms of r. r x r x 2 r r 2 r r r 2 2 r r x x r x r r x x x 2 2 2 r 2 r 2 r x 2 x 2 r x 2 Since r x 2 y 2 z 2 , r x 2 r 1 x 1 and x 2 x r r x x r x 1 x2 1 . r r 2 2 x 2 1 x 2 2 2 1 , and similarly for and . x 2 r r 2 r r 2 r z 2 y 2 Add ‘em up 2 2 2 2 2 2 2 2 2 x y z 2 1 x y z 3 x 2 y 2 z 2 r r r r 2 r r r r r 2 Therefore, 2 2 1 2 r r 2 r r r r 2 1 2 r 1 2 2 2 2 r r 2 c t 2 2 r 1 r 2 r 2 c t 2 We see that the wave function (the solution to the wave equation) has the form f r, t rr, t . For outgoing waves, f r, t f r ct , while for converging waves, g r, t g r ct . For harmonic waves, the profile is A A r , t cos k r ct or r , t e ik r ct . r r A surface of constant phase is defined by k r kr a constant . However, the amplitude of A is , so the profile is not constant. r 29 d. Cylindrical waves x cos y sin zz 1 1 2 We require that r . The wave equation is with solutions c 2 t 2 A A ik r ct . r , t cos k r ct or r , t e B. Fourier Analysis 1. Superposition of Waves As the wave equation is linear, the sum of two solutions of the wave equation is also a solution. Query: what is the net disturbance of the medium. a. General case Consider two waves, with the same v and . E1 Eo1 sin t 1 E2 Eo 2 sin t 2 where 1 k1x and 2 k2 x . The net disturbance is E = E1 + E2. E Eo1 sin t 1 Eo 2 sin t 2 E Eo1 cos 1 Eo 2 cos 2 sin t Eo1 sin 1 Eo 2 sin 2 cos t In general this is a funny, varying wave profile. b. Flux density The energy transported by the wave is related to the amplitude squared. 2 2 2 2 E E1 E2 Eo1 Eo 2 2 Eo1Eo 2 cos 2 1 The interference term, 2Eo1Eo 2 cos2 1 , depends on the phase difference 2 1 . c. Optical path difference The optical path difference is a phase difference arising from difference in path length traverses by two waves. 30 kx1 kx2 2 x1 x2 2 nx1 x2 ko o If constant , then the waves are said to be coherent. d. Phasor addition E1 Eo1 sin t 1 Graphically, we can represent this expression as a vector of length Eo1 rotating about the origin, thusly: Two such vectors are added on the phasor diagram by placing the tail of the second vector on the head of the first. We can imagine that both are rotating with the same angular frequency, or with different frequencies. e. Standing waves In this case, we have two waaves of the same frequency, but travelling in opposite directions. E1 Eo1 sin kx t and E2 Eo 2 sin kx t with Eo1 Eo 2 The net disturbance is E1 + E2. 31 E E1 E 2 Eo1 sin kx t sin kx t 2kx 2 t Eo1 2 sin cos 2 2 2 Eo1 sin kx cos t This expression is not of the form of a travelling wave, so it is a stationary of standing wave, not moving through space. If we fix x, then E A cos t . 3 At x 0, , , , sin kx 0 and E 0 for all t. These are the nodes. 2 2 3 At x , , sin kx 1 and E 2 E o1 for all t. These are the antinodes. 4 4 Every point (x) in the medium executes simple harmonic motion, of differing but constant amplitude. 2. Waves Having Different Frequencies a. Beats In this case, we superimpose two waves having different frequencies, but travelling in the same direction. E1 Eo1 cosk1 x 1t and E2 Eo1 cosk 2 x 2 t The sum is E E1 E 2 E o1 cosk1 x 1t cosk 2 x 2 t 1 1 2 E o1 cos k1 k 2 x 1 2 t cos k1 k 2 x 1 2 t 2 2 2 E o1 cosk m x m t cos k x t The energy transports is proportional to 32 Eo2 4 Eo21 cos 2 k m x m t 2 Eo21 1 cos2k m x 2 m t . The beat frequency is the frequency of the modulation E o2 , namely 1 2 m 2 1 2 1 2 .. 2 b. Group velocity Look at the modulation envelope Eo x ,t 2Eo1 cosk m x m t . This is a wave, with a speed d t x m 1 2 vg km k1 k 2 k dk x t dv dv 0 , then v g v . But in dispersive media, Since kv, v g v k . If indeed dk dk c kc dn c k dn v 1 , so that v g 2 kv k . [ nk is the index of refraction.] n n dk nk n dk 3. Fourier Sums a. Anharmonic waves If E1 and E2 have different frequencies and different amplitudes, in general the resultant wave form won’t be a harmonic wave, but will still be periodic. Fourier’s Theorem: A periodic function, f(x), having a spatial period , can be expanded as a series of harmonic functions whose wavelengths are integral submultiples if . I.e., A 2 f x o Am cos mkx Bm sin mkx, where k , and Am and Bm are expansion 2 m 1 m 1 coefficients. Now the sine and cosine functions have the orthogonality property. cos akx cos bkxdx 2 ab 0 sin akx sin bkxdx 2 ab 0 cos kx sin kxdx 0 0 We use this property to evaluate the expansion coefficients for a known f(x). 0 m 1 0 m 1 m 1 cos nkxf ( x )dx 0 cos nkxAm cos mkxdx Bm 0 Am nm An 33 Similarly, Bn sin nkxf ( x )dx . For n = 0, 0 For a travelling wave, f kx t f ( x )dx 0 0 Ao 2 dx 0 0 Ao f ( x )dx . 2 0 Ao Am coskx t Bm sin kx t . 2 m 1 m 1 b. Non-periodic waves What can we do if the waveform is aperiodic? We like the Fourier expansion so much that we still want to use it. 34 c. Band widths 35 36 C. Fraunhoffer Diffraction Diffraction occurs whenever a portion of a wave front is obstructed and a change in amplitude or phase occurs in that portion. 1. Far Field (Fraunhoffer) Diffraction a. Far field condition Fraunhoffer diffraction obtains when the incident and outgoing waves are (nearly) plane waves, as opposed to spherical. a2 The pattern observed at the screen, , changes only in size as the screen is moved back and forth. R b. Line source Consider a linear array of point sources of spherical wavelets. If R>>D, then the short source elements, yi , where EL EL for all ri R EL is the amplitude at P of the spherical wavelet ri emanating from yi . The phases of all the spherical wavelets are not the same, though. The contribution of each wavelet to the total electric field at P is approximately E dE L sin( t kr)dy . R The radius, r, is a function of y. We’ll have to obtain it from a Maclauren series. y2 r R y sin cos 2 2R Now, if R>>D, then the third term is negligible for all . In which case r is a linear function of y. Integration over the line source yields the total field at the point P. 37 D E E L R sin t krdy D D E EL R L L L sin t k R y sin dy D L Integration over y (t and are parameters, since P is fixed) yields kD D sin sin E 2 sin t kR E L R kD sin 2 2 The irradiance is <E > (average over time), so 2 kD sin 2 2 2 2 sin 1 EL D 1 EL D sin 2 1 EL D 2 2 2 I () sinc I (0)sinc 2 2 2 R 2 kD 2 R 2 R sin 2 Keep in mind that is measured in the plane containing the line source. When D>> , the D sin becomes very large, namely irradiance decreases rapidly with increasing , since D . This means that side fringes are very small, only the central maximum of I(0) remains. The result is, for a coherent line source with length D>> , the irradiance is approximately that of a point source radiating a circular wave in the xz-plane at 0 . 2. Appertures a. Single slit Essentially, we extend the development from the linear source to the second dimension. The slit is composed of many line sources of length D. Each of these acts like a point source lined up on the z-axis, radiating outward in the xz-plane. The calculation is the same as before, except now we integrate over z from 2 b b to , and is 2 2 sin kb measured in the xz-plane. I () I (o) where now sin . However, we are 2 assuming that b is small, so is not large. Therefore the side fringes cannot be neglected. We find that the I( ) maxima by setting dI 0. d 38 dI 2 sin ( cos sin ) I (0) d 3 This equals zero when sin 0 , or when ,2,3, What does this look like? For monochromatic light: The minima occur at m ; the maxima m 0 occur when cos sin 0 or when tan . Notice that the side fringes depend on wavelength. Thus a multi-chromatic light source gives side fringes at different locations for each colour. b. Rectangular Now, do not assume b is small. We have a rectangular opening a x b divided into square area elements, ds. Wavelets propagate outward from each area element. (X,Y,Z) are the coordinates of the field point, P. Let EA be the source strength per unit area, constant over the aperture. The contribution to the amplitude at P from ds is E dE A ei t kr ds , r where r X 2 Y y Z z In the Fraunhoffer regime, all the EA E A . In the phase, r R 2 2 1 2 . y 2 z 2 2Yy Zz 2 r X Y 2Yy y Z 2Zz z R 1 . R2 R2 2 2 y z 0 , leaving The limits on the y and z are the edges of the aperture, so R2 2 2 2 2 2 1 1 2 2Yy Zz 2 r R 1 . This goes in for r in the phase. Now we have “only” to integrate over R2 the aperture. 1 E E A R E b a 2 2 e Yy Zz i t k R 1 R 2 dzdy b a 2 2 a 2 E A i t kR e ae R 2 39 Zz ik R b 2 dz e b 2 ik Yy R dy bY ik bY ik e 2R e 2R kbY sin Now, e dy b , where . Similarly, b 2ikbY 2 R b 2 2R a aZ ik aZ ik Zz 2 2R 2R ik kaZ e e sin R , where . Therefore, we obtain e dy a a a 2ikaZ 2R 2 2R E sin sin E A ei t kR ab R 2 The irradiance at the point P is E . b 2 ik Yy R 2 sin sin I (Y , Z ) I (0) As before, the I(o) is the irradiance at the center of the diffraction pattern, (Y.Z) = (0,0). 2 D. Fresnel Diffraction Diffraction of spherical wavefronts. 1. Fresnel Zones a. Obliquity In Huygens’ construction, every point a primary wavefront acts as a point source of secondary wavelets. In order to provide the forward bias to these wavelets (alternatively, eliminate the unphysical backward wavelet propagation), we introduce a quantity called the obliquity or the 1 inclination factor, K () 1 cos ; K (0) 1; K () 0 . 2 b. Half-period zones Spherical wave fronts are emitted from the point source, S. In cross section the picture is: 40 The particular wave front is emitted at t = 0. The radius of the wave front at t = t’ the radius is E . At , the disturbance is E o cost k , where Eo is the source strength. Secondary wavelets from the primary wave at travel toward the point P. We define a differential area element, ds, such that all wavelets from ds travel the same distance, r, to reach P. These wavelets also are all coherent, in phase with each other. Thus, at time t, wavelets from ds arrive at P with phase t k r . Eo . Taking into account the obliquity, we have the contribution to the amplitude at P from ds as E dE K A cos t k r ds r We need ds in terms of r. ds d2sin The source strength per unit area on ds is E A Q Law of cosines r 2 2 ro 2 ro cos 2 Take d and solve for d . d dr 2 ro sin d rdr d ro sin 2r ds 22 sin rdr 2rdr ro sin ro So, from one zone, we integrate over ds. E r dE K r 1 r E A 2 cos t k r dr , ro r1 1 . where, r ro and r 1 ro 2 2 EA 1 r E 2K sin t k r r1 ro k 2 Now, k , so we get 41 E K E A 1 ) sin t k ( ro ) sin t k ( ro ro 2 2 E 1 1 2 K E A sin t k ( ro ). ro k 1 .] 2 Notice that E has opposite sign from E 1 , but K K 1 . Therefore wavelets from successive zones don’t quite cancel each other. [Used an identity and the fact that sin Next, we add up the contributions from the half-period zones, up to the mth zone at , at the backside of the wave front. E E1 E2 E3 Em The terms alternate signs, so make that explicit (let m be odd) E E1 E2 E3 E4 Em E E E E E E E E E1 2 E2 3 3 E4 5 m 2 Em 1 m m 2 2 2 2 2 2 2 E E 1 Since the zones are narrow, and K is slowly varying, E 1 . Therefore the 2 E E quantities in the parentheses are small, whence, E 1 m . But, for Em, K () 0 , so Em=0. 2 2 E E 1 !! 2 2. Fresnel Lens a. Zone plates E1 , where E1 that due 2 to the first half-period zone. This occurs because the odd and even indexed zones almost (not quite) cancelled each other. Suppose we made a device that blocked alternate zones—we would avoid the cancellations and obtain a greater optical disturbance. The problem is to create such an item. Let’s block the odd zones. Am outer edge of the mth zone. An unobstructed spherical wave gives a total disturbance at the point P of The zones are defined such that m rm o ro m . On the other hand, 2 42 m R 2 m 2 o 1 2 Rm2 o 2o rm Rm2 ro2 2 ro 1 o Rm2 2ro Rm2 R2 ro m o ro m 2o 2ro 2 1 1 m o ro Rm2 Rm is the radius of the mth zone on the zone plate. If the odd zones are blocked, the total disturbance at P is (let’s say including 40 zones) E E E1 E3 E5 E39 20 E1 1 2 Notice that the zones are about wide, so 40 zones is not very far from the OP line. 2 b. Focal length With a point source a distance o from a zone plate, waves converge at the point P, a distance ro from the zone plate. 1 1 m o ro Rm2 Define a primary focal length: f1 Rm2 . m E. Interference 1. Coherent Interference a. Coherent sources b. Phase difference between sources 2. Wave Front Splitting a. Young’s experiment 43 b. Fresnel double mirror c. Fresnel biprism d. Lloyd’s prism e. Multiple slits 3. Amplitude Splitting a. Double beam interference b. Near-normal interference—Newton’s Rings c. Michelson interferometer d. Fabry-Perot interferometer F. Vector Wave Theory G. Polarization 44 45 III. Extra A. Absorption and Emission of Radiation 1. Transition Coefficients a. Absorption b. Spontaneous emission c. Induced or stimulated emission 2. Transition Rates a. Level populations b. Transitions c. Dynamic equilibrium d. Induced transition rate 46 3. Population Inversion a. Equilibrium level population b. Inverting the population c. Metastable states B. Fourier Optics C. Lenses, Thick and Thin c. Effective focal length We would like to use the same Gaussian equation involving so , si , and f to express the net effect of a system of refracting surfaces. This can be done if the distances are redefined. Consider for instance a thick lens—two spherical refracting surfaces in sequence. We define principle planes and principle points H1 and H2, where the optical axis passes through the principle planes. The distance from the object to H1 is so. The distance from the image to H2 is si. The distance between the principle points is d. With these definitions, and where we call f the effective focal length of the combination of two surfaces. The principle planes are located relative to the vertices of the refracting surfaces. 47 d. Application to thin lenses For a thin lens,. So for one thin lens, A=R1R2. For two thin lenses, separated by a gap, d, . By inspection, the effective focal length is. We notice, too, that the locations of the principle points, H1 & H2, relative to the vertices of the two lenses, are D. Aberration 48