Optics Learning by Computing with Examples using MATLAB - K.D. Moller, Notas de estudo de Engenharia Elétrica

Baixe Optics Learning by Computing with Examples using MATLAB - K.D. Moller e outras Notas de estudo em PDF para Engenharia Elétrica, somente na Docsity! Karl Dieter Môller Second Edition EE A RE MathCad, MATLAB, Mathematica, and Maple FAR Optics To colleagues, staff, and students of the New Jersey Institute of Technology, Newark, New Jersey Preface The book is for readers who want to use model computational files for fast learning of the basics of optics. In the Second Edition, Matlab, Mathematica and Maples files have been added to the Mathcad files on the CD of the First Edition. The applications, given at the end of files to suggest different points of view on the subject, are extended to home work problems and are also on the CD of the Second Edition. While the book is suited well for self learning, it was written over several years for a one semester course in optics for juniors and seniors in science and engineering. The applications provide a simulated laboratory where students can learn by exploration and discovery instead of passive absorption. The text covers all the standard topics of a traditional optics course, includ- ing: geometrical optics and aberration, interference and diffraction, coherence, Maxwell’s equations, wave guides and propagating modes, blackbody radiation, atomic emission and lasers, optical properties of materials, Fourier transforms and FT spectroscopy, image formation, and holography. It contains step by step derivations of all basic formulas in geometrical and wave optics. The basic text is supplemented by over 170 Mathcad, Matlab, Mathematica and Maple files, each suggesting programs to solve a particular problem, and each linked to a topic in or application of optics. The computer files are dynamic, allowing the reader to see instantly the effects of changing parameters in the equations. Students are thus encouraged to ask “what . . . if” questions to asses the physical implications of the formulas. To integrate the files into the text, applications are listed connecting the formulas and the corresponding computer file, and problems for all 11 chapters are on the CD. The availability of the numerical Fourier transform makes possible an intro- duction to the wave theory of imaging, spatial filtering, holography and Fourier transform spectroscopy. vii viii PREFACE The book is written for the study of particular projects but can easily be adapted to a variation of related studies. The three fold arrangement of text, applications and files makes the book suitable for “self-learning” by scientists and engineers who would like to refresh their knowledge of optics. All files are printed out and are available on a CD, (Mathcad 7) (Mathcad 2000) (Matlab 6.5) (Mathematica 4.1) (Maple 9.5) and may well serve as starting points to find solutions to more complex problems as experienced by engineers in their applications. The book can be used in optical laboratories with faculty-student interaction. The files may be changed and extended to study the assigned projects, and the student may be required to hand in printouts of all assigned applications and summarize what he has been learned. I would like to thank Oren Sternberg and Assaf Sternberg for the translation of the files into Matlab, Mathematica and Maples, Prof. Ken Chin and Prof. Haim Grebel of New Jersey Institute of Technology for continuous support, and my wife for always keeping me in good spirit. Newark, New Jersey K.D. Möller CONTENTS xi 3.2.3 On Axis Observation for Circular Stop . . . . . . . . . . . . . . 135 3.3 Fresnel Diffraction, Far Field Approximation, and Fraunhofer Observation . . . . . . . . . . . . . . . . . . . . . . . . . . 136 3.3.1 Small Angle Approximation in Cartesian Coordinates . . . . . . 137 3.3.2 Fresnel, Far Field, and Fraunhofer Diffraction . . . . . . . . . . 138 3.4 Far Field and Fraunhofer Diffraction . . . . . . . . . . . . . . . . . . . 139 3.4.1 Diffraction on a Slit . . . . . . . . . . . . . . . . . . . . . . . . 140 3.4.2 Diffraction on a Slit and Fourier Transformation . . . . . . . . . 144 3.4.3 Rectangular Aperture . . . . . . . . . . . . . . . . . . . . . . . 145 3.4.4 Circular Aperture . . . . . . . . . . . . . . . . . . . . . . . . . 148 3.4.5 Gratings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 3.4.6 Resolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 3.5 Babinet’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 3.6 Apertures in Random Arrangement . . . . . . . . . . . . . . . . . . . . 169 3.7 Fresnel Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 3.7.1 Coordinates for Diffraction on a Slit and Fresnels Integrals . . . . . . . . . . . . . . . . . . . . . . . . . 172 3.7.2 Fresnel Diffraction on a Slit . . . . . . . . . . . . . . . . . . . . 173 3.7.3 Fresnel Diffraction on an Edge . . . . . . . . . . . . . . . . . . 175 A3.1.1 Step Grating . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 A3.2.1 Cornu’s Spiral . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 A3.2.2 Babinet’s Principle and Cornu’s Spiral . . . . . . . . . . . . . . 182 4 Coherence 185 4.1 Spatial Coherence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 4.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 4.1.2 Two Source Points . . . . . . . . . . . . . . . . . . . . . . . . 185 4.1.3 Coherence Condition . . . . . . . . . . . . . . . . . . . . . . . 189 4.1.4 Extended Source . . . . . . . . . . . . . . . . . . . . . . . . . 190 4.1.5 Visibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 4.1.6 Michelson Stellar Interferometer . . . . . . . . . . . . . . . . . 197 4.2 Temporal Coherence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 4.2.1 Wavetrains and Quasimonochromatic Light . . . . . . . . . . . 200 4.2.2 Superposition of Wavetrains . . . . . . . . . . . . . . . . . . . 201 4.2.3 Length of Wavetrains . . . . . . . . . . . . . . . . . . . . . . . 202 A4.1.1 Fourier Tranform Spectometer and Blackbody Radiation . . . . 203 5 Maxwell’s Theory 205 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 5.2 Harmonic Plane Waves and the Superposition Principle . . . . . . . . . 206 5.2.1 Plane Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 5.2.2 The Superposition Principle . . . . . . . . . . . . . . . . . . . . 208 5.3 Differentiation Operation . . . . . . . . . . . . . . . . . . . . . . . . . 208 xii CONTENTS 5.3.1 Differentiation “Time” ∂/∂t . . . . . . . . . . . . . . . . . . . 208 5.3.2 Differentiation “Space” ∇  i∂/∂x + j∂/∂y + k∂/∂z . . . . . . 208 5.4 Poynting Vector in Vacuum . . . . . . . . . . . . . . . . . . . . . . . . 209 5.5 Electromagnetic Waves in an Isotropic Nonconducting Medium . . . . . 210 5.6 Fresnel’s Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 5.6.1 Electrical Field Vectors in the Plane of Incidence (Parallel Case) . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 5.6.2 Electrical Field Vector Perpendicular to the Plane of Incidence (Perpendicular Case) . . . . . . . . . . . . . . . . . . . . . . . 214 5.6.3 Fresnel’s Formulas Depending on the Angle of Incidence . . . . . . . . . . . . . . . . . . . . . . . . 215 5.6.4 Light Incident on a Denser Medium, n1 < n2, and the Brewster Angle . . . . . . . . . . . . . . . . . . . . . . . . . . 216 5.6.5 Light Incident on a Less Dense Medium, n1 > n2, Brewster and Critical Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 5.6.6 Reflected and Transmitted Intensities . . . . . . . . . . . . . . . 222 5.6.7 Total Reflection and Evanescent Wave . . . . . . . . . . . . . . 228 5.7 Polarized Light . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 5.7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 5.7.2 Ordinary and Extraordinary Indices of Refraction . . . . . . . . 231 5.7.3 Phase Difference Between Waves Moving in the Direction of or Perpendicular to the Optical Axis . . . . . . . . . . . . . . . . . 232 5.7.4 Half-Wave Plate, Phase Shift of π . . . . . . . . . . . . . . . . 233 5.7.5 Quarter Wave Plate, Phase Shift π/2 . . . . . . . . . . . . . . . 235 5.7.6 Crossed Polarizers . . . . . . . . . . . . . . . . . . . . . . . . . 238 5.7.7 General Phase Shift . . . . . . . . . . . . . . . . . . . . . . . . 240 A5.1.1 Wave Equation Obtained from Maxwell’s Equation . . . . . . . 242 A5.1.2 The Operations ∇ and ∇2 . . . . . . . . . . . . . . . . . . . . . 243 A5.2.1 Rotation of the Coordinate System as a Principal Axis Transformation and Equivalence to the Solution of the Eigenvalue Problem . . . . . . . . . . . . . . . . . . . . . . . . 243 A5.3.1 Phase Difference Between Internally Reflected Components . . 244 A5.4.1 Jones Vectors and Jones Matrices . . . . . . . . . . . . . . . . . 244 A5.4.2 Jones Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 A5.4.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 6 Maxwell II. Modes and Mode Propagation 249 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 6.2 Stratified Media . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252 6.2.1 Two Interfaces at Distance d . . . . . . . . . . . . . . . . . . . 253 6.2.2 Plate of Thickness d  (λ/2n2) . . . . . . . . . . . . . . . . . . 255 6.2.3 Plate of Thickness d and Index n2 . . . . . . . . . . . . . . . . 256 6.2.4 Antireflection Coating . . . . . . . . . . . . . . . . . . . . . . . 256 CONTENTS xiii 6.2.5 Multiple Layer Filters with Alternating High and Low Refractive Index . . . . . . . . . . . . . . . . . . . . . . . . . . 258 6.3 Guided Waves by Total Internal Reflection Through a Planar Waveguide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 6.3.1 Traveling Waves . . . . . . . . . . . . . . . . . . . . . . . . . . 259 6.3.2 Restrictive Conditions for Mode Propagation . . . . . . . . . . 261 6.3.3 Phase Condition for Mode Formation . . . . . . . . . . . . . . . 262 6.3.4 (TE) Modes or s-Polarization . . . . . . . . . . . . . . . . . . . 262 6.3.5 (TM) Modes or p-Polarization . . . . . . . . . . . . . . . . . . 265 6.4 Fiber Optics Waveguides . . . . . . . . . . . . . . . . . . . . . . . . . 266 6.4.1 Modes in a Dielectric Waveguide . . . . . . . . . . . . . . . . . 266 A6.1.1 Boundary Value Method Applied to TE Modes of Plane Plate Waveguide . . . . . . . . . . . . . . . . . . . . . . . . . . 270 7 Blackbody Radiation, Atomic Emission, and Lasers 273 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 7.2 Blackbody Radiaton . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274 7.2.1 The Rayleigh–Jeans Law . . . . . . . . . . . . . . . . . . . . . 274 7.2.2 Planck’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . 275 7.2.3 Stefan–Boltzmann Law . . . . . . . . . . . . . . . . . . . . . . 277 7.2.4 Wien’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278 7.2.5 Files of Planck’s, Stefan–Boltzmann’s, and Wien’s Laws. Radiance, Area, and Solid Angle . . . . . . . . . . . . . . . . . 279 7.3 Atomic Emission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 7.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 7.3.2 Bohr’s Model and the One Electron Atom . . . . . . . . . . . . 282 7.3.3 Many Electron Atoms . . . . . . . . . . . . . . . . . . . . . . . 282 7.4 Bandwidth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 7.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 7.4.2 Classical Model, Lorentzian Line Shape, and Homogeneous Broadening . . . . . . . . . . . . . . . . . . . . 286 7.4.3 Natural Emission Line Width, Quantum Mechanical Model . . . 289 7.4.4 Doppler Broadening (Inhomogeneous) . . . . . . . . . . . . . . 289 7.5 Lasers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 7.5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 7.5.2 Population Inversion . . . . . . . . . . . . . . . . . . . . . . . 292 7.5.3 Stimulated Emission, Spontaneous Emission, and the Amplification Factor . . . . . . . . . . . . . . . . . . . . . . . 293 7.5.4 The Fabry–Perot Cavity, Losses, and Threshold Condition . . . 294 7.5.5 Simplified Example of a Three-Level Laser . . . . . . . . . . . 296 7.6 Confocal Cavity, Gaussian Beam, and Modes . . . . . . . . . . . . . . . 297 7.6.1 Paraxial Wave Equation and Beam Parameters . . . . . . . . . . 297 7.6.2 Fundamental Mode in Confocal Cavity . . . . . . . . . . . . . . 299 xvi CONTENTS Appendix A About Graphs and Matrices in Mathcad 435 Appendix B Formulas 439 References 443 Index 445 1C H A P T E R Geometrical Optics 1.1 INTRODUCTION Geometrical optics uses light rays to describe image formation by spherical surfaces, lenses, mirrors, and optical instruments. Let us consider the real image of a real object, produced by a positive thin lens. Cones of light are assumed to diverge from each object point to the lens. There the cones of light are transformed into converging beams traveling to the corresponding real image points. We develop a very simple method for a geometrical construction of the image, using just two rays among the object, the image, and the lens. We decompose the object into object points and draw a line from each object point through the center of the lens. A formula is developed to give the distance of the image point, when the distance of the object point and the focal length of the lens are known. We assume that the line from object to image point makes only small angles with the axis of the system. This approximation is called the paraxial theory. Assuming that the object and image points are in a medium with refractive index 1 and that the lens has the focal length f, the simple mathematical formula 1 −x0 + 1 xi  1 f (1.1) gives the image position xi when the object position x0 and the focal length are known. Formulas of this type can be developed for spherical surfaces, thin and thick lenses, and spherical mirrors, and one may call this approach the thin lens model. For the description of the imaging process, we use the following laws. 1. Light propagates in straight lines. 2. The law of refraction, n1 sin θ1  n2 sin θ2. (1.2) 1 2 1. GEOMETRICAL OPTICS The light travels through the medium of refractive index n1 and makes the an- gle θ1 with the normal of the interface.After traversing the interface, the angle changes to θ2, and the light travels in the medium with refractive index n2. 3. The law of reflection θ1  θ2. (1.3) The law of reflection is the limiting case for the situation where both refraction indices are the same and one has a reflecting surface. The laws of refraction and reflection may be derived from Maxwell’s theory of electromagnetic waves, but may also be derived from a “mechanical model” using Fermat’s Principle. The refractive index in a dielectric medium is defined as n  c/v, where v is the speed of light in the medium and c is the speed of light in a vacuum. The speed of light is no longer the ratio of the unit length of the length standard over the unit time of the time standard, but is now defined as 2.99792458 × 108m/s for vacuum. For practical purposes one uses c  3 × 108m/s, and assumes that in air the speed v of light is the same as c. In dielectric materials, the speed v is smaller than c and therefore, the refractive index is larger than 1. Image formation by our eye also uses just one lens, but not a thin one of fixed focal length. The eye lens has a variable focal length and is capable of forming images of objects at various distances without changing the distance between the eye lens and the retina. Optical instruments, such as magnifiers, microscopes, and telescopes, when used with our eye for image formation, can be adjusted in such a way that we can use a fixed focal length of our eye. Image formation by our eye has an additional feature. Our brain inverts the image arriving on the retina, making us think that an inverted image is erect. 1.2 FERMAT’S PRINCIPLE AND THE LAW OF REFRACTION In the seventeenth century philosophers contemplated the idea that nature always acts in an optimum fashion. Let us consider a medium made of different sections, with each having a different index of refraction. Light will move through each section with a different velocity and along a straight line. But since the sections have different refractive indices, the light does not move along a straight line from the point of incidence to the point of exit. The mathematician Fermat formulated the calculation of the optimum path as an integral over the optical path∫ P2 P1 nds. (1.4) 1.2. FERMAT’S PRINCIPLE AND THE LAW OF REFRACTION 5 With sin θ1  y/r1(y) and sin θ2  (y − yf )/r2(y) (1.14) we have sin θ1/v1  sin θ2/v2 (1.15) and after multiplication with c, the Law of Refraction, n1 sin θ1  n2 sin θ2. (1.16) FileFig 1.1 (G1FERMAT) Graph of the total time for travel fromP1 toP2, through medium 1, with velocities v1, and medium 2, with v2. For minimum travel time, the light does not travel along a straight line between P1 and P2. Changing the velocities will change the length of travel in each medium. G1FERMAT Fermat’s Principle Graph of total travel time: t1 is the time to go from the initial position (0, 0) to point (xq, y) in medium with velocity v1. t2 is the time to go from point (xq, y) to the final position (xf, yf ) in medium with velocity v2. There is a y value for minimum time. v1 and v2 are at the graph. xq : 20 xf : 40 yf ≡ 40 y : 0, .1 . . . 40. Time in medium 1 Time in medium 2 t1(y) : 1 v1 · √ (xq)2 + y2 t2(y) : 1 v2 · √ (xf − xq)2 + (yf − y)2 T (y) : t1(y) + t2(y). 6 1. GEOMETRICAL OPTICS v1 ≡ 1 v2 ≡ 2.5. Changing the parameters v1 and v2 changes the minimum time for total travel. Application 1.1. 1. Compare the three choices a. v1 < v2 b. v1  v2 c. v1 > v2 and how the minimum is changing. 2. To find the travel time t1 in medium 1 and t2 in medium 2 plot it on the graph and read the values at y for T (y) at minimum. FileFig 1.2 (G2FERMAT) Surface and contour graphs of total time for traversal through three media. Changing the velocities will change the minimum position. G2FERMAT is only on the CD. Application 1.2. Change the velocities and observe the relocation of the minimum. FileFig 1.3 (G3FERREF) Demonstration of the derivation of the law of refraction starting from Fermat’s Principle. Differentiation of the total time of traversal. For optimum time, the expression is set to zero. Introducing c/n for the velocities. G3FERREF is only on the CD. 1.3. PRISMS 7 1.3 PRISMS A prism is known for the dispersion of light, that is, the decomposition of white light into its colors. The different colors of the incident light beam are deviated by different angles for different colors. This is called dispersion, and the angles depend on the refractive index of the prism material, which depends on the wavelength. Historically Newton used two prisms to prove his “Theory of Color.” The first prism dispersed the light into its colors. The second prism, rotated by 90 degrees, was used to show that each color could not be decomposed any further. Dispersion is discussed in Chapter 8. Here we treat only the angle of deviation for a particular wavelength, depending on the value of the refractive index n. 1.3.1 Angle of Deviation We now study the light path through a prism. In Figure 1.3 we show a cross- section of a prism with apex angle A and refractive index n. The incident ray makes an angle θ1 with the normal, and the angle of deviation with respect to the incident light is call δ. We have from Figure 1.3 for the angles δ  θ1 − θ2 + θ4 − θ3 A  θ2 + θ3 (1.17) and using the laws of refraction sin θ1  n sin θ2 n sin θ3  sin θ4 (1.18) we get for the angle of deviation, using asin for sin−1 δ  θ1 + asin {( √ n2 − sin2(θ1)) sin(A) − sin(θ1) cos(A)} − A. (1.19) In FileFig 1.4 a graph is shown of δ (depending on the angle of incidence). A formula may be derived to calculate the minimum deviation δm of the prism, depending on n and A. From the Eq. (1.17) and (1.18) we have δ  θ1 − θ2 + θ4 − θ3, A  θ2 + θ3, (1.20) FIGURE 1.3 Angle of deviation δ of light incident at the angle θ1 with respect to the normal. The apex angle of the prism is A. 10 1. GEOMETRICAL OPTICS to be in a medium with index n1, the image point P2 in the medium with index n2. We assume that n2 > n1, and that the convex spherical surface has the radius of curvature r > 0. For our derivation we assume that all angles are small; that is, we use the approximation of the paraxial theory. To find out what is small, one may look at a table of y1  sin θ and compare it with y1  θ . The angle should be in radians and then one may find angles for which y1 and y2 are equal to a desired accuracy. We consider a cone of light emerging from pointP1.The outermost ray, making an angle α1 with the axis of the system, is refracted at the spherical surface, and makes an angle α2 with the axis at the image pointP2 (Figure 1.4). The refraction on the spherical surface takes place with the normal being an extension of the radius of curvature r , which has its center at C. We call the distance from P1 to the spherical surface the object distance xo, and the distance from the spherical surface to the image point P2, the image distance xi . In short, we may also use xo for “object point” and xi for “image point.” The incident ray with angle α1 has the angle θ1 at the normal, and pene- trating in medium 2, we have the angle of refraction θ2. Using the small angle approximation, we have for the law of refraction θ2  n1θ1/n2. (1.27) From Figure 1.4 we have the relations: α1 + β  θ1 and α2 + θ2  β. (1.28) For the ratio of the angles of refraction we obtain θ1/θ2  n2/n1  (α1 + β)/(β − α2). (1.29) We rewrite the second part of the equation as n1α1 + n2α2  (n2 − n1)β. (1.30) The distance l in Figure 1.4 may be represented in three different ways. tan α1  l/xo, tan α2  l/xi, and tan β  l/r. (1.31) Using small angle approximation, we substitute Eq. (1.31) into Eq. (1.30) and get n1l/xo + n2l/xi  (n2 − n1)l/r. (1.32) FIGURE 1.4 Coordinates for the derivation of the paraxial imaging equation. 1.4. CONVEX SPHERICAL SURFACES 11 The ls cancel out and we have obtained the image-forming equation for a spher- ical surface between media with refractive index n1 and n2, for all rays in a cone of light from P1 to P2: n1/x0 + n2/xi  (n2 − n1)/r. (1.33) So far all quantities have been considered to be positive. 1.4.2 Sign Convention In the following we distinguish between a convex and a concave spherical surface. The incident light is assumed to travel from left to right, and the object is to the left of the spherical surface. We place the spherical surface at the origin of a Cartesian coordinate system. For a convex spherical surface the radius of curvature r is positive; for a concave spherical surface r is negative. Similarly we have positive values for object distance x0 and image distance xi , when placed to the right of the spherical surface, and negative values when placed to the left. Using this sign convention, we write Eq. (1.33) with a minus sign, and have the equation of “spherical surface imaging” (observe the minus sign), − n1 x0 + n2 xi  n2 − n1 r . (1.34) The pair of object and image points are called conjugate points. We may define ζo  xo/n1, ζi  xi/n2, and ρ  r/(n2 − n1) and have from Eq. (1.34) −1/ζo + 1/ζi  1/ρ. (1.35) This simplification will be useful for other derivations of imaging equations. 1.4.3 Object and Image Distance, Object and Image Focus, Real and Virtual Objects, and Singularities When the object point is placed to the left of the spherical surface, we call it a real object point. When it appears to the right of the spherical surface, we call it a virtual object point. A virtual object point is usually the image point produced by another system and serves as the object for the following imaging process. To get an idea, of how the positions of the image point depend on the positions of the object point, we use the equation of spherical surface imaging −n1/xo + n2/xi  (n2 − n1)/r (1.36) or xi  n2/[(n2 − n1)/r + n1/xo], 12 1. GEOMETRICAL OPTICS and plot a graph (FileFig 1.6). We choose an object point in air with n1  1, a spherical convex surface of radius of curvature r1  10, and refractive index n2  1.5. We do not add length units to the numbers. It is assumed that one uses the same length units for all numbers associated with quantities of the equations. When the object point is assumed to be at negative infinity, we have the image point at the image focus xif  n2r/(n2 − n1). (1.37) Similarly there is the object focus, when the image point is assumed to be at positive infinity xof  −n1r/(n2 − n1). (1.38) We see from the graph of FileFig 1.6 that there is a singularity at the object focus (at xo  −20). To the left of the object focus all values of xi are positive. To the right of the object focus the values of xi are first negative, from the object focus to zero, and then positive to the right to infinity. When xo  0 we have in Eq. (1.36) another singularity, and as a result we have xi  0. One may get around problems in plotting graphs around singularities t by using numerical values for xo that never have values of the singular points. In FileFig 1.7 we have calculated the image point for four specifically chosen object points, discussed below. FileFig 1.6 (G6SINGCX) Graph of image coordinate depending on object coordinate for convex spherical surface, for r  10, n1  1 and n2  1.5. There are three sections. In the first and third sections, for a positive sign, the image is real. In the middle section, for a negative sign, the image is virtual. G6SINGCX Convex Single Refracting Surface r is positive, light from left propagating from medium with n1 to medium with n2. xo on left of surface (negative). Calculation of Graph for xi as Function of xo over the Total Range of xo Graph for xi as function of xo over the range of xo to the left of xof . Graph for xi as function of xo over the range of xo to the right of xof . r ≡ 10 n1 : 1 n2 : 1.5. 1.4. CONVEX SPHERICAL SURFACES 15 Image focus Object focus xif : n2 · r n2 − n1 xif  30 xof : n1 · r n1 − n2 xof  −20. 1. x1o : −100 x1i : n2( n2−n1 r ) + n1 x1o x1i  37.5 mm1 : x1i · n1 x1o · n2 mm1  −0.25. 2. x2o : −10 x2i : n2( n2−n1 r ) + n1 x2o x2i  −30 mm2 : x2i · n1 x2o · n2 mm2  2. 3. x3o : −10 x3i : n2( n2−n1 r ) + n1 x3o x3i  15 mm3 : x3i · n1 x3o · n2 mm3  0.5. 4. x4o : 100 x4i : n2( n2−n1 r ) + n1 x4o x4i  25 mm4 : x4i · n1 x4o · n2 mm4  0.167. Application 1.7. 1. Calculate Table 1.1 for refractive indices n1  1 and n2  2.4 (Diamond). 2. Calculate Table 1.1 for refractive indices n1  2.4 and n2  1. 1.4.4 Real Objects, Geometrical Constructions, and Magnification 1.4.4.1 Geometrical Construction for Real Objects to the Left of the Object Focus We consider an extended object consisting of many points. A conjugate point at the image corresponds to each point. When using a spherical surface for image formation, a cone of light emerges from each object point and converges to the conjugate image point. Let us present the object by an arrow, parallel to the positive y axis. The corresponding image will also appear at the image parallel to the y axis, but in the opposite direction (Figure 1.5). The image position and size can then be determined by a simple geometrical construction. In Figure 1.5a we look at the ray connecting the top of the object arrow with the center of curvature of the spherical surface. We call the light ray corresponding to this line the C-ray (from center). A second ray, the PF-ray, starts at the top of the object arrow and is parallel to the axis along the distance to the spherical surface. It is refracted and travels to the image focal point Fi on the right side of the spherical surface (Figure 1.5c). The paraxial approximation 16 1. GEOMETRICAL OPTICS FIGURE 1.5 (a) The C-ray and conjugate points for extended image and object; (b) for the calculation of the lateral magnification we show the C-ray, and the ray from the top of y0, refracted at the center of the spherical surface, connected to the top of yi ; (c) geometrical construction of image using the C-ray and the FP-ray. requires that all C-rays and PF-rays have small angles with the axis of the system. The C-ray and the PF-ray meet at the top of the image arrow. 1.4.4.2 Geometrical Construction for Real Object to the Right of the Object Focus We place the object arrow between the object focus and the spherical surface. From FileFig 1.7, with the input data we have used before, we find that the image position is at −30, when the object position is at −10. The geometrical construction is shown in Figure 1.7b. The C-ray and the PF-ray diverge in the forward direction to the right. However, if we trace both rays back they converge on the left side of the spherical surface. We find the top of an image arrow at the image position, at −30. We call the image, obtained by tracing the diverging rays back to a converging point, a virtual image. A virtual image may serve as a real object for a second imaging process. We have listed in Table 1.1 the image positions for real object positions discussed so far and have indicated for images and objects if they are real or virtual. 1.4. CONVEX SPHERICAL SURFACES 17 FIGURE 1.6 (a) The C-ray and the PF-ray diverge in the forward direction; (b) they are traced back to the virtual image. 1.4.4.3 Magnification If we draw a C-ray from the top of the arrow representing the object, we find the top of the arrow presenting the image (Figure 1.5). The lateral magnification m is defined as m  yi/yo. (1.39) It is obtained by using the proportionality of corresponding sides of right triangles, and taking care of the sign convention −yi/(xi − r)  yo/(−xo + r). (1.40) For m  yi/yo we have m  −(xi − r)/(−xo + r). (1.41) Rewritten, eliminating the radius of curvature, one gets with Eq. (1.36), m  yi/yo  (xi/xo)(n1/n2). (1.42) 1.4.5 Virtual Objects, Geometrical Constructions, and Magnification In Figure 1.7 we have made geometrical constructions of virtual objects to the left and right of the image focus. The objects are placed before and after the 20 1. GEOMETRICAL OPTICS first and third sections, for a negative sign, the image is virtual. In the middle section, for a positive sign, the image is real. G8SINGCV is only on the CD. Application 1.8. 1. Observe the singularity at the object focus, which is on the “other side” in comparison to the convex case. 2. Change the refractive index and look at the separate graphs for the sections to the left and right of the object focus. To the left of the object focus, xi is negative to the left of zero, positive to the right. To the right of the object focus it is negative. What are the changes? 3. Change the radius of curvature, and follow Application 2. FileFig 1.9 (G9SINGCV) Concave spherical surface. Calculation of the image and object foci, and image coordinate for four specifically chosen object coordinates. G9SINGCV is only on the CD. Application 1.9. 1. Calculate Table 1.2 for refractive indices n1  1 and n2  2.4 (Diamond). 2. Calculate Table 1.2 for refractive indices n1  2.4 and n2  1 (Diamond). The results are listed in Table 1.2, together with the labeling of the real and virtual objects and image. The geometrical constructions of the four cases calculated in FileFig 1.9 are shown in Figures 1.8a to 1.8d. 1. and 2. Real objects. A real object is positioned to the left of the spherical surface. The C-ray and PF-ray diverge in a forward direction. The PF-ray is traced back through the image focus (it is on the left). The C-ray and PF-ray meet at an image point. We have virtual images for both positions of the real object. TABLE 1.2 Concave Surface. r  −10, xif  −30, xof  20a xo xi m Image Object −100 −25 .167 vi r −20 −15 .5 vi r 10 30 .2 r vi 100 −37.5 −.25 vi vi a Calculations with C9SINGCV. 1.5. CONCAVE SPHERICAL SURFACES 21 FIGURE 1.8 Geometrical construction of images for the concave spherical surface. The images of real objects are constructed in (a) and (b), for virtual objects in (c) and (d). The light converges to real images in (c). The light diverges in (a), (b), (d), and a virtual image is obtained by “trace back.” 3. Virtual object between spherical surface and object focus. We draw the C-ray and have to trace back the PF-ray to the surface and through the image focus. From there, we extend the ray in a forward direction. The rays converge in a forward direction and we have a real image. 4. Virtual objects to the right of object focus. The C-ray is drawn through C in a forward direction. The PF-ray is traced back to the surface and then drawn backwards through the image focus. In the backward direction the two rays meet at a virtual image. Comparing Figures 1.7 and 1.8, one finds that the regions of appearance of real and virtual images are dependent upon the singularities: one when the object distance is equal to the focal length, and the other when the object distance is zero. A virtual image is always found when the C-ray and PF-ray diverge in a 22 1. GEOMETRICAL OPTICS forward direction. If we could place a screen into the position of a virtual image, we could not detect it because the rays toward it are diverging. The case where n1 > n2 and r is positive is very similar and is discussed as an application in FileFig 1.9. Applications to Convex and Concave Spherical Surfaces 1. Single convex surface. A rod of material with refractive index n2  1.5 has on the side facing the incident light a convex spherical surface with radius of curvature r  50 cm. a. What is the object distance in order to have the image at +7 cm? b. What is the object distance in order to have the image at −7 cm? c. Assume r  25 cm; make a graph of xi as a function of xo for n1  1, n2  1.33, and do the graphical construction of the image (i) for real objects before and after the object focal point, and (ii) for virtual objects before and after the image focal point. 2. Rod sticks in water, calculation of image distance. A plastic rod of length 70 cm is stuck vertically in water. An object is positioned on the cross-section at the top of the rod, which sticks out of the water and faces the sun. On the other side in the water, the rod has a concave spherical surface, with respect to the incident light from the sun, with r  −4 cm. The refractive index of the rod is n1  1.5 and of water n2  1.33. Calculate the image distance of the object. 3. Single concave surface. A rod of material with refractive index n2  1.5 has on one side a concave spherical surface with radius of curvature r  −50 cm. a. What is the object distance in order to have the image at +5 cm? b. What is the object distance in order to have the image at −5 cm? c. Assume r  25 cm; make a graph of xi as a function of xo for n1  1, n2  1.33, and do the graphical construction of the image (i) for real objects before and after the image focal point, and (ii) for virtual objects before and after the object focal point. 4. Plastic film on water as spherical surface. A plastic film is mounted on a ring and placed on the surface of water. The film forms a spherical surface filled with water. The thickness of the film is neglected and therefore we have a convex surface of water of n2  1.33. Sunlight is incident on the surface and the image is observed 100 cm deep in the water. Calculate the radius of curvature of the “spherical water surface.” 1.6. THIN LENS EQUATION 25 FIGURE 1.10 Graph of C-ray connecting object and image arrows. The length of the object arrow y0 and image arrow yi and their distances from the thin lens x0 and xi are also indicated. 1.6.3 Magnification In Figure 1.10 we consider the case of a real object and real image and draw a line from the top of the object arrow through the center of the lens to the top of the arrow on the image arrow. The corresponding light ray is called the chief ray and is again referred to as the C-ray. It passes the lens at the center and therefore is not deviated by refraction. From the two “similar” triangles shown in Figure 1.10 we define the magnification m as m  yi/yo  xi/xo. (1.52) 1.6.4 Positive Lens, Graph, Calculations of Image Positions, and Graphical Constructions of Images In FileFig 1.10 we show a graph of the thin lens equation. The image distance xi is plotted as a function of xo for positive f . There is a singularity at the object focus at −f . To the left of the object focus, xi is positive. To the right between the object focus and lens, xi is negative, and on the right of the lens it is positive. As a result, we have three sections. In the first and third sections, for a positive sign, the image is real. In the middle section, for a negative sign, the image is virtual. In FileFig 1.11 we have chosen four specific values of object distances and calculate the corresponding image distances and magnifications. FileFig 1.10 (G10TINPOS) Graph of image coordinate xi , depending on the object coordinate xo for the thin lens equation with f  10. G10TINPOS Positive Lens Focal length f is positive, light from left propagating from medium with index 1 to lens of refractive index n. xo on left of surface (negative). 26 1. GEOMETRICAL OPTICS Calculation of Graph for xi as Function of xo over the Total Range of xo Graph for xi as function of xo over the range of xo to the left of f . Graph for xi as function of xo over the range of xo to the right of f . f ≡ 10 Image focus: f Object focus: −f xo : −100.001,−99.031 . . . 100 xi(xo) : 1( 1 f ) + 1 xo . xxo : −50.001,−49.031 . . .− 11 xxi(xxo) : 1( 1 f ) + 1 xxo . 1.6. THIN LENS EQUATION 27 xxxo : −9.001,−8.031 . . . 50 xxxi(xxxo) : 1( 1 f ) + 1 xxxo . Application 1.10. 1. Observe the singularity at the object focus, which has the same absolute value as the focal length but with a negative sign. The image focus has a positive sign. Note that they play different roles in the geometrical construction of the image. 2. Change the refractive index and describe what happens. 3. Change the focal length and describe what happens. 30 1. GEOMETRICAL OPTICS and both are traced back to the left. They meet at the virtual image. A virtual image is always found when the C-ray and the PF-ray diverge in a forward direction. If we could place a screen into the position of a virtual image, we could not see it, because the rays toward the virtual image are diverging. 3. and 4. Virtual object and real images. In Figures 1.11c and 1.11d we place the object to the right of the lens. We are considering virtual objects. A virtual object may be produced by the image formed by another optical imaging system. The virtual objects are placed between the lens and the image focus and to the right of the image focus. In both cases we draw the C-ray in a forward direction. The PF-1 ray is drawn first backward to the lens and then forward through the image focus. The C-ray and the PF-ray converge to real images for all positions of the virtual object. The results of the calculations of the positive thin lens with f  −10 are listed in Table 1.3. 1.6.5 Negative Lens, Graph, Calculations of Image Positions, and Graphical Constructions of Images In FileFig 1.12, we show graphs of the thin lens equation, plotting xi as a function of xo for negative f . We see the singularity is at the object focus f , which is now to the right of the lens. To the left of the lens, xi is negative. Between the lens and the object focus, xi is positive. To the right of the object focus, xi is negative. As a result, we have three sections. In the first and third sections, for a negative sign, the image is virtual. In the middle section, for a positive sign, the image is real. In FileFig 1.13, we have calculated for four specific values of object distance the corresponding image distances and the magnification. In Figure 1.12 we have the geometrical construction of the images for the values calculated in FileFig 1.13. TABLE 1.3 Positive Lens. f  10, Image Focus 10, Object Focus −10 xo xi m Image Object −30 15 −.5 r r −5 −10 2 vi r 5 3.3 .67 r vi 30 7.5 .25 r vi 1.6. THIN LENS EQUATION 31 FIGURE 1.12 Geometrical construction of the images for a diverging lens with negative f . Real objects for (a) and (b) and virtual objects for (c) and (d). The light converges to real images in (c). The light diverges in (a), (b), (d), and a virtual image is obtained by “trace back.” FileFig 1.12 (G12TINNEG) Graph of image coordinate xi , depending on object coordinate xo for the thin lens equation with f  −10. G12TINNEG is only on the CD. Application 1.12. 1. Observe the singularity at object focus, which has the same absolute value as the focal length but with a positive sign. The image focus has a negative sign. Note that they play different roles in the geometrical construction of the image. 2. Change the refractive index and describe what happens. 3. Change the focal length and describe what happens. 32 1. GEOMETRICAL OPTICS FileFig 1.13 (G13TINNEG) Calculation of image focus and object focus for negative lens. Calculation of image distances xi and magnification for four specific values of object distance xo. G13TINNEG is only on the CD. Application 1.13. The distance between the chosen object coordinate and resulting image coordinate changes with the choice of the object coordinate. 1. Modify the analytical calculation done in Application FF11 for the condition of the shortest distance between image and object. 2. Make a sketch. The geometrical construction of the images for the values calculated in FileFig 1.13 are shown in Figures 1.12a to d. 1. and 2. Real object to the left of the lens and virtual image. The object is presented by an arrow of length yo, placed at the object point xo to the left of the negative lens. The image point and the length of the arrow presenting the image can be geometrically determined using the C-ray and the PF-ray. The C-ray is drawn from the top of the object arrow through the center of the thin lens. The PF-ray is drawn from the object arrow parallel to the axis to the lens, and then diverges in a forward direction. It is traced back to the image focus. The two rays meet at the positions of the image arrow. In Figures 1.12a and 1.12b, we obtain for a real object a virtual image. A virtual image is obtained when the C-ray and the PF-ray diverge in a “forward” direction. 3. Virtual object between lens and object focus. In Figure 1.12c, we place the virtual object between the object focus and the lens and draw the C-ray. The PF-ray is first traced back to the lens, then connected to the image focus, and extended in the forward direction. The two rays meet in the forward direction at a real image. 4. Virtual object on the right side of the object focus. In Figure 1.12d, we place the virtual object to the right of the object focus and draw the C-ray. The PF-ray is first traced back to the lens, then connected to the image focus and extended further in the backward direction. The two rays meet in the backward direction for a virtual image. The results of the calculations of the negative thin lens with f  −10 are listed in Table 1.4 For the geometrical construction, we note that the size of the lens does not matter. One uses a plane in the middle of the lens with sufficient extension in the y direction; see Figure 1.13a. 1.7. OPTICAL INSTRUMENTS 35 Application 1.14. Consider the case n2 > n1. What is the result when interchanging n1 and n2? Applications for the Sections on Positive and Negative Lenses 1. Air lens in plastic. A plastic rod is flat on one side and has a spherical surface on the other side. The spherical surface is concave with respect to the incident light, which comes from the flat side. An identical second rod is taken and the two curved ends are put together, forming an air lens by the ends of the two rods. The cross-section of this lens has its thinnest point in the middle. Assume that the radii of curvatures of the spherical surfaces are r  −r ′  10 cm and the refractive index of the rod is n  1.5. Sunlight is incident on an object on the face of the first rod at 20 cm from the air lens. Find the image distance. 2. Thin lens on water. A lens of refractive index n  1.5 is put on water, one surface in air, the other in water. The lens is a symmetric biconvex lens and has a focal length of f  10 cm in air. The refractive index of water is n  1.33. a. Calculate the radii of curvature of the lens in air and the focal length to be used in the above position. b. Sunlight is shining on the lens; calculate the image distance in the water. 1.7 OPTICAL INSTRUMENTS Optical instruments, such as magnifiers, and microscopes, enlarge tiny objects, making it possible to observe objects we can barely see with the naked eye. The magnifier gives us a modest magnification, in most cases less than ten times. The microscope makes it possible to observe objects of about 1 micron diameter, and the telescope enables us to see objects at a far distance in more detail. Our eye is a one lens system and may produce a real image of a real object, like a positive lens (Figure 1.11a). The real image of a real erect object of a positive lens is 36 1. GEOMETRICAL OPTICS inverted. However, our brain makes a “correction” (another inversion) and we “see” the object erect, as it is. In discussing optical instruments, we have to take this fact into account when making statements about image formation. For a microscope or astronomical telescope it does not matter much if the final image is erect or inverted. However, for the telescope of a sharpshooter it is important. From Figures 1.11 and 1.12, we read a simple rule: If the image appears at the same side of the lens as an erect object, it is erect. If it appears on the other side of the lens, it is inverted. 1.7.1 Two Lens System To obtain the final image distance of a two-lens system, one first applies the thin lens equation to the first lens and determines the image distance. The object distance for the second lens is calculated from the distance between the two lenses and the image distance of the first lens. The thin lens equation is then applied to the second lens and the final image distance for a two-lens system is obtained as the distance from the second lens. The formulas for this procedure are listed in FileFig 1.15. For graphical constructions one proceeds in the same way. Using C- and PF- rays, one constructs the image of the first lens. The image is taken as an object for the second lens, and C- and PF-rays are used to construct the image formed by the second lens. The existence of the first lens is ignored when going through the second process. The magnification of the system is the productm of the magnification of each of the two lenses. One has m  m1m2 with m1  xi1/xo1 and m2  xi2/xo2, where m1 is calculated with respect to the first lens and m2 with respect to the second lens. FileFig 1.15 (G15TINTOW) Calculation of the final image distance of a two-lens system, for a given object distance of the first lens, focal length, and separation of the two lenses. G15TINTOW Two Thin Lenses, Distance Between Lenses: D 1. First lens, xo1, xi1, f 1 xo1 : −5 f 1 : 6 xi1 : 1( 1 f 1 ) + 1 xo1 xi1  −30. 1.7. OPTICAL INSTRUMENTS 37 2. Second Lens, xo2, xi2, f 2, and Distance D (Positive Number) D : 10 f 2 : 1.85. The image distance of the first process is given with respect to the first lens. (Let us assume it is positive.) The object distance must be given with respect to the second lens, taking the distanceD between the two lenses into account. (D is negative when counted from the second lens.) Therefore we have xo2 : −D + xi1 xo2  −40 xi2 : 1( 1 f 2 ) + 1 xo2 xi2  1.94. 3. Magnification for each lens and product for the magnification of the system m1 : xi1 xo1 m1  6 m2 : xi2 xo2 m2  −0.048 System m1 ·m2  −0.291. Application 1.15. 1. Distance between the lenses is larger than 2f . Calculate the final image distance for two lenses at distance D  50. Assume that the object distance from the first lens is −20. Give the magnification, and make a sketch of object and image, assuming that the object is erect. Consider the following cases. a. First lens f1  10; second lens f2  10. b. First lens f1  10; second lens f2  −10. c. First lens f1  −10; second lens f2  10. d. First lens f1  −10; second lens f2  −10. 2. Distance between the lenses is smaller than 2f . Calculate the final image distance for two lenses at distance D  6. Assume that the object distance from the first lens is −20. Give the magnification, and make a sketch of object and image, assuming that the object is erect. Consider the following cases. a. First lens f1  10; second lens f2  10. b. First lens f1  10; second lens f2  −10. c. First lens f1  −10; second lens f2  10. d. First lens f1  −10; second lens f2  −10. 1.7.2 Magnifier and Object Positions The size of an image on the retina increases when placed closer and closer to the eye. There is a shortest distance at which the object may be placed, called the near 40 1. GEOMETRICAL OPTICS FIGURE 1.15 Angular magnification; (a) object at the near point, seen with the eye lens; object at the near point, seen with magnifier and eye lens. 1.7.2.3 Angular Magnification or Magnifying Power To avoid the difficulties we encountered in Section 1.7.2.2, where we calculated meaningless numbers for the magnification, we take a different approach and use angular magnification. We compare the angles at the eye by looking at the object with and without a magnifier (Figure 1.15). The object is positioned at the near point because that gives the largest mag- nification without a lens. First the eye looks at the object without a magnifier, (Figure 1.15a), where angle α is α  yo1/xo1  yo1/(−25). (1.62) Then we introduce the magnifier and have for the angle β, as shown in Figure 1.15b, β  yi1/xi1  yo1/xo1β  yo1(1/xi1 − 1/f1), (1.63) where xo1β is the object distance when calculating the angle β, and the thin-lens equation was used to eliminate xo1β . We define the angular magnification or magnifying power as MP  β/α  −25(1/xi1 − 1/f1). (1.64) We now discuss the applications of angular magnification to the cases where the virtual image is at the near point and at infinity. 1.7. OPTICAL INSTRUMENTS 41 1. Near point. The object is at the near point, and assumingD  0 we have xo1  xi1  −25 and get MP  1 + 25/f1. (1.65) This is the same expression we obtained in Section 1.7.2.1; for the case of the Near point, see Eq. (1.61). FileFig 1.17 (G17MAGNP) Calculations of the magnifier in the near point configuration. Assume D  0. First step: Determination of object point for image point at −25 for first lens with f1  12, result xo1  −8.108. Second step: Determination of xi2 for xo2  −25 and eye lens f2  1.85, result xi2  2. Calculation of magnification. G17MAGNP is only on the CD. Application 1.17. Find the resulting magnifications for three choices of f1. 2. Virtual image at infinity. We consider the virtual image of lens 1 as the real object of lens 2. We have xi1  −∞, and have for the angular magnification MP  −25(1/xi1 − 1/f1) (1.66)  25/f1. This value is marked on magnifiers as MP times x. Example: for f1  5 we would have MP  5x. In both cases the object is placed at the near point of the eye without a magnifier, and the resulting angular magnification depends on the focal length of the magnifier. FileFig 1.18 (G18MAGIN) Calculations of the magnifier for the “virtual image at infinity” configuration. Assume D  0. First step: Determination of object point for image point xi1  −1010 , that is, at (−∞) for the first lens with f1  12, result is xo1  −12. Second step: Determination of xi2 for xo2  (−∞), for the eye lens f2  1.85, result is xi2  1.85. Calculation of magnification. G18MAGIN is only on the CD. Application 1.18. Study several resulting magnifications for three choices of f1. 42 1. GEOMETRICAL OPTICS FIGURE 1.16 Microscope as three-lens system of objective, magnifier (ocular), and eye. The object is close to the focal length of the objective lens L1 and the image is yi1. The magnifier L2 and eye lens act in the magnifier configuration on the image y02 produced by L1. The image yi2 is the object y0e for the magnifier and the image yie appears on the retina erect; we see it therefore upside down. 1.7.3 Microscope 1.7.3.1 Microscope as Three-Lens System In a compound microscope, the first lens L1 (objective lens) has a short focal length and forms a real inverted image of a real erect object. Then the magnifier configuration is applied, which is the second lens L2 (ocular lens) plus the eye lens. See Section 1.7.2 above and Figure 1.16. The final image on the retina is erect, but we see it upside down. We ignore the eye lens and calculate the final image of a two lens system, using for the image distance xi1 the fixed value of tube length 16 cm plus Fi1 (in cm), see Figure 1.16. The magnification is the product ofm1 of the objective lens, times m2 of the ocular lens (magnifier). We discuss the following cases where the magnifier is used in (1) the near point configuration, and (2) the virtual image at infinity configuration. 1.7. OPTICAL INSTRUMENTS 45 FIGURE 1.17 Optical diagram of a Kepler telescope: (a) the object is far away from the objective lens L1 and the image is yi1, located close to xi1  f1; (b) the image yi1 is the object for the magnifierL2 and eye lens in the magnifier configuration and produces the virtual image yi2  y0e. The final image yie appears on the retina erect, we therefore see it upside down. The distance f1 + f2 is approximately the length of the telescope. magnification we have m  m1m2  −f1/f2. (1.73) Note that this is a negative number since f1 and f2 are both positive, and the object is “seen” inverted. To get a large magnification, we need a large value of f1 and a small one of f2. The large value of the focal length of the first lens makes powerful telescopes “large.” FileFig 1.21 (G21TELK) The Kepler telescope is treated as a two-lens system, assuming for xo1 and xi2 the same large negative numerical values. The magnification is calculated from 46 1. GEOMETRICAL OPTICS m  (xi1/xo1)(xi2/xo2) and results inm  m1m2  −f1/f2. Lens L1: f1  30; xo1  −1010; xi2  30. Lens L2: f2  6; distance a  f1 + f2; xi2  −1010, xo2  −6. Calculation of magnification. G21TELK is only on the CD. Application 1.21. Study magnifications of 2 and 4 by changing f1 and f2 and make a sketch. 1.7.4.2 Galilean Telescope The Galilean telescope is the combination of a positive lens L1 and a negative lens L2. The positive lens forms a real inverted image of a far-away real erect object (Figure 1.18a). The negative lens replaces the magnifier. The image of lens 1 is the object for lens 2 and is virtual inverted, see Fig. 1.12(d). Lens 2 forms a virtual erect image of it, at negative infinity (Figure 1.18b). The eye looks at the virtual erect image of lens 2 as a real erect object and forms a real inverted image on the retina (we see it erect). The calculation is shown in FileFig 1.22. FIGURE 1.18 Optical diagram of a Galileo telescope: (a) the object is far away from the objective lens L1 and the image is yi1, located close to xi1  f1; (b) the image of lens 1 is virtual inverted object for lens 2, and lens 2 forms a virtual erect image of it. This virtual erect image is the object of the eye lens and the image yie appears on the retina upside down, therefore we see it erect. 1.7. OPTICAL INSTRUMENTS 47 For the magnification one gets: m  (xi1/xo1)(xi2/xo2), (1.74) where m1  xi1/xo1 is approximately f1/xo1 because the image of lens 1 is close to the focal point. The magnification of the second lens, m2  xi2/xo2, is approximately −xi2/f2, because the object of lens 2 is close to the focal point and to the right side of the lens, and f2 is negative. Since xo1 and xi2 are both large numbers, of the same order of magnitude, they cancel each other out and we have for the magnification m  m1m2  −f1/f2. (1.75) Note that this is a positive number since f2 is a negative lens, and the object is seen erect. The Galilean telescope is used for many terrestrial applications in theaters and on ships. FileFig 1.22 (G22TELG) The Galilean telescope is treated as a two-lens system with the first lens having a positive focal length and the second lens a negative focal length. For xo1 and xi2 the same large negative numerical values are assumed. The magnification is calculated as m  (xi1/xo1)(xi2/xo2) and results in m  m1m2  −f1/f2. (Note that the numerical value is positive.) Lens L1: f1  30, xo1  −1010, xi1  30 Lens L2: f2  −29.99, xi2  −9 · 104; xo2  30. G22TELG is only on the CD. Application 1.22. Go through all the stages and study magnifications by changing f1 and f2. Applications to Two- and Three-Lens Systems 1. Magnifier. A magnifier lens of f1  12cm is placed 8 cm from the eye. a. Find the position of xo1 for i. the near point configuration; and ii. the infinite configuration. b. Give the magnification and the angular magnification. 2. Microscope.A microscope has a first lens (objective) with focal length .31 cm, a magnifier (ocular) lens of 1.79 cm, and the eye lens is assumed to be fe  2 cm. The focal length of the objective lens has been chosen so that the image is at about 16 cm. The distance between the lenses is 18 cm and we assume that the eye is in near point configuration. Calculate the magnification of the 50 1. GEOMETRICAL OPTICS FIGURE 1.20 Coordinates for vector and matrix formulation: (a) the coordinates 11 and α1 are used to form the vectors I1  (l1, α1), and the coordinates l2 and α2 are used to form the vectors I2  (l2, α2); (b) translation, the dependence of d on α1, l1, and l2. We define the vectors I1 of object coordinates and I2 of image coordinates using for I1 the coordinates l1 and α1, and for I2 we using l2 and α2, I1  ( l1 α1 ) I2  ( l2 α2 ) . (1.79) The two equations (1.77) and (1.78) may be written in matrix notation as( l2 α2 )  ( 1 0 −(1/r)(n2 − n1)/n2 n1/n2 ) ( l1 α1 ) . (1.80) For a proof, we may multiply the matrix with the vector and arrive back at Eqs. (1.76) to (1.78). In short notation we may also write I2  R12I1. The matrix R12 is called the refraction matrix of a single spherical surface R12  ( 1 0 −(1/r)(n2 − n1)/n2 n1/n2 ) . (1.81) For a plane surface, that is, for an infinite large radius of curvature, the matrix of Eq. (1.81) reduces to the refraction matrix of a plane surface R  ( 1 0 0 n1/n2 ) . (1.82) We get the translation matrix T , that is, the translation from one vertical plane to the next over the distance d, by taking into account that l2  l1 + α1d; see 1.8. MATRIX FORMULATION FOR THICK LENSES 51 Figure 1.20b. T  ( 1 d 0 1 ) . (1.83) 1.8.2 Two Spherical Surfaces at Distance d and Principal Planes 1.8.2.1 The Matrix For a thick lens we use the refraction and translation matrices. We apply the refraction matrix corresponding to the first spherical surface, the translation matrix corresponding to the thickness of the lens, and the refraction matrix corresponding to the second spherical surface. We again assume that the light comes from the left, and realize that the sequence of the matrices is the sequence of action on I1. In other words, the first surface is represented by the matrix on the far right. First operation: Refraction on first surface: Matrix on the right Second operation: Translation between the surfaces: Matrix in the middle Third operation: Refraction on the second surface: Matrix on the left. For the refraction matrix of a thick lens of thickness d and two different spherical surfaces, we obtain( 1 0 −(1/r2)(n3 − n2)/n3 n2/n3 )( 1 d 0 1 )( 1 0 −(1/r1)(n2 − n1)/n2 n1/n2 ) . (1.84) Multiplication of the three matrices will give us one matrix representing the total action of the thick lens. To do this we define some abbreviations, called refracting powers P12, P23, and P , where P is related to the focal length of the thick lens. P12  −(1/r1)(n2 − n1)/n2 (1.85) P23  −(1/r2)(n3 − n2)/n3, and (1.86) P  −1/f  P23 + dP12P23 + (n2/n3)P12. (1.87) (From the 2,1 element P we get the focal length of the system.) We obtain the thick-lens matrix as( 1 + dP12 d(n1/n2) P d(n1/n2)P23 + (n1/n3) ) . (1.88) FileFig 1.23 (G23SYMB3M) Symbolic calculation of the product of three matrices corresponding to a thick lens of refractive index n2 and thickness d. The light is incident from a medium 52 1. GEOMETRICAL OPTICS with refractive index n1 and transmitted into a medium with refractive index n3. The case of the thin lens is derived by setting d  0 and n1  n3; one obtains the thin-lens matrix. G23SYMB3M Thin-Lens Matrix Special case of the thin-lens matrix. We start with the symbolic calculation of two surfaces at distance d P 12  (−1/r1)(n2 − n1)/n2 P 23  (−1/r2)(n3 − n2)/n3⎡ ⎣ 1 0 P 23 n2 n3 ⎤ ⎦ · [ 1 d 0 1 ] · ⎡ ⎣ 1 0 P 12 n1 n2 ⎤ ⎦ ⎡ ⎢⎣ 1 + d · P 12 d · n1 n2 (P 23 · n3 + P 12 · P 23 · d · n3 + P 12 · n2) n3 (P 23 · d · n3 + n2) n3 · n1 n2 ⎤ ⎥⎦ P  P 23 + dP12P 23 + (n2/n3)P 12. We go to the thin lens and set d  0⎡ ⎣ 1 0 P 23 n2 n3 ⎤ ⎦ · [ 1 0 0 1 ] · ⎡ ⎣ 1 0 P 12 n1 n2 ⎤ ⎦ ⎡ ⎣ 1 0(P 23 · n3 + P 12 · n2) n3 1 n3 · n1 ⎤ ⎦ . Since n3 and n1 are set to 1 we have[ 1 0 (P 23 + P 12 · n2) 1 ] . We set P  (P 23 + P 12 · n2) and P  −1 f ; f is the focal length of the lens. With P 12  (−1/r1)(n2 − n1)/n2 P 23  (−1/r2)(n3 − n2)/n3 1.8. MATRIX FORMULATION FOR THICK LENSES 55 FileFig 1.24 (G24SYMBH) Symbolic calculations of the general transformation for a thick lens. Calculation of the two-spherical-surface matrix and displacement matrix with parameters −h and hh. A numerical example is presented for n1  1, n2  1.5, n3  1, r1  10, r2  −10, and d  20. G24SYMBH Symbolic Calculations of the Product of Three Matrices Corresponding to a General Thick Lens 1. Symbolic calculation of the matrix for the thick lens⎡ ⎣ 1 0 P 23 n2 n3 ⎤ ⎦ · [ 1 d 0 1 ] · ⎡ ⎣ 1 0 P 12 n1 n2 ⎤ ⎦ P 12  (−1/r1)((n2 − n1)/n2) P 23  (−1/r2)((n3 − n2)/n3)⎡ ⎢⎣ 1 + d · P 12 d · n1 n2 (P 23 · n3 + P 12 · P 23 · d · n3 + P 12 · n2) n3 (P 23 · d · n3 + n2) n3 · n1 n2 ⎤ ⎥⎦ . 2. Determination of h and hh. For simpler calculation we define the matrix[ M0,0 M0,1 M1,0 M1,1 ] M0,0  1 + d · P 12 M0,1  d · n1 n2 M1,0  (P 23 · n3 + P 12 · P 23 · d · n3 + P 12 · n2) n3 M1,1  (P 23 · d · n3 + n2) n3 · n1 n2 and determine h and hh,[ 1 hh 0 1 ] · [ M0,0 M0,1 M1,0 M1,1 ] · [ 1 −h 0 1 ] [ M0,0 + hh ·M1,0 − h ·M0,0 −h · hh ·M1,0 +M0,1 + hh ·M1,1 M1,0 −M1,0 · h+M1,1 ] . 3. The results for h, hh, and f are hh  1 −M0,0 M1,0 h  −(1 −M1,1) M1,0 f  −1 M1,0 . 56 1. GEOMETRICAL OPTICS 4. Numerical calculation P 12 : −1 r1 · n2 − n1 n2 P 23 : 1 r2 · n3 − n2 n3 P 12  −3.333 · 10−11 P 23  −0.05 M0,0 : 1 + d · P 12 M0,1  d · n1 n2 M0,0  1 M0,1  6.667 M1,0 : (P 23 · n3 + P 12 · P 23 · d · n3 + P 12 · n2) n3 M1,1 : (P 23 · d · n3 + n2) n3 · n1 n2 M1,0  −0.05 M1,1  0.667. 5. The result for h, hh, and f hh : 1 −M0,0 M1,0 h : −(1 −M1,1) M1,0 f : −1 M1,0 hh  −6.667 · 10−9 h  6.667 f  20. 6. The input values are globally defined n1 ≡ 1 n2 ≡ 1.5 n3 ≡ 1 r1 ≡ 1010 r2 ≡ −10 d ≡ 10. The transformation using the two matrices( 1 hh 0 1 ) ( 1 −h 0 1 ) (1.99) has the effect that we have to count xo from the point on the axis determined by h, and xi from the point on the axis determined by hh. We do not count from the vertex of the spherical surfaces. If we call the vertex of the first surface V1 and the vertex of the second surface V2, we have a similar sign convention as we have used before: 1. if h > 0, the point to start calculating xo is to the right of V1; otherwise to the left; and 2. if hh > 0, the point to start calculating xi is to the right of V2; otherwise to the left. The calculation is shown in FileFig 1.25. The planes perpendicular to the axis ath andhh are called principal planes.As a check one finds, that in the approximation of the thin lens, the difference hh− h  0. We state the general procedure for using the thin-lens equation with the matrix method: One calculates hh  (1 −M0,0)/M1,0 and −h  (1 −M11)/M1,0. The 1.8. MATRIX FORMULATION FOR THICK LENSES 57 focal length f is obtained from P  −1/f  M1,0. One measures xo from h and xi from hh and applies the thin-lens equation. FileFig 1.25 (G25SYMBGTH) Calculation of the general transformation for a thin lens. Calculation of the product of the two-spherical-surface matrix, and the displacement matrix. De- termination of the parameters −h and hh. Specialization for the case of the thin lens. Numerical example for n1  1, n2  1.5, n3  1.3, r1  120, and r2  −10. G25SYMBGTH is only on the CD. 1.8.2.4 Application to the Hemispherical Thick Lens We consider a thick lens of hemispherical shape (see Figure 1.22). In FileFig 1.26 we present the calculations and for the choice of parameters: n2  1.5, n1  n3  1, r1  20, and r2  ∞. If we set n2  n  1.5, n1  n3  1, r1  r  d, and r2  ∞, we have the result that P12  −1/3r , P23  0, P  −1/2r; that is, f  2r , h  0, and hh  −2r/3. FIGURE 1.22 Coordinates for a hemispherical thick lens of index n. The principal planes are indicated as H and H ′. 60 1. GEOMETRICAL OPTICS For the application to calculate the image from a given object point, focal lengths, and distance between the lenses, we measure xo from h, xi from hh, and get the focal length from −1/f  P . FileFig 1.28 (G28SYST2LTI) Calculation for a system of two thin lenses. For the numerical values we have chosen f1  10, f2  10, and a  100. G28SYST2LTI Symbolic Calculation to Determine the Principal Planes for Two Thin Lenses at Distance a The matrix (M) as the product of the two lenses and the displacement between them ⎡ ⎣ 1 0 − 1 f 2 1 ⎤ ⎦ · [ 1 a 0 1 ] · ⎡ ⎣ 1 0 − 1 f 1 1 ⎤ ⎦ ⎡ ⎢⎢⎣ (f 1 − a) f 1 a −(f 1 − a + f 2) (f 2 · f 1) −(a − f 2) f 2 ⎤ ⎥⎥⎦ . Special case a  0, two thin lenses in contact ⎡ ⎣ 1 0 − 1 f 2 1 ⎤ ⎦ · [ 1 0 0 1 ] · ⎡ ⎣ 1 0 − 1 f 1 1 ⎤ ⎦ ⎡ ⎣ 1 0−(f 1 + f 2) (f 2 · f 1) 1 ⎤ ⎦ . Principal planes with h and hh, and P  (−1/f 2)(1 − a/f 1) − 1/f 1 [ 1 hh 0 1 ] · ⎡ ⎢⎢⎣ −(−f 1 + a) f 1 a P −(a − f 2) f 2 ⎤ ⎥⎥⎦ · [ 1 −h 0 1 ] 1.8. MATRIX FORMULATION FOR THICK LENSES 61 [ (f 1−a+hh·P ·f 1) f 1 (−h·f 2·f 1+h·f 2·a−h·f 2·hh·P ·f 1+f 1·a·f 2−f 1·hh·a+f 1·hh·f 2) (f 1·f 2) P −P ·h·f 2+a−f 2) f 2 ] . If the (0, 0) and (1, 1) elements are one, we have for hh  a/P · f 1 and h  a/P · f 2, P is always −1/f P  (−1/f 2)(1 − a/f 1) − 1/f 1 P : (−1 f 2 ) · ( 1 − a f 1 ) − 1 f 1 hh : a P · f 1 h : −a P · f 2 M : [ (f 1−a+hh·P ·f 1) f 1 (−h·f 2·f 1+h·f 2·a−h·f 2·hh·P ·f 1+f 1·a·f 2−f 1·hh·a+f 1·hh·f 2) (f 1·f 2) P −(P ·h·f 2+a−f 2) f 2 ] f 1 ≡ 10 f 2 ≡ 10 a ≡ 100 M  [ 1 0 0.8 1 ] f : −1 P hh  12.5 h  −12.5 f  −1.25. Application 1.28. Consider the case where a  0, and compare the resulting focal length f with 1/(1/f1 + 1/f2). 1.8.3.2 System of Two Thick Lenses We consider two thick lenses and assume that lens 1 has the refractive index n lens 2 the index nn. We also assume that the radii of curvature of the four spherical surfaces are labeled r1 to r4 and that the distance between lens 1 and lens 2 is a. The matrix for the system is obtained from the sequence of three matrices (Figure 1.25). We start on the right with the thick-lens matrix of the first lens, then the translation matrix, and then to the left the thick-lens matrix of the second lens. The calculation is shown in FileFig 1.29 and one obtains( 1 + d2P34 d2/nn P2 d2(P45/nn) + 1 )( 1 a 0 1 )( 1 + d1P12 d1/n P1 d1(P23/n) + 1 ) (1.103) with P12  −(1/r1)(n− 1)/n P23  −(1/r2)(1 − n) P34  −(1/r3)(nn− 1)/nn 62 1. GEOMETRICAL OPTICS FIGURE 1.25 Coordinates for two thick lenses in air with corresponding matrices. P45  −(1/r4)(1 − nn) P1  P23 + P12P23d1 + P12n P2  P45 + P34P45d2 + P34nn. To determine the principal planes of this system, we callM the product of the three matrices in Eq. (1.103), and have to calculate (see FileFig 1.29) ( 1 hh 0 1 ) M ( 1 −h 0 1 ) . (1.104) We have to set in the product matrix the (0,0) and (1,1) elements equal to one, and it follows that the (0,1) element is 0. The result of the transformation is: h  −(1 −M1,1)/M1,0 hh  (1 −M0,0)/M1,0 1/f  −M1,0. To calculate the image distance for a given object distance, we have to measure xo from h, xi from hh, and apply the thin-lens equation with focal length f calculated from −1/f  M1,0. For a specific example of a system of two thick lenses we choose a system of two hemispherical lenses. Each lens is one-half of a sphere, and we assume that the distance a is zero. The results are the same as we found in Section 1.8.2.4 for a sphere. In Figure 1.26, we show the two hemispherical lenses with their refract- ing powers P12 and P45, each of thickness d and a refractive index n with the corresponding matrices. The details of the calculation are shown in FileFig 1.29. 1.8. MATRIX FORMULATION FOR THICK LENSES 65 We define M as[ M0,0 M0,1 M1,0 M1,1 ] . For the determination of h and hh we multiply by the two translation matrices[ 1 hh 0 1 ] · [ M0,0 M0,1 M1,0 M1,1 ] · [ 1 −h 0 1 ] [ M0,0 + hh ·M1,0 − h ·M0,0 −h · hh ·M1,0 +M0,1 + hh ·M1,1 M1,0 −M1,0 · h+M1,1 ] hh : 1 − (M0,0) M1,0 h : 1 − (M1,1) (−M)1,0 f : − 1 M1,0 hh  −10 h  10 f  15. Input Data n ≡ 1.5 nn ≡ 1.5 d1 ≡ 10 d2 ≡ 10 a ≡ 0 r1 ≡ 10 r2 ≡ 1010 r3 ≡ 1010 r4 ≡ −10. Check the form of the final matrix product MM : [ 1 hh 0 1 ] · [ M0,0 M0,1 M1,0 M1,1 ] · [ 1 −h 0 1 ] MM  [ 1 −1.776 · 10−15 −0.667 1 ] . Applications to Matrix Method 1. An exercise for matrix multiplication. Draw two cartesian coordinate systems x, y and x ′, y, the second rotated by the angle θ with respect to the first. Identify the matrix A  ( cos θ − sin θ sin θ cos θ ) with the rotation of x, y into x ′, y ′. a. Is this a rotation in the mathematical positive or negative sense? b. The matrix for rotation in the opposite direction A−1 is obtained by substituting for θ the negative value −θ . c. Show that A A−1 is the unit matrix. d. The transposed matrix AT is obtained from A by interchanging the 2,1 and 1,2 elements. In our case the AT is equal to A−1 and AAT is the unit matrix. 66 1. GEOMETRICAL OPTICS e. Show that A A is the same matrix if we substitute into A the angle 2θ . 2. Noncommutation of matrices. In general two matrices A and B may not be commuted; that is, AB is not equal to BA. We show this in the following example for a different sequence of the same matrices. We consider two hemispherical thick lenses where light is coming from the left. The light hits the first lens L1 at a spherical surface of radius of curvature r , then traverses the thickness d, and emerges from a plane surface. The second, L2, has the reverse order; first the plane surface, then thickness d, and then the curved surface with the same radius of curvature r . The refractive indices of the lenses are n2 and outside we assume n1  n3  1. Make a sketch. See how the two lenses are different. The product matrices for lens 1 and lens 2 are different for the two cases. Compare the position of the principal planes. Compare for the case where r  ∞. 3. Calculate, using the matrix method, the position of the two principal planes for a system of two thin lenses, both of focal length f , and a distance f . 4. Consider a convex-concave lens. The first surface has a radius of curvature r1  20 cm, the second, a radius of curvature r2  −10 cm with thickness of d  5 cm. a. Calculate the principal planes and focal length and find the image of an object positioned at 5 cm to the left of the first surface. b. Find the same result by using twice the imaging equation of a single surface. 5. Thick concentration lens. A thick lens of radius of curvature −r1  r2  −5 mm and thickness of 4 mm is used to concentrate incident parallel light on a detector. Using the matrix method, find the position with respect to the detector plane. 6. Plane-convex and convex-plane lens. The radii of curvature for the convex surface is r  10 cm and for the concave surface r  −10 cm and the thickness is 4 cm. a. Compare h, hh, and f for both lenses. b. An object is placed 100 cm to the left of the first surface. Find the image point for both lenses. See also on the CD PG1. Single convex Surface. (see p. 22) PG2. Single concave Surface. (see p. 22) PG3. Rod Sticks in Water, calculation of Image Distance. (see p. 22) PG4. Plastic Film on Water as Spherical Surface. (see p. 22) PG5. Air Lens in Plastic. (see p. 35) PG6. Positive thin Lens on Water. (see p. 35) PG7. Magnifier. (see p. 47) PG8. Microscope (Near Point). (see p. 48) 1.9. PLANE AND SPHERICAL MIRRORS 67 PG9. Microscope (negative infinity). (see p. 48) PG10. Kepler Telescope.(see p. 48) PG11. Galilean Telescope. (see p. 48) PG12. Laser Beam Expander. (see p. 48) PG13. An Exercise for matrix Multiplication. (see p. 65) PG14. Non commutation of Matrices. (see p. 66) PG15. System with Focal Length f. (see p. 66) PG16. Convex-concave Lens. (see p. 66) PG17. Plane-Concave lens. PG18. Convex-Concave Lens. PG19. Comparison of Plane-Concave and Convex-Plane Lens. (see p. 66) PG20. Convex-Concave Lens. (equal xov, xiv) PG21. Glass Sphere. (see p. 58) PG22. Short Focal Length Lens. PG23. Thick Concentration Lens. (see p. 66) PG24. Surface Corrections of hot Laser Rod. PG25. Laser burning of Image PG26. Corner Mirror. (p. 71) PG27. Reflectivity and Eigenvalues. (see p. 76) 1.9 PLANE AND SPHERICAL MIRRORS 1.9.1 Plane Mirrors and Virtual Images A two-dimensional object appears in a flat mirror as a virtual and left–right inverted image. First we look at one reflected ray (Figure 1.27a). We observe the law of reflection, which says the angle of incidence has the same absolute value as the angle of reflection. In Figure 1.27b we show the reflection of a cone of light, emerging from a point source. The object point appears to us as it is on the “other side” of the mirror. Now we look at a three-dimensional object, represented by the arrows of a right-handed coordinate system. The virtual image produced by the flat mirror appears as a left-handed coordinate system. This may be seen by comparing the image of our left hand as it appears in a mirror, with our right hand placed before the mirror. Similarly one finds “Ambulance” written on the front of an ambulance truck written in letters from left to right. A driver in a car before the truck can read it “normally” in the rear view mirror. 1.9.2 Spherical Mirrors and Mirror Equation Spherical concave mirrors of diameters of a few meters are used in astronomical telescopes, replacing the first lens, as discussed in Section 1.7 on optical instru- ments. A real inverted image is produced by a real erect object. Spherical convex 70 1. GEOMETRICAL OPTICS FIGURE 1.30 Geometrical construction of images for concave spherical mirror. The object is: (a) to the left of the focal point; (b) to the right of the focal point. 2. The ray incident through the top of the arrow and then going to the center of curvature C is reflected back onto itself (Figure 1.30). Both may be extended to the other side of the mirror when xo is between the focus and mirror. Graph of xi as Function of xo A concave spherical mirror has a negative radius of curvature. In FileFig 1.30 we calculate the image points for given object points and radius of curvature. The light comes from the left. For xif  −∞, the focus xof  r/2, and since r is negative for a convex mirror, it is to the left of the mirror. This is the only focus we have for a concave mirror. The focus is also a singularity. We obtain real images for xo to the left and virtual images for xo to the right. FileFig 1.30 (G30MIRCV) Concave spherical mirror. Calculation of image positions from given object positions. Graph for image positions depending on object positions for radius of curvature r  −50, that is, r/2  −25, and xo from −100 to −0.1. 1.9. PLANE AND SPHERICAL MIRRORS 71 G30MIRCV Concave Mirror Raduis of curvature is negative; xo is on left, and is negative. To get around the singularity at −xo  f one chooses the increments such that the value for the singularity does not appear. r : −50 xo : −60 xi : 1( 1 2 ) − 1 xo xi  −42.857 m : −xi xo m  −0.714. Graph xxo : −100,−99.1 . . .− .1 xxi(xxo) : 1( 1 r 2 ) − 1 xxo . 1.9.5.2 Convex Spherical Mirror Geometrical Construction 1. Choose yo and draw the PF-ray to the mirror and trace it forward to the focus F ′. 2. The ray from the top of the arrow to the center of curvature C is reflected back onto itself (Figure 1.31). Graph of xi as Function of xo A convex spherical mirror has a positive radius of curvature. We show in FileFig 1.31 a graph of the image points as a function of the object points for xo from 72 1. GEOMETRICAL OPTICS FIGURE 1.31 Geometrical construction of image for convex spherical mirror. The image yi of object yo is for any object distance to the right of the mirror (always virtual). −100 to −.1. When the light comes from the left, there is no singularity at r/2  xo, and we obtain virtual images for all positions of xo. FileFig 1.31 (G31MIRCX) Convex spherical mirror. Calculation of image position from given object posi- tion. Graph for image position depending on the object position coordinate for the radius of curvature r  50, that is, r/2  25, and xo from −100 to −0.1. G31MIRCX is only on the CD. A summary of the image formation and the dependence on the various parameters is given in Table 1.5. Applications to Spherical Mirrors 1. A corner mirror is made of two flat mirrors, joined together at an angle of 90 degrees. Show that the light incident on one mirror is parallel to the light leaving the other mirror for any angle of incidence. 2. Do the geometrical construction of: TABLE 1.5 Concave Concave Concave Convex Object xo Left of f at f Right of f Any Image xi Negative - Infinity Positive Positive Magnification Negative Positive Positive Real Virtual Virtual Inverted Upright Upright 1.10. MATRICES FOR A REFLECTING CAVITY AND THE EIGENVALUE PROBLEM 75 λ2 : −1 + 2 · g1 · g2 − 2 √ −g1 · g2 + g12 · g22 λ1  1 λ2  1. We set the product g1g2  x and plot it over the range from −1 to 2. x : −1,−9 . . . 2 y(x) : |(2 · x − 1) + √ (2 · x − 1)2 − 1| − 1 yy(x) : |(2 · x − 1) − √ (2 · x − 1)2 − 1| − 1 We obtain from FileFig 1.32 the matrix product of the matrices of Eq. (1.111),( −1 + 2g1g2 2dg2 2g1(−1 + 2g1g2)/d −1 + 2g1g2 ) . (1.113) The round trip in the cavity must have the symmetry of the path of the rays at the beginning and end of a unit cell of the lens line. This corresponds to a mode of oscillation of the cavity. The eigenvalues of this oscillation are obtained from the eigenvalues of the product matrix and the calculation is shown in FileFig 1.32. First the product of the five matrices is calculated using the symbolic method. Then the eigenvalues are obtained λ1  (2g1g2 − 1) + [(2g1g2 − 1)2 − 1]1/2 (1.114) λ2  (2g1g2 − 1) − [(2g1g2 − 1)2 − 1]1/2. (1.115) The coordinates, used for setting up the matrices, may now be transformed into a new coordinate system. In this coordinate system the matrix describing the round trip in the cavity is a diagonal matrix. When the diagonal matrix is a unit 76 1. GEOMETRICAL OPTICS matrix, the light may pass through many round trips and no light will escape. One calls such a resonator stable, and the condition for stability is where the magnitudes of the eigenvalues are equal to 1. |λ1|  |λ2|  1. (1.116) We may write for Eq. (1.114), λ1  (2g1g2 − 1) + [(2g1g2 − 1)2 − 1]1/2 (1.117) or λ1  (2g1g2 − 1) + i[1 − (2g1g2 − 1)2]1/2. (1.118) The real and imaginary parts of Eq. (1.118) must be on a circle of radius 1; that is, |(2g1g2 − 1)| ≤ 1, or 0 ≤ g1g2 ≤ 1. (1.119) in agreement with the imaginary part and plotted in FileFig 1.32. In FileFig 1.33 we show a repetition of the calculations, starting from the five matrices of the cavity in Eq. (1.111), but now in terms of r1, r2, and d. In Figure 1.33, we show schematics of the Fabry–Perot, a focal, a confocal and a spherical cavity for values of the parameters r1, r2, and d, and also of g1 and g2. For both representations one finds that the absolute values of the eigenvalues λ1 and λ2 are always 1. FileFig 1.33 (G33RESCY) Calculation of the eigenvalues of the cavity with two reflecting mirrors using r1, r2, and d. Numerical calculation with r1  1, r2  1, and d  2. G33RESCY is only on the CD. Application 1.33. Use the values of the parameters r1, r2, and d for the Fabry– Perot, focal, confocal, and spherical cavities, and find that all are stable cavities. 1.10. MATRICES FOR A REFLECTING CAVITY AND THE EIGENVALUE PROBLEM 77 FIGURE 1.33 Schematic of light path for four cavities with different values of radii of curvature and length of cavity. The corresponding values of g1 andg2 are indicated: (a) Fabry–Perot; (b) focal; (c) confocal; (d) concentric. 2.2. HARMONIC WAVES 81 λ1 λ2 λ1 FIGURE 2.1 Change of wavelength as the wave enters and leaves a dielectric medium. B3 have been assumed to have the same value, and the three phase constants φ1 to φ3 and 1 to 3 are assumed to be different. Comparing the graphs, one observes equivalence of the dependence of the cosine function on x/λ and t/T . Changing the range of variable from x to t , the family of curves depending on x is similar to the one depending on t . We may modify x/λ and t/T in such a way that they contain phase constants. Then, in the “net” expression cos[2π (x/λ−t/T +φ)] we can not distinguish if φ belongs to the space part or the time part. We show below that for our discussions on interference we do not need the time dependence and it is eliminated. The product of the frequency ν and wavelength λ/n is equal to the phase velocity v  ω/k of the wave propagating in the medium of refractive index n. The angular frequency ω  2πν, and the wave vector k  2πn/λ, where λ is the wavelength in vacuum. We may then write Eq. (2.3) as u  A cos(kx − ωt) (2.4) or u  A cos k(x − (ω/k)t)  A cos k(x − vt). The phase velocity in vacuum is c, and in an isotropic medium with refractive index n it is c/n. The wavelength of “free space” λ is reduced in the medium to λ/n (see Figure 2.1). FileFig 2.1 (I1COSWS) Cosine functions depending on space and time coordinate and one additional phase constant. Graphs are shown for cosine functions depending on the space coordinates for three time instances. This may be interpreted as graphs of the 82 2. INTERFERENCE same wave at three consecutive snapshots. Graphs are shown for cosine functions depending on the time coordinates for three points in space. I1COSWS is only on the CD. Application 2.1. 1. One may change the phase φ and the space coordinate and choose both so there is no resulting change in the graph. Choose φ  2, 4, 6. 2. One may change the phase φ and the time coordinate and choose both so there is no resulting change in the graph. Choose φ  2, 4, 6. 3. Change φ in such a way that there is a shift to smaller values of the position coordinate. 4. Change φ in such a way that there is a shift to larger values of the time coordinate. 2.3 SUPERPOSITION OF HARMONIC WAVES 2.3.1 Superposition of Two Waves Depending on Space and Time Coordinates We describe the interference of two waves in a simple way, using the superpo- sition of two harmonic waves u1 and u2. Both waves will propagate in the x direction and vibrate in the y direction. u1  A cos 2π [x/λ− t/T ] u2  A cos 2π [(x − δ)/λ− t/T ]. (2.5) We assume that the two waves have an optical path difference δ. At time instance t  0, the wave u1 has its first maximum at x  0, and u2 at x  δ (Figure 2.2). Adding u1 and u2 we have u  u1 + u2  A cos 2π [x/λ− t/T ] + A cos 2π [(x − δ)/λ− t/T ]. (2.6) Using cos(α) + cos(β)  2 cos{(α − β)/2} cos (α + β)/2 (2.7) we get u  [2A cos{2π (δ/2)/λ}][cos{2π (x/λ− t/T ) − 2π (δ/2)/λ}]. (2.8) In FileFig 2.2 we show graphs of the square of Eq. (2.8) for the same time instant t1 and wavelength λ. We choose a number of optical path differences δ1  0, δ2  0.1, δ3  0.2, δ4  0.3, δ5  0.4, δ6  0.5, corresponding to the ratios of the optical path difference to the wavelength between 0 and 12 . One observes that the height of the maxima decreases with increasing δ1 to δ6, and shifts to larger values of x. 2.3. SUPERPOSITION OF HARMONIC WAVES 83 FIGURE 2.2 Two waves with magnitude A and wavelength λ. We have u1  A for x  0 and u2  A for x  δ. We now discuss the two factors of Eq. (2.8).The first factor 2A cos{2π (δ/2)/λ} depends on δ and λ, but not on x and t . One obtains for δ equal to 0 or a multiple integer of the wavelength [2A cos{2π (δ/2)/λ}]2 is 4A2 (2.9) and for δ equal to a multiple of half a wavelength [2A cos{2π (δ/2)/λ}]2 is 0. (2.10) The first factor in Eq. (2.8) may be called the amplitude factor and is used for characterization of the interference maxima and minima. One has maxima for δ  mλ, where m is 0 or an integer (2.11) minima for δ  mλ, where m is 12 plus an integer (2.12) and m is called the order of interference. The second factor is a time-dependent cosine wave with a phase constant depending on δ and λ. For the description of the interference pattern this factor is averaged over time and results in a constant, which may be factored out and included in the normalization constant (see below). In Figure 2.3 we show schematically the interference of two water waves with a fixed phase relation. When the interference factor is zero one has minima, indicated by white strips. They do not depend on time. The maxima oscillate and appear and disappear along the line in the observable direction. Maxima and minima are shown in FileFig 2.3 as 3-D graphs. The maxima are shown for δ  λ, and in the second graph, for δ  λ/2, there is just one minimum. The maxima show the time dependence of the second factor for each of the space coordinates. One can estimate that a time average will result in half the maximum value. The minimum is zero. It is zero for all time.