1 Introduction and Ray Optics

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1 Introduction Ray Optics Optics is the study of light its interaction with matter. Light is visible electromagnetic radiation, which transports energy momentum (linear angular) from source to detector. Photonics includes the generation, transmission, modulation, ampli cation, frequency conversion detection of light.,methods of studying light, in historical order are: -ray optics -wave optics -electromagnetic optics -photon optics (E & M elds are wavefunctions of photons). In this course, we will focus on electromagnetic optics. Maxwell s equations give accurate descriptions of most optical phenomena. However, for pedagogical reasons, we begin, with ray optics.. Ray Optics Ray optics is the simplest theory of light. Rays travel in optical media according to a set of geometrical rules; hence ray optics is also called geometrical optics. Ray optics is an approximate theory, but describes accurately a variety of phenomena. Ray optics is concerned with the locations directions of light rays, which carry photons light energy (They also carry momentum, but the direction of the momentum may be di erent from the ray direction). It is useful in describing image formation, the guiding of light, energy transport. Optical systems are often centered around an axis, called the optical axis. If rays are nearly parallel to such an axis, they are called paraxial rays. The study of paraxial rays is called paraxial optics..2 Postulates of Ray Optics. Light travels in the form of rays (can think of rays as photon currents). Rays are emitted by light sources, can be observed by light detectors. 2. n optical medium (through which rays propagate) is characterized by a real scalar quantity n, called the refractive index. The speed of light in vacuum is c = m=s.the speed of light in a medium is v = c=n; this is the de nition of the refractive index. The time taken by light to cover a distance d is t = nd=c; it is proportional to nd, which is called the optical path length.

2 3. In an inhomogeneous medium, the refractive index n(r) varies with position; hence the optical path length OP L between two points B is OP L = n(r)ds where ds is an element of length along the path. The time t taken by light to go from to B is t = OP L=c. 4. Fermat 0 sprinciple Light rays between the points B follow a path such that the time of travel, relative to neighboring paths, is an extremum (minimum). This means that the variation in the travel time, or, equivalently, in the optical path lenght, is zero. That is, n(r)ds = 0 () Usually, the extremum is a minimum; then light rays travel along the path of least time. If there are many paths with the minimum time, then light rays travel along all of these simultaneously. Why should Fermat s principle work? Were is the physics? (Fermat s principle is the main principle of quantum electrodynamics! It is a consequence of Huygen s principle: waves with extremal paths contribute the most due to constructive interference.).2. Propagation in a homogeneous medium In a homogeneous medium, the refractive index n is the same everywhere, so is the speed of light v. Therefore the optical path length of least time is the shortest one - that is, a straight line. Proof: Suppose the path taken by light is along the curve described by y(x). The optical path length is OP L = nds = n ds (2) 2

3 We want to minimize the OP L. Suppose y 0 (x) is the shortest path. Then we must have that any other path is longer. Consider the path y(x) = y 0 (x)+"(x), where "(x) is small. We note that ds = p r dx 2 + dy 2 = dx + ( dy dx )2 = dx p + y 02 (x) (3) Since y 0 (x) = y0(x) 0 + " 0 (x), we can write q ds = dx + y y0 o" 0 + " 02 (4) If " 0 is also small, we have, approximately ds = dx p s + y y0 o" 0 + y0 0 2 ' dxp + y0 0 2 " 0 y0 0 + dxp + y (5) the optical path length which we varied, OP L v is OP L v = n ( p + y "0 y0 0 p )dx (6) + y Now we said that y 0 (x) is the shortest path, that is, the shortest opical path length OP L S OP LS = n ( p + y )dx (7) is a minimum. This means that OP L v cannot be less than OP L S. If we could choose " 0 freely, then we could argue that the coe cient of " 0 must be equal to zero, otherwise, by choosing " 0 appopriately, we chould make OP L v less than OP L S, in violation of our original assumption. However, we cannot choose " 0 freely, since we have the constraint that "(x) = 0 at the end points. (This is because all paths must go through the end points, since y 0 (x) certainly does, " 0 (x) must vanish there.) We therefore recall that write udv = uvj B y0 0 y p + y 0 2 "0 0 0 dx = p 0 + y 0 2 "jb 0 vdu (8) " d dx ( y 0 0 p + y ) (9) Since " vanishes at the end points, the rst term on the rhs is zero, we have OP L v = n ( p + y n " d dx ( y 0 0 p + y ) (0) 3

4 Now we can argue that if the coe cient of " is not zero, then, by choosing "(x) appropriately, we can make OP L v less than OP L S, in violation of our original assumption. We must therefore have or or d dx ( y 0 0 p + y ) = 0 () y 0 0 p + y = C (2) y 0 0 = m (3) y 0 = mx + b (4) So if y 0 (x) is the shortest path, giving the minimum OP L, we must have y(x) = mx + b (5) the path must be a straight line. This illustrates the method for nding the function which makes the variation vanish - it is the basic idea of variational calculus. We conclude therefore that in a homogeneous medium, light rays travel in straight lines. The principle that light in a homogeneous medim takes the shortest path is Hero s Principle.2.2 Re ection from a mirror Mirrors are madefrom metallic surfaces, or dielectric lms, or other highly re- ective surfaces. re ective surfaces is shown below. 4

5 The plane of incidence is de ned by the surface normal the incident ray. (Note: the plane of incidence may be di erent when we talk about waves!) The angle is the angle of incidence, 0 is the angle of re ection. Consider the plane mirror shown below. ccording to Fermat s principle, B must be such that the OP L = B +BC is a minimum. (ssume there is no delay associated with the re ection) If C C 0 lie on the normal to the mirror, are equidistant from it,. then it is clear that B 0 C 0 is the shortest distance between C 0. Since B 0 C = B 0 C, it is clear that B 0 C is the shortest OP L. This means that = 0. The re ected ray lies in the plane of incidence; the angle of re ection is equal to the angle of incidence..2.3 Re ection Refracton ant the boundary between two media t the boundary between two media with refractive indices n n 2, the incident ray is split into a re ected ray, a refracted (transmitted) ray. 5

6 The re ected ray obeys the law of re ection. The refracted ray obeys the law of refraction: The refracted ray lies in the plane of incidence; theangle of refraction 2 is related to the angle of incidence by Snell 0 s Law: n sin = n 2 sin 2 Proof: Consider This will be a homework assignment..3 Simple Optical Components These can be understood in terms of propagation in straight lines (in homogeneous media) the laws of re ection refraction. 6

7 .3. Planar Mirror Re ects light originating from point P such that the re ected rays appear to originate from a point P 2 behind the mirror. P 2 is called the image of P. Planar Mirror.3.2 Paraboloidal Mirror Surface is paraboloid of revolution; it focuses all incident rays parallel to its axis to a single point called the focus. the distance P F is the focal length f. They are often used as light-collecting elements in telescopes, re ectors to make parallel beams in ashlights. Parabolic Mirror (Recall: parabola is the set of points equidistant from a point (focus) a straight line.) Time to reach focus is the same for all rays. If y = x 2, what is f? 7

8 z y α 2y β 2z β f α Figure : Imaging with a paraboloidal mirror.3.3 Image formation by Parabolic Mirror We now consider image formation by a parabolic mirror. When we see an object, we see light coming to our eyes from each point on the surface of the object. The light can originate in the object (hot lament of a light bulg), or it can be re ected light when the object is illumated by outside sources. Imagine placing an object on axis (shown by arrow of height y in the Figure below), we consider paths of rays of light leaving the object. Light rays will be leaving the object, will be re ected by the mirror. Consider the ray travelling horizontally from the tip to the mirror, re ected through angle, passing through the focus. Consider next the ray travelling from the tip to the focus,passing through the focus, continuing to the mirror, then re ected through the angle travelling horizontally away from the mirror. These two rays come together at the tip of the image (shown by the arrow of height y 2 ) Consider a third ray, showed by the dashed line, travelling from the tip of the object to point where the mirror the axis intersect. It will be re ected, also pass through the point at the tip of the image. In this way, the parabolic mirror collects the rays from the tip of the object, collects them at th tip of the image. The same is true of every other point on the object image. This is image formation. We are interested in knowing where the image will be, how large will it be. 8

9 First, we consider the location. We begin by noting that (z f) tan = y We also note that Substituting for y (z 2 f) tan = y 2 y = f tan y 2 = f tan y 2, we get Multiplying these gives simplifying, we get (z f) tan = f tan (z 2 f) tan = r tan (z f)(z 2 f) = f 2 z + z 2 = f This is exact; it is an important general result. Note that z is measured from the mirror, both image object distances must be greater than the focal length f. The image is inverted. The magni cation is de ned as M = y 2 y Since the triangles above below the axis are similar, we immediately have M = z 2 z This is another important general result. So if the location of the object is known, the location size of the image can be immediately determined. 9

10 Figure 2: Spherical mirror in paraxial approximation.3.4 Elliptical Mirrors ll rays emitted from one of two foci, say P, are focussed (imaged) at the other, P 2. Elliptical mirror. (Recall: n ellipse is the set of points such that the sum of the distance from two points (foci) is constant.) Spherical Mirrors Does not focus in general, however, for paraxial rays, a spherical mirror with radius R it resembles a parabolic mirror with focus f = R=2. ll paraxial rays from point P on axis are focused to point P 2 on axis. Now (ignoring signs in the illustration) = 0 (6) 2 = 0 + (7) 0

11 so + 2 = 2 0 (8) Since for small angles ' tan, we can write the above as tan + tan 2 = 2 tan 0 (9), to a good approximation y z + y z 2 = 2 y R z + z 2 = 2 R = f (20) (2).3.5 Imaging by a spherical mirror Spherical mirrors form images the way lenses do. Here, again, The magni cation is Image formation by spherical mirror z + z 2 = f M = y 2 y = z 2 z (22) (23) (consider ray from P to 0, it s re ection, which also goes through P 2 ).

12 .4 Planar Boundaries Use Snell s Law n sin = n 2 sin 2 (24) for small angles, n ' n 2 sin 2 Refraction by planar boundary Total internal re ectiion (TIR): 2 = =2; then c = sin n 2 n (25) For c, all the light is re ected. Examples are shown below. TIR at planar boundary; re ecting prism optic ber 2

13 .4. Beamsplitters & Beam combiners Partially re ecting mirror, thin glass plate (splitter combiner).5 Spherical Boundaries Lenses Consider the spherical interface below: Spherical interface It is straightforward to show that (ignoring signs in diagram) for paraxial rays, n + n 2 2 = (n 2 n ) y (26) R n + n 2 = n 2 n z z 2 R there is image formation, as for a spherical mirror:the magni cation is (27) M = y 2 y = z 2 z (28).5. Thin Lenses The surfaces of thin lenses may be thought of as the intersection of two spheres, with radii R R 2 : 3

14 Figure 3: Image formation Ray bending image formation by a thin lens. It is straightforward to show that, for paraxial rays, + 2 = y f (29) where f = (n )( R + R 2 ) (30) z + z 2 = f (3) M = y 2 y = z 2 z (32).5.2 Light Guides Light may be quided from one location to another in a variety of ways. Some are shown below. 4

15 Schemes for guiding light Lenses are partially re ective mirrors are partially absorptive, a: b are not particularly e cient. Total internal re ection (TIR) is e cient inexpensive. Optic bers rely on TIR. stepped index ber is shown below. Stepped index optical ber utilizing TIR.5.3 Numerical perture ngle of cceptance of Optical Fiber For propagation via TIR, the angle of incidence between the core cladding must be no greater than the critical angle for TIR. 5

16 cceptance angle of optical ber If the angle of incidence is n 0 sin a = n sin t (33) we must have or We also have n sin c = n 2 (34) sin c = n 2 n 0 t + c = 2 (35) We would like to eliminate c t, have an expression for a. We write cos( t + c ) = cos t cos c sin t sin c = 0 (36) Now tan t = tan c (37) tan t n 0 sin a = n p = n p = n cos c = n s + tan 2 t + /tan2 c n 2 2 n 2 (38) nally n 0 sin a = q n 2 n 2 2 (39) The quantity n 0 sin a is known as the numerical aperture N of the ber. q N = n 0 sin a = n 2 n 2 2 (40) The angle a is the angle of acceptance; rays with angles of incidence greater than a will not undergo TIR, will not propagate. 6

17 .6 Graded Index (GRIN) Optics Fermat s principle tells us that light takes the path of least time; hence the optical path lenght is an extremum. That is, OP L = n(r)ds (4) is an extremum. Let us see what this tells us about ray propagation. Suppose that the location of points on the path the ray follows is given by the vector r(s) from an arbitrary reference point in space. The quantity s is distance measured along the path, from to B. Now p dr2 = ds (42) we can write OP L = n(r) )2 ds (43) Let us suppose that the path which gives the extremum of the OPL is r 0 (s); s is the distance measured along the path r 0 (s). )2 =. Now suppose we consider a small perturbation "(s) to the path; s is still measured along the path r 0 (s). Now the new path will be r(s) = r 0 (s) + "(s) (44) (s is still measured as before, along r 0 (s).) The OPL along this new path will be r OP L = n(r 0 + ") @s )2 ds (45) Now we do Taylor s series expansion of both terms in the integral r OP L = [n(r 0 ) + rn(r 0 ) ")][ )2 + We recall )2 =,, rearranging, we get OP L = n(r 0 )ds )2 ]ds (46) (rn(r 0 ) " + n(r 0 0 )ds Now if the extremal path is r 0 (s), then we must guarantee that the second integral vanishes - otherwise r 0 (s) would not be the extremal path. s before, we integrate by parts, then n(r 0 ds = n(r n(r 0 ds 7

18 Since " vanishes at the points B, we have OP L = n(r 0 )ds + "(rn(r n(r 0 )ds we must rn(r 0 n(r = 0 (50) so in order to extremize the OP L, the ray must follow the path n(r)@r = rn This is the Ray Equation. If the refractive index pro le the location direction of the ray is known, the Ray Equation can be solved to determine how the ray propagates. This is a very important powerful result. (Valid only in isotropic media!).6. Paraxial approximation of the Ray Equation If the rays are nearly on axis (z-direction), we can let s ' z; then the Ray n@y(z) @z @x Usually, this is easier to solve than the general equation Eq. 5. (53) Graded Index Slab If the refractive index n = n(y) depends only on y, then we have, in the paraxial 2 2 (54) This is often quite straightforward to solve. 8

19 Slab with parabolic index pro le Suppose that the refractive index n(y) has the form n = n 0 ( 2 2 y 2 ) (55) where y << in the regions of interest. Materials with such a pro le have the trade name of SELFOC. We have Solution: Now so we can write the = n 0 2 y = n 0 2 y n 0 ( 2 y 2 ) ' 2 y 2 2 = 2 y (58) y = cos z + B sin z (59) y(0) = (60) y 0 (0) = B (6) y = y 0 cos z + 0 sin z (62) where y 0 = y(0) 0 = y 0 (0). pitch is The ray oscillates about the optic axis, the p = (63) the amplitude is y M = r y ( 0 )2 (64) Ray trajectories in SELFOC slab 9

20 SELFOC cylinder - a GRIN optical ber In the cylidrical geometry, the index is given by n(x; y) = n 0 ( 2 2 (x 2 + y 2 )) (65) In this case, we 2 ' 2 y (66) with the 2 ' 2 x (67) y = y 0 cos z + y0 0 sin z (68) x == x 0 cos z + x0 0 sin z (69) In general, the solution corresponds to helical ray propagation. a. Meridional b. helical rays in GRIN ber GRIN Cylinders & Lensing act as lenses. Grin cylinders with parabolic index pro le can 20

21 SELFOC cylinder used as a lens. Calculating the focal length will be a homework assignment..7 Matrix Optics Matrix optics is a way of tracing paraxial rays in a single plane. (Planar geometry, or meridional rays in cylindrically symmetric geometries.) Rays are described by their position direction (angle with respect to the optic axis). In the paraxial approximation, the positions angles of rays at the entrance exit surfaces of an optical system are described by two linear equations. Thus an optical system is described by a 2 2 matrix, called the transfer matrix. Optical system 2

22 The relation betwen the position orientation of the input ray (y, ) the position orientation of the output ray (y 2, 2 ) can be written in the form y2 B y = (70) C D where ; B; C D are real numbers. The matrix B M = C D 2 (7) is known as the transfer matrix. (Note that if y 2 ; 2 ; y ; are given, then we can calculate ; B; C D. This justi es the assumption of the form of the relationship.) Important Convention: We did not worry about signs of quantities when deriving ray equations before. However, now we have to follow convention. The two rules are as follows:. If rays are propagating downward, the sign of the angle is negative. 2. The radius of concave concave surfaces (mirrors lenses) is negative. Free space propagation We know that in free space, We therefore have for the propagation matrix d M = 0 y 2 = y + d (72) 2 = (73) (74) 22

23 Refraction at a planar boundary Snell s law, in the paraxial approximation is it follows that n = n 2 2 (75) 0 M = 0 n n 2 Refraction at a spherical boundary Recalling the relation in Eq. 26 n n 2 2 = (n 2 n ) y R (76) (note that we are using the sgn convention) we see that, at the interface, y 2 = y (77) 2 = n n 2 n 2 n R y (78) M = 0 n 2 n n 2R n n 2 (79) Transmission through a thin lens Here we use our result from Eq.29 2 = y f (80) we get at once that y 2 = y (8) 23

24 2 = y f M = 0 f (82) (83) Re ection from a planar mirror Here again we note that (with the sign convention) the propagatiomn matrix is M = y 2 = y (84) 2 = (85) 0 0 (86) Re ection from a spherical mirror Here again y 2 = y (87), from Eq. 8 2 = 2 0 = 2y 2 R 24 (88)

25 (again using the previous notation convention,) 2 = + 2 R y 2 (89) the propagation matrix is 0 M = 2 R (90) Matrices of cascaded optical components If n optical components are cascaded, the total propagation matrix is just the product: M T = M n M n ::: M 3 M 2 M (9) Nb: note the order of the matrices! Using this approach, many complex optical systems can be analyzed in the geometrical optics approximation. 25