光电子学课件 Lesson2

1、1.3 GRADE-INDEX OPTICS,A. The Ray Equation Fermats principle: to determine the trajectories of light rays in an in homogeneous medium with refractive index n(r), the trajectory is described by the function x(s),y(s),and z(s),s is the length of trajectory.,(1.3-1),If,Derived from Fermats principle ,

2、according to calculus of variations, the partial equations that satisfy the trajectories of light rays will be easily achieved:,(1.3-2),The ray equation,In the paraxial approximation, the trajectory is almost parallel to the z axis ,so that ds=dz,The ray equations then simplify to,(1.3-4),The Paraxi

3、al Ray Equation,B. Graded-Index Optical Components,Graded-Index Slab Consider a slab of material, refractive index n=n(y) is uniform in the x, z direction but varies in the y directions(Fig.1.3-3). The trajectories of paraxial rays in the y-z plane are described by the paraxial ray equation,(1-3.5),

4、from which,Slab with Parabolic Index Profile. An important particular For the graded refractive index is,(1.3-9),Usually,is so small that,1 for all y of interest. Thus,Because,the fractional change of refractive index,is very small, Taking the derivative of(1.3-9), the right-hand side of (1.3-6) is,

5、so that (1.3-6) becomes,(1.3-10),Assuming an initial position y(0)=y0 and an initial slope dy/dz=0 at z=0 inside the GRIN medium,EXAMPLE 1.3-1,(1.3-11),From which the slope of the trajectory is,(1.3-12),The ray oscillates about the centre of the slab with a period (distance) 2/ known as the pitch, a

6、s illustrated 1.3-4,Graded-Index Fibers A graded-index fiber is a glass cylinder with a refractive index n that varies as a Function of the radial distance from its axis. In the paraxial approximation, the ray Trajectories are governed by paraxial ray equations(1.3-4).Consider, for example, the dist

7、ribution,(1.3-14),Substituting(1.3-14) into (1.3-4) and assuming that,for all x and y of,Interest, we obtain,(1.3-15),Because of the circular symmetry, there is no loss of generality in choosing xy =0. The solution of (1.3-15) is then,(1.3-16),If x0=0,the ray continues to lie in that plane following

8、 a sinusoidal trajectory similar To that in the GRIN slabFigure.1.3-7(a) On the other hand, if y0=0, and x0=y0 ,then,(1.3-17),so that the ray follows a helical trajectory lying on the surface of a cylinder of radius y0Fig.1.3-7(b),In both cases the ray remains confined within the fiber, so that the

9、fiber serves as a light guide. Other helical pattern are generated with different incident rays,1.4 MATRIX OPTICS,A. The Ray-Transfer Matrix,The system is characterized completely by its effect on incoming ray of arbitrary Position and direction(y1,1).It steers the ray so that it has new position(y2

10、,2) at the output plane(Fig.1.4-2) In the paraxial approximation, when all angles are sufficiently small so that sin , the relation between (y2,2) and (y1,1) is linear and can be generally be written in the form,(1.4-1),(1.4-2),Where A,B ,C, and D are real numbers, equations(1.4.1) and (1.4.2) may b

11、e conveniently written in matrix form as,(1.4-3),It is known as the ray-transfer matrix,Free-Space Propagation,y2=y1+1d and 1=2 The ray-transfer matrix is,(1.4-4),(1),(2),n1sin1=n1sin1, y2=y1 The ray-transfer matrix is,(1.4-5),B. Matrices of Simple Optical Components,The ray-transfer matrix is,The r

12、ay-transfer matrix is,A cascade of N optical components or systems whose ray-transfer matrices are M1,M2,MN is equivalent to single optical system of ray-transfer matrix,D. Periodic Optical Systems Difference Equation for the Ray Position A periodic system is composed of a cascade of identical unit

13、systems(stages), Each with a ray-transfer matrix (A,B,C,D), as shown in Fig.1.4-5. A ray enters the system with initial position y0 and slope 0 .To determine the position and slope(ym,m) of the ray ,we apply the ABCD matrix m times,(1.4-18),C. Matrices of Cascaded Optical Components,We also iterativ

14、ely apply the relations,(1.4-19),(1.4-20),Using a software routine with 1.4-19,1.4-20 can resolve every value,From (1.4-19),(1.4-21),Replacing m with m+1 in (1.-21) yields,(1.4-22),Substituting (1.4-21) and (1.4-22) into (1.4-20) gives,(1.4-23),Where,(1.4-24),(1.4-25),We use a trial solution of geometric form,(1.4-26),where h is a constant. Substituting (1.4-26) into (1.4-23) immediately shows that the trial solution is suitable