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BEAM OPTICS 3.1 THE GAUSSIAN BEAM EXERCISE 3.1-1 Parameters of a Gaussian Laser Beam Given: 9 3 3 0633 633 10 ; 10 ; 0.05 10nm m P W W m         )a   30 0 4.03 10 4.03W rad mrad       0 0 0 0.012z W   Depth of focus 02 0.025 2.5z m cm   At 5 83.5 10 3.5 10z km m    ,   2 6 0 0( ) 1 1.21 10W z W z z m    Diameter= 2821km )b At 0,z R  At 0 0, 2 2.5z z R z cm   At   2 0 02 , 1 0.031 3.1z z R z z z m cm       )c At beam center, 2 5 2 2 0 02 2.546 10 25.46I I P W W m W cm     On beam axis at   2 2 0 0 0 0 0, 2 12.73z z I I W W z I W cm      A spherical wave of power 100P W at 0 2.5z z cm  has intensity  2 4 2 24 5.169 10 5.169I P z W m W cm    EXERCISE 3.1-2 Validity of rhe Paraxial Approximation for a Gaussian Beam The condition(2.2-21) is A z kA   . In accordance with (3.1-4), 2 1 exp 2 A jk A q q        where 0q z jz  .Therefore, 2 2 ' 2 '1 1 2 exp exp 2 2 2 A AA jkp jk q jk q z q q q q q                                  ' 2 ' 22 q A jk q A q q          Where ' 1 q q z     . The condition A z kA   is therefore equivalent to 2 22A q jk q A kA     ,or 2 21 2 1kq j q     . Substituting 2 21 1q R j kW  we have         2 2 2 2 2 2 2 2 2 21 1 2 2 1 2 1kR W j k W W W R W             Assuming that  is not much greater than W, i.e.,at points not much outside the beam width, This condition is satisfied if )a 1kR  , )b 1kW  ,and )c R W condition )a means that the Radius of curvature R  .Because the minimum radius of curvature is 0z ,condition )a is satisfied if 0z  .similarly, condition )b is satisfied if 0W  ,or 0 0 1W    .We now show that condition )c is also satisfied if 0 1  . for small 0,z R W .For 0 0 0, 2z z R z W W    because 0 0 0 1W z   .For large ,z R z and 0W z so that 01 1R W   . In summary,the conditions 0z  ,and 0 1  ,guarantee that A z kA   and, Therefore, that paraxial approximation is satisfied. EXERCISE 3.1-3 Determination of a Beam with Given Width and Curvature Use   22 2 0 01W W z z     1   2 01R z z z     2 to obtain     2 20 0 0 0W R z z W z z z    from which   20z z W R   3 Substituting  3 into  1 and  2 ,we obtain(3.1-26)and(3.1-25). EXERCISE 3.1-4 Determination of the Width and Curvature at One Point Given the Width and Curvature at Another Point Given: 610 m  ; At position 1, 1R m and 31 10W m  . Find: 2R and 2W at position 2, 2 1 , 0.1z z d d m   . We use the relations: 2 1 2 1 1 1 2 2 2 2 1 1 1 1 q q d q R j W q R j W           Thus, 11 1 0.32q j  and 1 0.91 0.29q j  。 therefore 2 1.01 0.29q j  and 21 0.92 0.26q j  ,so that 2 1 0.92 1.09R m  and 2 2 0.26W   ,from which 3 2 1.11 10 1.11W m mm    EXERCISE 3.1-5 Identification of a Beam with Known Curvatures at Tow Points Using(3.1-9)and 2 1z z d  ,we obtain   2 1 1 0 11R z z z    From which 2 2 1 1 1 0 0z R z z    1 We also obtain     22 1 0 11R z d z z d      From which     2 2 2 1 2 1 0 0z d R z d z      2 Equations  1 and  2 are two equations in two unknowns, 0z and 1z .They can be manipulated Algebraically to obtain(3.1-27)and(3.1-28). 3.2 TRANSMISSION THROUGH OPTICAL COMPONENTS EXERCISE 3.2-1 Beam Relaying Considering a lens and substituting ' 2z z d  in(3.2-6) we obtain 1M  . From(3.2-9a),  0 2r z d f  and  2rM f d f  . Putting 1M  into(3.2-9)we obtain 2 21rM r  so that     222 02 1 2f d f z d f       From which   22 2 02f d f z   ,or     2 22 2 0 2 2z f d f fd d     i.e.,  20 4z d f d  . Since 0z is real, this is only possible if 4f d or 4d f . EXERCISE 3.2-2 Beam Collimation )a Substituting(3.2-9)and(3.2-9a)into(3.2-6),             2 2 ' 2 22 2 0 01 z f f z f z f f z f z z f z f z                   From which     ' 2 2 0 1 1 1 z z f f z f z f      follows.  1 )b Let 0 , 1a z f x z f   and ' 1y z f  Then  1 become: 2 2y x x a    . For a fixed value of a and varying ,x y is maximum if 2 22 2 2 2 1 0 dy x dx x a x a        This occurs at 2 2 22x a x    or x a , i.e., 01z f z f  ,or 0z f z  . )c 0 01 , 50 , 0.02z cm f cm a z f    Optimum 0 51z f z cm   , Distance 'z : 2 2 1 51 50 1 0.02 1 2 25 x z f a y x x a x             But ' 1y z f  . Therefore,  ' 1 50 26 1300z f y cm     . Magnification:   1 50rM f z f x     0 2 1 1 2 50 2 35.4r r r z z f a x M M r M          Width: ' 0 0 035.4W MW W  ' 0 0 056 , 2W z m W mm     EXERCISE 3.2-3 Beam Expansion Imaging at the first lens: Since 1z f and 1 0z f z  ,applying(3.2-11)and(3.2-12)we obtain:         1 1 1 1 '' 0 1 1 0 1 0 2'' 2 0 1 0 1 0 1 1 M f z f f z W f z f W f z W z M z f z z z f            Imagining at the second lens: Based on the result of Exercise(3.2-2),the optimal distance is '' 2 0 2z z f  ,so that   2'' 1 2 1 0 2 1 1 0 2d z z z z f f f z z f        Also the magnification at this optimal distance is   '' 22 2 2 2 2 0 2 1 02 2 2M f z f f z f M z      . The overall magnification of the system is   1 2 2 1 0 2 1 02 2M M M f M z f f z z   . This is a large magnification since 2 1f f and 0z z . EXERCISE 3.2-4 Variable-Reflectance Mirrors The complex amplitude reflectance of this mirror is    2 2 2exp exp mjk R W   . Therefore,upon reflection,the phase of a Gaussian beam increases by 2k R ,so that The radius of curvature becomes 2R where 2 11 1 2R R R  . In addition the amplitude if the beam is multiplied by the factor  2 2exp mW and becomes  2 22exp W ,where 2 11 1 2 mW W W  . The reflected beam remains Gaussian and has width 2W and radius of curvature 2R ,given by the above equations. EXERCISE 3.2-5 Transmission of a Gaussian Beam Through a Transparent Plate From(1.4-11),elements of the ABCD matrix of the plate are: 1, , 0, 1A B d n C D    . Therefore,    2 1 1 1q Aq B Cq D q d n     ,from which 2 02 1 02z jz z jz d n    i.e., 02 01z z and 2 1z z d n  .It follows that the transmitted beam has the same depth of focus and its center is displaced by a distanced n ,as illustrated in the figure. PHOTONIC-CRYSTAL OPTICS 7.1 OPTICS OF DIELECTRIC LAYERED MEDIA EXERCISE 7.1-1 Quarter-Wave Film as an Anti-Reflection Coating The M mateix is a product of the M matrix for a single dielectric boundary(Example 7.1-3)and Another for propagation followed by a boundary(Example7.1-4):                 2 1 2 13 2 3 2 2 1 2 13 2 3 23 2 1 1 2 2 j j j j n n n nn n e n n e M n n n nn n e n n en n                        , where 2 0 2n k d d    and 0 2n  .The B element of this matrix is      2 3 2 1 3 2 2 1 2 3 1 4 j jB n n n n e n n n n e n n          The reflection coefficient can be made to vanish if 0B  ,i.e.,       22 3 2 1 3 2 2 1 0 jn n n n n n n n e       This requires that 2je  is real, i.e., 2 is an integer multiple of  .The value 2  ,leads to 4d  and      2 3 2 1 3 2 2 1 0n n n n n n n n      , From which we obtain the condition 2 1 3 2n n n . The choice 2 2  , or an even multiple of  leads to      2 3 2 1 3 2 2 1 0n n n n n n n n      , Which gives the trivial solution 1 3n n GUIDED-WAVE OPTICS 8.1 PLANAR-MIRROR WAVEGUIDES EXERCISE 8.1-1 Optical Power The power flow is determined by the Poynting vector   *1 2S E H  .For the TE mode 0y z xE E H   .The component of S in the z direction is therefore:   *1 2z x yS E H  From Maxwell’s equation, 0E jw H   ,we have 0 .y xjw H E z    Therefore,   *01 2 .z x xS jw E E z   Substituting    expx m m mE a u y j z  ,we obtain     22 02z m m mS w a u y  . The total power flow in the z direction is the integral of zS with respect to y .Since the Integral of   2 mu y is unity, the power flow   2 02m mw a  . Because  cos cosm m mk w c    , the power flow=     2 2 01 2 cos 1 2 cosm m m mc a a    . EXERCISE 8.1-2 Optical Power in a Multimode Field From Exercise(8.1-1),the power flow in the z direction is the integral of   *01 2 .z x xS jw E E z   with respect to y .Substituting    expx m m m mE a u y j z  ,we obtain          * *02 exp expz m m m m m n n n mS w a u y j z a u y j z       Because the integral of    *m nu y u y with respect y is unity for n m and zero otherwise, the total power is     2 2 02 1 2 cosm m m m m mw a a     . 8.2 PLANAR DIELECTRIC WAVEGUIDES EXERCISE 8.2-1 Confinement Factor Since the waveguide is symmetric we will consider confinement for only 0y  . For    2, cos sin ,m m my d u y A k y m  even  1a  sin sin ,m mA k y m odd For    2, expm m my d u y B y    1b Where   2 2 2 0 1 2 cos 1m mn k n n    2 Because  mu y must be continuous at 2y d    cos sin 2 exp 2 ,m m m mA k d B d m   even  3a    sin sin 2 exp 2 ,m m m mA k d B d m   odd  3b Power in the region   2 2 1 0 2 d my d P u y dy    Substituting from  1a and integrating,      21 4 1 1 sin sin sin m m m mP A d kd kd       4 Similarly the power in the region 2y d is:    22 1 2 expm m mP B d   .  5 The confinement ratio 1 1 2 2 1 1 1 m P P P P P       6 can be obtained by substituting from  4 and  5 and using  3 to substitute for m mA B :           2 1 1 1 1 cos sin 1 1 sin sin sin m m m m m m d kdP P kd kd             7 It is convenient to write the result in terms of the variable sin cM d    ,  8 by writing 2 sin ckd d M     ,  9     2 2 2 1 1 2 cos 1m md kd n n n n   2 2 2 2cos cos sin sinm c c mkd kd       2 21 sin sinm cM    .  10 It is also convenient to define the ratio: sin sinm m cs   And write 1m md M s   .  11 Substituting  10 and  11 into  7 ,         2 1 1 1 cos 1 1 sin m mm m m m m MssP P s Ms Ms           12 This giver an expression for the confinement ration  2 11 1m P P   as a function of The parameter M ,which represents the number of modes, and the parameter sin sinm m cs   ,which is determined by the normalized angles of the modes. As an example ,take the case 8M  .The parameters ms are determined from the characteristic equation(8.2-4),which can be weitten in terms M of and ms as:   2tan 2 2 1 1m mMs m s    Solutions of this equation are shown in Figure(8.2-2)for 8M  .For 0m  ,the first intersection point occurs at  0sin 0.933 2d  ,or 0 0.993s M . Similarly, 1 1.86s M and 2 2.778s M and so on. Substituting these valres in  12 and  6 ,we obtain the confinement ratios: 0 0.999,  1 20.996, 0.990.    The lowest-order mode therefore has the highest conginement. EXERCISE 8.2-2 The Asymmetric ;anar Waceguide Let the complements of the critical angles for reflection from the substrate and the cover be  12 2 1cosc n n  and  13 3 1cosc n n  ,respectively.since 2 3n n , 2 3c c  . Therefore,a guided ray must be inclined at an angle smaller than the smaller of 3c and 2c , i.e., 2c  . )a Since the numerical aperture is governed by 2c , 2 2 1 2NA n n  . )b The self-consistency condition in the symmetric waveguide(8.2-1)is modified here to: 2 3 2 2 sin 2r rd m          where 2r and 3r are respectively the phase shifts introduced by total internal reflection at the substrate and cover boundaries . These phases are given by the general formula in (8.2-3),with the appropriate critical angles 2c and 3c used. )c The number of modes is governed by the critical angle of reflection at the substrate.It is Therefore given by  02M d NA ,where 2 2 1 2NA n n  . 9 FIBER OPTICAL 9.3 ATTENUATION AND DISPERSION EXERCISE 9.3-1 Optimal Grade Profile Parameter The group velocities are   1 q qv d dw   ,where  1 0 1 s q n k q M      ; 2 2 2 1 0M sn k a  ;  2s p p  ; 0 0k w c . To simplify the process of taking the derivative we write  1 0 1q qn w c x     Where   s qx q M  .        0 1 0 1 01 1q qd dw c d n w dw x n w c dx dw          1 0 0 1 01 qN c x n w c dx dw   1 0 1 qN c x     Where  1 1N d n w dw is the group refractive index and   1 11 q qn N w x dx dw     1 If qx  is small,the group velocity       1 1 0 1 0 11 1q q q qv d dw c N x c N x                 2 Let us now determine      1 21s sqdx dw s q M q M dM dw q M d dw         1 1q qsx M dM dw x d dw     ;  3     22 2 1 0 1 0 1 02dM dw sn k d n k dw a s n k a d dw        1 0 1 02 1M n k d n k dw M d dw      1 0 1 02 1M n k N c d dw       4 Substituting into  3 :        1 0 1 01 2 1 1q qx dx dw s n k N c d dw d dw            1 12 1 1sN n w s d dw      Substituting into  1 :  1 2 1 2ss s p      ，where   1 12sp n N w d dw   Thus        1 2 2 2 2 2s sp p p p p p p           , which ,when substituted into  2 ,gives(9.3-10). EXERCISE 9.3-2 Differential Group Delay in a Tow-Segment Fiber )a If 500L m is the length of a segment, then the group delays of the x and y components at the end of the first segment are: 2.4367x xT LN c s  and 2.4383y yT LN c s  each of these components can be analyzed into two components of equal magnitudes along the principal axes 'x and 'y of the second segment. These components travel to the end of the second segment with group delays 'xT and 'yT . The overall delay may therefore take four value: 'x xT T , 'y yT T 'x yT T .and 'y xT T . Since 'x xT T and 'y yT T ,we actually have three possible delays: 2 4.8733xT s and 2 4.8767yT s .and 4.873x yT T s  . Since the pulse with the delay x yT T results from two possibilities, its amplitude depends on the phase shifts encountered,which are sensitive to the phase velocities and of the fiber segments ,and is sensitive to any slight disturbance in the system. This middle pulse will therefore have random polarization. The differential delays between the fastest pulse and the slowest pulse is 2 2 3.4y xT T vs  .To determine whether this differential delay will be visible, we examine the pulse broadening due to GVD.For a single segment, the GVD broadening is 20 50 0.5 500D L ps     , so that the width of each pulse is broadened from an initial valve of100ps to a value of 1ns . The shape of the received the shape of the received pulses will therefore be as shown below. )b The two fiber segments are equivlent to two cascaded identical retarders with their principal axes rotated by 045 .The Jones matuix of this systerm is the product of three matrices cos sin 1 0 cos sin 1 0 sin cos 0 sin cos 0j j T e e                                   where  2x yN N L    is the retardation introduced by a segment and is the angle of rotation.Since 045  , 1 1 1 0 1 1 1 01 1 1 1 0 1 1 02 2 j j T e e                           and therefore     1 1 1 2 1 1 j j j j j j e e e T e e e                      The eigenvalues and eigenvectors of this matrix may be determined for any value of  . Since the matrix is unitary,the eigenvalues will always be phase factors.For example, If  then the eigenvalues are j and the eigenvetors are 1 j      i.e.,are circularly polarized modes. In any case a pulse in one of the polarization modes travels with a single group velocity so that it arrives as a single pulse instead of two. 11.4 PARTIAL POLARAIZATION EXERCISE 11.4-1 Partially Polarized Light The coherency matrix for a superposition of unpolarized light with intensity   1x yI I P  and polarized light with intensity  x yI I P at angle is     2 2 1 0 cos cos sin 1 0 12 cos sin sin x y x y I I G P P I I                           . The four elements of this matrix are      21 2 cosxx x y x yG I I P I I P       1      21 2 sinyy x y x yG I I P I I P       2   cos sinxy yx x yG G I I P      3 We wish to show that for some  xx xG I  4 yy yG I  5   2 xy yx x y xyG G I I g   6 From  4 and  1      2 1 2 cos x x y x y I I I P I I P        7 From  5 and  2      2 1 2 sin y x y x y I I I P I I P       .