By Keigo Iizuka; John Wiley & Sons, Inc
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Additional resources for Elements of Photonics, Volume II : For Fiber and Integrated Optics
Sample text
Calculate both V0 and V2 from Eqs. 139). 2. Use 2 d1 D V22 K1 d1 2 to calculate 2 d1 for a given K1 d1 . 3. Insert 2 and the given value of K1 d1 into Eqs. 132) to find A and B and hence the corresponding value of 0 d1 for the given K1 d1 from Eq. 131). Repeat the same with the other values of K1 d1 to complete such curves as shown in Fig. 12. 4. The intersections between the above curves and the circle of Eq. 136) are the desired solutions. 5. From Eq. 133), ˇ is calculated for the obtained value of K1 .
The unbounded field in the bottom layer is the evanescent wave Hy0 x , which is expressed by the bottom equation of Eq. 11). The bottom equation of Eq. 11) is inserted into Eq. 83 The unbounded field Hy2 x in the top layer is also an evanescent wave and is expressed by the top equation of Eq. 11). Inserting the top equation of Eq. 11) into Eq. 9 Geometry of the three-layer slab guide. 85 The minus sign on the right-hand side of Eq. 84) should be noted. 86 The Ez 2d and Ez 0 in Eq. 81) are rewritten in terms of Hy 2d and Hy 0 , applying the boundary conditions to Eqs.
The conditions for a single mode guide. The field distributions for the first few mode orders. The correlation of the region of ˇ with the field pattern and directions of propagation of the component plane waves. Solution The change in the index of refraction at the air–film interface is much greater than that at the film–substrate interface. 14 Geometry of an asymmetric slab optical guide. 640 PLANAR OPTICAL GUIDES FOR INTEGRATED OPTICS internal reflection for the film–substrate is satisfied, that for the air–film is automatically satisfied.