By Professor Koichi Shimoda (auth.)
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71) where k is equal to k1 = 'fJlwlc in the interior of the film (\z\ < d) and k2 = 'fJ2W/c outside the film (\z\ > d). 6 Planar Waveguide 55 (Izl < d) whieh can be expressed as a harmonie wave. 71) should satisfy these conditions, we must have k;-Jq=y>O, k;-ki=-ß2< O. , H x = 0, while Ex =1= 0 for the electrie field. We shall deal first with the TM wave. 74) outside (Izl > d), where we have dropped the factor exp (iwt - ik,x). Since A, B, and C are constants and Ex is continuous on the boundary plane, we have C = A eyd sinßd or B eyd cosßd .
Since different atoms are emitting at different points of the source, they will bring about no interference. Thus, as the width of the slit Po is widened the spatial coherence decreases because light waves emitted from different points of such a wide light source are incoherent. Now, if we use a laser, even though the width of the slit Po is widened to the extent of there being virtually no slit at all , we still find interference fringes of high visibility. Not only does laser light have high temporal coherence, it also has very high spatial coherence, so that light from any two separate points on a laser can interfere distinct1y.
In an actual experiment it is customary to scatter the light from the source by putting a ground-glass plate between the monochromatic light source and the interferometer, and to observe the interference fringes through the interferometer with a low-magnification telescope or to photograph them with a camera focused at infinity. The reflection coefficients r of the two reflecting surfaces at z = 0 and z = L facing each other are assumed to be equal, as are the transmission coefficients d. Although both rand d are functions of the angle of incidence and direction of polarization, they are practically constant5 when the angle of incidence is small, as can be seen from the Fresnel formula.