( 380 ) 



way, and very firmly attaelied to the glass, the values /= 72'34'. 8, 

 H = 42' 21'. 7 were found for pi-iiicipal angle of incidence / and [)i-inci})al 

 azimuth H^) (// being the angle which the plane of polarization of 

 the reflected light, being restored by compensation to plane polarization 

 makes with the plane of incidence) from which follows for: 



/=:0° 20° 40' 60' 80' 90' 



?i = 0,295 0,450 0,800 0,928 0,990 1,03 

 /^■=2,88 2,90 2,95 3,01 3,04 3,05 



In the same way I found for a steel mirror ^), 1 = 77°23'.5, 

 H=2Q"W, so that for: 



i = 0" i —I i = 90^ 



n = 2,684 2,794 2,799 

 k = 3,404 3,491 3,496 

 As follows from (9) and (10), h and n increase with the angle 

 of incidence /. From (8) follows, that always n^ '^ sin"^ i. Media 

 which absorb the light, can never reflect the light totally. 



4. Normal to the planes of equal amplitude the amplitude decreases 

 in ratio 1 : «?"' over a distance ). : 2jr /•. In the planes of equal phase 

 the points whose anqilitudes stand in the same ratio, lie at a dis- 

 tance A ; 2.T k sin («J — «J. 



According to (6) and (7) n depends on h. The velocity of propa- 

 gation depends therefore on the way, in which the amplitude in a 

 plane of equal phase varies. If « = 0, it follows from (6) and (7) 

 that k = k^ ,n = )ia- The planes of equal phase and amplitude can 

 therefore only coincide with a propagation normal to the bounding 

 plane. If this took place in every direction, k would be zero according 

 to (8), so the substance would have to be perfectly transparent. 



When the ■ planes of equal phase and anqilitude are normal to 

 each other, « = 90'. For light that penetrates into the metal from 

 outside, the planes of equal amplitude are parallel to the bounding 

 plane, so for a =^ 90^ those of equal phase are perpendicular to it. 

 The propagation then takes place parallel to the bounding plane. 

 This is in harmony with what follows from (7) and (8). According 

 to (7) k„ n„ = for « = 90' and so according to (8) either k == or 

 n = sin i. The first case leads us back to perfectly transparent media. 

 For n =i= sin i there is total reflection. This however, can only be 

 the case with light absorbing media, if ko n^ = or, as ??,„ > 0, if 



1) SissiNGH, Thesis for the doctorate, p. 88, 1885, Arch. Néerl., 20, 207, 1886. 

 ~) SissiNGH, Verb. Akad. v. Wetensch., Amsterdam, deel 28, 1890; Wied. Ann., 

 42, 132, 1891, 



