10 



POLARISATION OF LIGHT. 



ferent angles of polarisation, and upon 

 making the experiment with homogene- 

 ous light he found that the unpolarised 

 portion disappeared. When the blue 

 light, therefore, is polarised, and disap- 

 pears to an eye at E, fig. 11, the red 

 light is not polarised, and consequently 

 is partly reflected to the eye at E. In 

 like manner, when the angle is such as 

 to polarise the red, the blue is not po- 

 larised, and is consequently partly re- 

 flected to the eye at E. This will be 

 obvious from the following table : 



WATER. 



Polarising 



Fie. 



53 4') 



53 11 > 



53 19 J 



15' 



Indices of ^ 



retraction. angle. Variation. 



1-330 red rays 

 T336 mean rays 

 1'342 violet rays 



PLATE GLASS. 



1-515 red rays 56 36] 

 1-525 mean rays 57 45 > 19' 

 1*535 violet rays 57 55J 



OIL OF CASSIA. 

 1-597 red rays 57 57) 

 1-641 mean rays 58 39>1 24' 

 1-687 violet rays 59 21J 



A number of important conclusions 

 may be drawn from the law of the tan- 

 gents now explained. Let MN,y?-. 13, 

 be a surface of any transparent body, 

 and having drawn E A K perpendicular 



Fig. 13. 



to MN, describe "round A as a centre 

 the circle M E N K. Draw E F a tan- 

 gent to the circle at E, and making A E 

 = 1 , set the index of refraction of the 

 body M N, 1-525 for glass, from E to F, 

 and join FA, which will be the direction 

 of a ray which will be polarised after 

 reflexion from MN in the direction AD. 

 The angle EAB-E AD = 56 45' is the 

 polarising angle of the substance M N, 

 or it might be found at once by finding 

 the angle in a table of natural tangents 

 corresponding to 1-525 in the column of 



tangents. If we now calculate the an- 

 gle of refraction C A K corresponding to 

 the index 1.525, and to the incident ray 

 BA, we shall find it to be 32 15', or 

 equal to the complement B A N of the 

 angle of incidence, and, consequently, to 

 D A M. Hence, since M A K is a right 

 angle, DAG will be a right angle, or 

 the reflected ray is perpendicular to the 

 refracted ray. That these properties are 

 general may be thus shown : 



Since tang. B A E m, or index of re- 

 T? P 1 TJ C* 



fraction, we have C L = = ^- But 



m iii r. 



T> /~< 



HB = - because in the similar trian- 

 gles ABH, AEF, AH or BG : HB 

 = EF ; Rad. consequently CL = HB, 

 andBAN = CAK, that is, 



The complement of the polarising an- 

 gle is equal to the angle of refraction. 



But since E AB + BAN = 90, we 

 have EAB+CAK= 90, that is, 



At the polarising angle the sum of 

 the angles of incidence and refraction is 

 a right angle. 



And since D AM = B AK = C A K, 

 the angle D A C = M A K, that is, 



When a ray of light is polarised by 

 reflexion^ the reflected ray forms a right 

 angle with the refracted ray. 



Polarisation of Light at the second 



Surfaces of Bodies. 



Hitherto we have considered only what 

 takes place at the first surface of bodies ; 

 but we shall find that the same law is 

 applicable also to the second surfaces 

 of bodies. 



Let MNP Q, fig. 14, be a plate of 

 glass, AB a ray incident on the first 

 Fig. 14. 



surface at the polarising angle," AD the 

 polarised ray ; and A C the refracted ray. 

 It is found by experiment, that the ray 

 C M reflected at the second surface is 

 polarised. In this case, too, the angle 

 M C F formed by the refracted and re- 

 flected ray is a right angle. For since 

 D A C is a light angle, M N parallel to 

 P C, and B A to C F, the angle F C P is 

 equal to D A M, but M C P is equal to 

 MAC: hence the whole M C F is equal 

 to the whole D A C or to a right angle. 



