PROBLEMS ON REFRACTIVE INDEX 7 



will be totally reflected from the surface C D back into the denser 

 medium. 



A simple illustration of this is shown in fig. 3. It represents 

 a glass of water so held that the surface of the water is above the 

 eye. If we look obliquely from below at this surface, it appears 

 brighter than polished silver, and an object placed in the water has 

 the upper portion of it brightly reflected. 



The action on all light incident on C D in the denser medium 

 (fig. 2) at an angle greater than the critical angle is precisely the 

 same in fact as if C D were a silvered mirror. 



A critical angle can only exist in a denser medium, for obviously 

 there can be no critical angle in the rarer medium, since a ray of 

 any angle of incidence can enter. 



When the relative or absolute refractive index of the deruser 

 medium is given, the critical angle for that medium can be found, 

 thus: The absolute refractive index of water is l'33 = /u; find its 

 critical angle 9. 



(Problem) IV. : 11 



Sin 6 = = - ='<): 



fJL 1'33 



6 = 481 (found by table). 



So the sine of the critical angle is the reciprocal of the refractive 

 index. 



The connection between the path of an incident ray in a first 

 medium and its refracted ray in a second medium is established by 

 the formula 



H sin = ;u' sin 0'. 



where p. is the absolute refractive index of the first medium, the 

 angle of the incident ray in it, / the absolute refractive index of 

 the second medium, and 0' the angle of the refracted ray in it. 



The angle <p = 45 of the incident ray in the first medium A F" E 



(fig. 2) and p = 1. p.' = _, the absolute refractive indices of both the 



_ 



media, air and glass respectively, being given, find <//, the angle of 

 the refracted rav in glass. 

 (Problem) V. 1 : 



. ,, n sin 1 x sin 45 1 x '707 , 71 . 

 &m * : -7- IT, T5 



0' = 28 (found by table). 



To put another case. Suppose the angle 0' = 28 (fig. 2, B F G) 

 is given ; find 0, the refractive indices remaining the same as before. 

 (Problem) V. 2 : 



u' sin a/ 1-5 x sin 28 1-5 x '741__. 7O ,, 



>M11 t\\ ' ' . _._ i \J\J'J. 



OI11 (JJ .. -, 



U. 



= 45 (found by table). 



Now, suppose the A side of CD (fig. 2) is crown glass, M = 1'5, 

 and the B side of C D is flint glass, / = 1'6. The angle of the 

 incident ray A F E = 45, find" the angle of the refracted ray 0' or 

 B FG. 



