3Q2 OBJECTIVES. EYE-PIECES, THE APEKTOMETER 



problem is to find the angle this pencil of light will make with the 

 perpendicular in the water. 



To do this we must remember that n sine u on the air side is 

 equal to n' sine u' on the water side. Thus on the air side n=l. 

 ^=30, and by the tables of sines sine 30 = '5; consequently on 

 the air side we have n sine ^='5. 



On the water side ri=l'33, and u' is to be found. But as 



, . . , n sine u '5 

 n sine u == n sine u, therefore sine n' = 7 =_ = -376 ; 



*" 1ft I'Ou 



which (as the tables show) is the natural sine of an angle of 22 

 (nearly) ; consequently u' = 22 ; so the pencil of light in passing 

 out of air into water has been bent 8 from its original direction. 

 Conversely a pencil in water, making an angle of 22 with the 

 perpendicular, would on emerging from the water be bent in air 8 

 further away from the perpendicular, and so make an angle of 30 

 with it. 



Now if we suppose that these pencils of light revolve round the 

 perpendicular, cones would be described, and we can readily see that 

 a solid cone of 60 in air is the exact equivalent of a solid cone of 

 44 in water. 



If we further suppose that the water in the vessel is replaced by 

 cedar oil, the pencil in air, remaining the same as before, will, when 

 it enters the oil, be bent more than it was in the water, because the 

 oil has a higher refractive index than water ; n in this case is equal 

 to 1-52. 



The exact position of the pencil can be determined in the 

 same manner as in the previous case. On the air side, as before, 

 n sine u='5 ; on the oil side n' sine u'n sine u ; sine u'= 



7 =TT^;O ='329, which (by the tables) is the natural sine of 



n 1 *OZ 



19J. It follows that the pencil has been bent in the cedar oil lOf 

 out of its original course, and a cone of 60 in air becomes a cone of 

 38^ in cedar oil or crown-glass. 



Finally, it is instructive to note the result when an incident pencil 

 in air makes an angle of 90 with the perpendicular : n sine u becomes 

 unity, and u in water 48|, in oil 41 (nearly) ; consequently a cone 

 of either 97^ in water, or 82 J in oil or crown glass, is the exact 

 equivalent of the whole hemispherical radiant in air. In other words, 

 and to vary the mode in which this great truth has been before 

 stated, the theoretical maximum aperture for a dry lens is equiva- 

 lent to a water-immersion of 97^ and an oil-immersion of 82J 

 angular aperture. 



The last problem that need occupy us is to find the angular 

 aperture of an oil-immersion which shall be equivalent to a water- 

 immersion of 180 angular aperture. 



On the water side n = T33, u 90, sine 90 = 1, n sine u= 

 1-33. On the oil side n' = T52 and u r has to be found. 



As n' sine u' = n sine u, therefore sine u' 7 ^ m ^i =: l 



n' 1-52 



= -875; u' = 61 (nearly) by the tables; 2w' = 122 (nearly), 

 the angle required. 



