THE USE OF THE APERTOMETER 395 



screens ; thus we get an angle in glass the exact equivalent of the 

 aperture of the objective. As the numerical apertures of these arcs 

 are engraved on the apertometer they can be read off by inspection. 

 Nevertheless a difficulty is experienced, from the fact that it is not 

 easy to determine the exact point at which the edge of the screen 

 touches the periphery of the back lens, or, as we prefer to designate 

 it, the limit of aperture, for, curious as this expression may appear, 

 we have found at times that the back lens of an objective is larger 

 than the aperture of the objective requires. In that case the edges 

 of the screen refuse to touch the periphery. 



On the whole we have found that a far better way of employing 

 this instrument is to use it in Connection with a graduated rotary 

 stage, the edge of the flame of ar paraffin lamp being the illumi- 

 nator. 



Thus : Set the lamp in a direction at right angles to the chord 

 of the apertometer, and suppose that the index of the stage is at 0. 

 The edge of the flame will be seen in the centre of tne bright band. 

 The sliding screens being dispensed with, rotation of the stage will 

 cause the image of the flame to travel towards the edge of the 

 aperture ; rotation is continued until the image of the flame is half 

 extinguished by the edge of the aperture, the arc is then read, and 

 the same thing is repeated on the other side, and the mean of the 

 readings is taken. 



If the stage rotates truly, and if the instrument is properly set 

 up, the reading on the one side ought to be identical with that on 

 the other. 



Suppose that the sum of the readings on both sides = 60, the 

 mean reading is consequently 30, which is the semi-angle of aperture 

 of the lens in glass. From this datum we have to determine the N.A. 

 of the dry J-inch as well as its angular aperture in air. 1 



(i) As before, N.A. = n sine u, and n sine u = n f sine u r ; 

 which means that the aperture on the air side is equal to the aperture 

 on the glass side ; n = 1 for air ; n' = T615, the refractive index of 

 the apertometer ; u' is the mean angle measured, which in this case 

 is 30 ; and n sine u has to be found. 



Now sine 30 = '5 (by the tables) ; n' sine u' 1-615 X sine 30 

 = 1-615 x *5 = '8 = n sine u = the N.A. required. 



(ii) Again, to find the angular aperture or 2u. As before, n sine u 



' sine ' and sine = _' = L 61 : 5 x J 5 = -8 ; u = 53" 



n 1 



nearly (by the tables) ; 1u = 106, which is the angle required. 



(iii) If it be a water-immersion we have to deal with, suppose 

 the mean angle = 45 = u f ; sine 45 = 707 (by the tables) ; n 

 = 1-33; and w' = 1-615. 



n sine u = n f sine u' = 1'615 x '707 = 1-14, the N.A. required. 



(iv) Again, sine = ' sine u ' = 1 ' 615 x " 707 = -86 ; u = 59| 



n 1'33 



(by the tables) ; and 2u = 118^, the angle required. 



(v) In the case of an oil- immersion, suppose the mean angle 



1 Vide p. 2 et seq. 



