Manchester Memoirs, Vol. lv.{\g\\\ No. 21. 3 



The horizontal scale in Fig: i represents a metre scale 

 divided into tenths, whilst the vertical scale is one of any 

 equal parts, say 40, so that the line passing through D (this 

 may be called the dioptric line) passes through 1/40 on 

 the horizontal scale. From the zero of the vertical scale 

 a line is drawn to the principal focus F of a lens supposed 

 to be at the zero of the horizontal scale. The length of 

 the intercept at D — in the diagram =4 — gives the focal 

 power. It will be noted that the focal lengths and the 

 corresponding dioptrics are co-ordinates of points on the 

 hyperbola xy- i. 



A second graphical method is shown in Fig. 2. The 

 horizontal scale is divided into ten equal parts. They 

 represent, but need not be equal to tenths of a metre. On 

 this scale AF represents the focal length. A line is drawn 

 through the centre of the lens and at right angles to the 

 axis. From this is cut off any arbitrary length AB to 

 serve as a unit. At the other end of the horizontal scale 

 a scale of dioptrics is set up with AB as a unit. This scale 

 is numbered as shown. On drawing a line from B through 

 F the value of the intercept at D on the vertical scale will 

 give the dioptrics. In the diagram this value is 2*5, which 

 is equal i/0"4. The method is easily proved for 



AB/AF = tanAFB = tanDBO = OD/OB = D-AB/unity 

 or i/AF = D. 



The actual determination of/" for a thin converging 

 lens presents no difficult}', for several optical-bank methods 

 give the value directly. The use of real conjugate foci 

 is involved in other methods. If the distances of object 

 and real image from the lens be u and v metres respectively, 



then numerically : — 



i/«-f- \lv= i//=D. 



A graphical construction for finding D from u and v 

 is shown in Fig: 3. 



