273 



1. One division of a Thoma Zeiss square = "05 milli- 

 metre. 



2. So that if the diameter happened to cover eight divi- 

 sions the diameter would be '4 millimetres, or the radius '2 

 millimetres. 



3. The area of the field will therefore be tt/^, and the 

 corresponding volume of blood Trr* x ^ (-^ = depth between 

 coverglass and chamber, i.e., irr'' ,-^ = ~ cubic millimetres. 



4. So that to get the number of leucocytes in the cubic 

 millimetre we must multiply by 79-51, but it would be much 

 more convenient to multiply by 100. In that case where 



■Trr^ X ^ :z= ~, r = m/S; millimetres, or diameter of field is 

 357 millimetres. We require, therefore, to arrange our micro- 

 scope so that the diameter is of this value, and this is readily 

 done. 



Two observations only are necessary. 



1. Draw the tube of the microscope out so 

 that diameter of field covers exactly eight divisions 

 of the Thoma Zeiss chamber ; note length of tube. 

 Let this = x. 



2. Draw tube out so that diameter covers 

 exactly seven divisions. Note length of tube = }'. 



3. Therefore an increase in tube length of}' —x 

 reduces diameterfrom eight divisions ("4 millimetre) 

 to seven divisions ("35 millimetre) ; difference = "05 

 millimetre. 



4. Required to calculate what increase in 

 length will cause reduction from "4 to '357 

 (diffei"ence = '043). 



This will be ^ — - x '043 

 •05 



This calculation is made, once for all, by the 

 observer for his microscope, and the tube is drawn 

 out the required amount, using, of course, the 

 same eye-piece and objective. 

 T 



