CH.I] 



MICROSCOPE AND ACCESSORJES 



Flt >s. 33, 34, 35 are somewhat modified from Ellenberger, and are 

 introduced to illustrate the relative amount of utilized light, with dry, zvaler 

 immersion and homogeneous immersion objectives of the same equivalent 

 focus. The point from which the rays emanate is in air in each case. 

 If Canada balsam were beneath the cover-glass in place of the air 



there would be practically no refraction 

 of the rays on entering the cover-glass 



(§21). 



Fig. 33. Showing the course of the 

 rays passing through a cover- glass from 

 an axial point of the object, and the num- 

 ber that finally enter the front of a dry 

 objective. 



^ f? 



ISp^ 



~-^\\\ 



i//s^ 



\\\\1 



\ll//cove,v 







Pig. 34. Rays from the axial point 

 of the object traversing a cover of the 

 same thickness as in Fig. 33, and entering 

 the front lens of a water immersion 

 objective. 



Fig. 35. Rays from an axial point 

 of the object traversing a cover-glass and 

 entering the front of a homogeneous im- 

 mersion objective. 



next lower sine whose angle is known. Add this number of minutes to the 

 angle of the next lower sine and the sum will represent the desired angle. 

 Or if the sine whose angle is to be found is nearer in size to the sine 

 just greater, proceed exactly as before, getting the difference in the sines, 

 but subtract the number of minutes of difference and the result will give the 

 angle sought. For example take the case in Section 10S where the sine of the 

 angle of 28° 54' is given as 0.48327. If one consults the table the nearest sines 

 found are 0.48099, the sine of 28° 45', and 0.4848 r, the sine of 29°. Evidently 

 then the angle sought must lie between 28 45', and 29 . If the difference 

 hetween o 48481 and 0.48099 is obtained, 0.48481 — 0.48099—0.00382, and if this 

 increase for 15' be divided by 15 it will give the increase for 1 minute ; 

 0.00382^-15^0.000254. Now the difference between the sine whose angle is to 

 be found and the next lower sine is 0.48327 — 0.48099=0.00228. If this differ- 

 ence be divided by the amount found necessary for I minute it will give the 

 total minutes above 28 45', 0.00228-^-0.000254=9. That is, the angle sought is 

 •9 minutes greater than 28° 45'=28° 54'. 



