i8 



MICROSCOPE AND ACCESSORIES. 



\CH. I. 



Figs. 27-29 are somewhat modified from 

 Ellenberger, and are introduced to illustrate 

 the relative amount of utilized light , with dry, 

 water immersion and homogeneous immer- 



27 sion objectives of the same equivalent focus. 

 The point from which the rays emanate is in 

 air in each case. If Canada balsam were in 

 place of the air beneath the cover glass there 

 would be practically no refraction of the 

 rays on entering the cover glass ($ 16). 



Fig. 27. Showing the course of the rays 

 passing through a cover glass from an axial 

 point of the object, and the number that 



28 finally enter the front of a dry objective. 



Fig. 28. Rays from the axial point of the 

 object traversing a cover of the same thick- 

 ness as in Fig. 2j, and entering the front 

 lens of a water immersion objective. 



FiG. 29. Rays from an axial point of the 



29 object traversing a cover glass and entering 

 the front of a homogeneous immersion 

 objective. 



With the water immersion objective the 

 medium in front is water, audits index of 

 refraction is 1.33, whence;/ = 1.33. Half 

 the angular aperture is -V° =47°, and by the table the sine of 47 is 



corresponding to a given angle ; or if one has an angle which does not correspond 

 to any sine or angle given in the table, the sine or angle may be closely approxi- 

 mated by the method of interpolation, as follows : Find the sine in the table nearest 

 the sine whose angle is to be determined. Get the difference of the sines of the an- 

 gles greater and less than the sine whose angle is to be determined. That will give 

 the increase of sine for that region of the arc for 15 minutes. Divide this increase by 

 15 and it will give with approximate accuracy the increase for 1 minute. Now 

 get the difference between the sine whose angle is to be determined and the sine 

 just below it in value. Divide this difference by the amount found necessary for 

 an increase in angle of 1 minute and the quotient will give the number of minutes 

 greater the sine is than the next lower one whose angle is known. Add this num- 

 ber of minutes to the angle of the next lower sine and the sum will represent the 

 desired angle of the sine. 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 the last section 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 0.48481, the sine of 29 . Evidently 

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

 0.48481 and 0.48099 be obtained, 0.48481 — 0.48099 =0.00382, and this increase for 

 i5 / be divided by 15 it will give the increase for 1 minute ; 0.00382-=-! 5 = 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 difference is divided by the 

 amount found necessary for 1 minute it will give the total minutes above 28 45 r ; 

 0.00228 -f- 0.000254 = 9. That is, the angle sought is 9 minutes greater than 

 28° 45' = 28° 54'. 



