1 8 MICROSCOPE AND ACCESSORIES [CH. I 



With the oil immersion in the same way N. A.= n sin u ; n or the 

 index of refraction of the homogeneous fluid in front of the objective 

 is 1.52, and the semi-angle of aperture is j y ) -°=45°. The sine of 45 

 is 0.707, whence N. A.=n or 152 X sin 21 or 0.707=1.074. 



By comparing these numerical apertures : Dry 0.799, water 0.972, 

 homogeneous immersion 1.074, the same idea of the real light efficiency 

 and image power of the different objectives is obtained, as in the graphic 

 representations shown in Figs. 27-29. 



If one knows the numerical aperture (N. A.) of an objective the 

 angular aperture is readily determined from the formula ; and one 

 ■can determine the equivalent angles of objectives used in different 

 media (z. e., dry or immersion). For example, suppose each of three 

 objectives has a numerical aperture (N. A.) of 0.80, what is the an- 

 gular aperture of each ? Using the formula of N. A.=« sin u, one has 

 N. A.= 0.80 for all the objectives. 

 For the dry objective n = 1 (Refractive index of air). 



" water immersion objective 72 = 1.33 (Refractive index of water), 

 homogeneous immersion objective 72=1. 52 (Refractive index 

 of homogeneous liquid ). And 2 21 is to be found in each case. 



For the dry objective, substituting the known values the formula 

 becomes 0.80 = 1 sin 22, or sin it = 0.80. By inspecting the table of 

 natural sines (3d page of cover; it will be found that 0.80 is the sine 

 of 53 degrees and 8 minutes. As this is half the angle the entire 

 angular aperture of the dry objective must be 53 8'X2 = 106 16'. 



For the water immersion objective, substituting the known values 



in the formula as before : 0.80= 1.33 sin 72, or sin 22 = -- - =0.6015. 



Consulting the table of sines as before, it will be found that 0.6015 is 

 the sine of 36 59' whence the angular aperture (water angle) is 36 

 5 9 'X2 = 73° 58'. 



For the homogeneous immersion objective, substituting the known 

 values, the formula becomes: 0.80 = 1.52 sin 72 whence sin it = 



— — = 0.5263. And by consulting the table of sines it will be found 



then the angle sought must lie between 2S 45', and 29 . If the difference between 

 0.484S1 and 0.48099 be 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 ; 0.003S2 -.- 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 difference be divided by. the 

 amount found necessary for 1 minute it will give the total minutes above 2S 45'; 

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

 28°45' 28=54'. 



