22 MICROSCOPE AND ACCESSORIES \_CH. I 



With the water immersion objective the medium in front is water, and its 

 inde'x of refraction is 1.33, whence «= 1.3-3. Half the angular aperture is 

 ^°=47 , and by the table the sine of 47° is found to be 0.731, i. e., sin u= 

 0.731, whence N. A.=» or 1.33XSH1 « or 0.731=0.972. 



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

 of refraction of the homegeneous fluid in front of the objective is 1.52, and the 

 semi-angle of aperture is \°-°=45°. The sine of 45° is 0.707, whence N. A.=n 

 or i.52Xsin u or 0.707=1.074. 



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

 geneous 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. 33-35. 



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 {i. e., dry or immer- 

 sion). For example, suppose each of three objectives has a numerical aper- 

 ture (N. A. ) of 0.80, what is the angular aperture of each ? Using the formula 

 of N. A.=ra sinw, one has N. A=o.8o for all the objectives. 

 For the dry objective «=i (Refractive index of air). 



For the water immersion objective #=1.33 ( Refractive index of water). 



For the homogeneous immersion objective 72=1.52 (Refractive index of 

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



For the dry objective, substituting the known values the formula becomes 

 0.80=1 sin u, or sin «=o.8o. By inspecting the table of naturalsines (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=io6° 16'. 



For the water immersion objective, substituting the known values in the 



0.80 



formula as before : 0.80=1.33 s ' n u % or s ^ n u ~ =0.6015. 



i-33 

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

 01 36° 59' whence the angular aperture (water angle) is 36 59^X2=73° 58'. 



For the homogeneous immersion objective, substituting the known values, 



0.80 



the formula becomes : 0.80=1.52 sin u whence sin «= =0.5263. And by 



1-52 

 consulting the table of sines it will be found that this is the sine of 31° 45^ / 

 whence 2 u or the entire angle (balsam or oil angle) is 63°3i/. 



That is, three objectives of equal resolving powers, each with a numerical 

 aperture of 0.80 would have an angular aperture of 106° 16' in air, 73° 58' in 

 water and 63 31' in homogeneous immersion "liquid. 



For the apparatus and method of determining aperture, see Ch. X. 



I 39. Table of a group of Objectives with the Numerical Aperture (N. A.) 



and the method of obtaining it. Half the angular aperture is designated by u 

 and the index of refraction of the medium in front of the objective by n. For 



