CH. /.] MICROSCOPE AND ACCESSORIES. 19 



found to be 0.731, /. e., sin u = 0.731, whence N. K. —nor 1.33 X sin?/ 

 or 0.731 = 0.972. 



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 --^-° = 45 . The sine of 45 is 

 0.707, whence N. A. = n or 1.52 x sin u or 0.707 = 1.074. 



~&y 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 an- 

 gular aperture is readily determined from the formula ; and one can de- 

 termine the equivalent angles of objectives used in different media (J. e. , 

 dry or immersion). For example, suppose each of three objectives has 

 a numerical aperture (N. A.) of 0.80, what is the angular aperture of 

 each. Using the formula of N. A. = n sin u one has N. A. = 0.80 for 

 all objectives. 

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



" water immersion objective n = 1.33 (Refractive index of water). 

 " homogeneous immersion objective n = 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 ti, or sin u — 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' X 2 = 106 16'. 



For the water immersion objective, substituting the known values 



in the formula as before : 0.80 = 1.33 sin u, or sin u = — — = 0.6015. 



i-33 

 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 59' 

 X 2 = 73° 58'. 



For the homogeneous immersion objective, substituting the known 

 values, the formula becomes : 0.80 = 1.52 sin u whence sin u = 



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



that this is the sine of 31 45^2' whence 2 u or the entire angle (balsam 

 or oil angle) is 63° 31'. 



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

 merical 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 appendix. 



