SIMPLE ILLUSTRATIONS OF THE USE OF N SINE U 



391 



to enable the .student to work out any of the problems which are 

 likely to arise in his practical work. 



We can best accomplish this by illustration. 



(i) If a certain dry objective has an angular aperture of 60, what 

 is its X.A. (i.e. numerical aperture) ? 



All that is needful is to find the value of n sine u ; in this case 

 ?i=the refractive index of the medium, which is air, is 1 ; and u, 

 which is half of (50 = 30 opposite 30 in a table of natural sines, 1 is 

 -") : sine K. therefore ='5, which multiplied by 1 gives '5 as the 

 N.A. of a dry objective having (30 of angular aperture. 



(ii) What is the X.A. of a water-immersion whose angular 

 aperture=44? 



n here=l'33, the refractive index of water ; and u, or half 44, 

 is 22. Sine 22 from tables=-375, which multiplied by 1'33 = '5 

 (nearly), which is the X.A. required. 



(iii) What is the X.A. of an oil-immersion, objective having 38V 

 of angular aperture ? 



n the refractive index of oil, which is equal to that of crown 

 glass, is 1'52; u = 19j and sine u from table= - 329, which multi- 

 plied by V52 = -5. 



Thus it is seen that a dry objective of 60, a water-immersion of 

 44, and an oil-immersion of 38^ all have the same X.A. of "5. 



It will be well, perhaps, to give the converse of this method. 



(iv) If a dry objective is '5 X.A., what is its angular aper- 

 ture 'I 



Here because n sine = - 5, sine "= the objective being 



n ' 



dry, ?t=l, therefore sine = - 5. Opposite '5 in the table of natural 

 sines is 30 ; hence =30. But as.u is half the angular aperture 

 of the objective, 2 or 60=the angular aperture required. 



(v) What is the angular aperture of a water-immersion objective 

 whose X.A. = '5? 



;") '5 



Here w-==l'33. n sine u='5 ; sine " = = ;, .oq = '37r> : 



% 1 " o o 



.=22 (nearly) from tables of sines ; .*. 2?(=44, the angle re- 

 quired. 



(vi) What is the angular aperture of an oil-immersion objective 

 of -5 X.A. ? 



5 '5 



Here ?i=l'52, n sine ='5 ; sine it= =,-^-=-329 ; 



/ *. JL *J *J 



/f=19-} (by tables of sines) ; and 2 = 38V the angle required. 



We may yet further by a simple illustration explain the use of 

 n sine u. 



In the accompanying diagram, fig. 333, let n' represent a vessel 

 of glass ; let the line A be perpendicular to the surface of the water 

 C I) ; suppose now that a pencil of light impinges on the surface of 

 the water at the point where the perpendicular meets it, making an 

 angle of 30 with the perpendicular. This pencil in penetrating the 

 water will be refracted or bent towards the perpendicular. The 



1 Vide Appendix A to this volume. 



