202 PROGRESS IN MICROSCOPY 



In Fig. 7.28. the path difference 'J and ad add up. In Fig. 7.29, they 

 are substracted from one another and ad— J should be written in equa- 

 tion (7.12), This is not very important but it should be determined 

 whether the wave surface shows a peak (Fig. 7.30) or a trough (Fig. 7.31). 

 Merely substituting the specimen for an object whose slope and index 

 are known provides the answer. For instance, let us consider an object 



Fig. 7.29. The path difference at M is ad — J. 



with the slope d =20°. Let us assume n = 1-515 and n = 1-526 

 and that the duplication brought about by the interference micro- 

 scope in the specimen plane d = 25 ft. The microscope is adjusted 

 so that J = 0, i.e. Bb^ is dark. At BB', the tint is, for instance, 

 lavender grey (0 097 //). Both waves are arranged as shown at BB'. 

 Now the adjustment is altered so as to increase J : the tint becomes 

 white at BB'. This means that the path difference increases there. 

 Hence, it is the wave most shifted towards the right by duplication 



B 



1. 



b2 



J>, 



^2 B/ j \C ^ 



B C' 



Fig. 7.30. Wave surface when /;' > //. FiG. 7.31. Wave surface when /;' < «. 



(wave Zg) that moves away from X^ as it drops. Or, which amounts 

 to the same thing, it is the wave most shifted to the left by dupHcation 

 (27j3 that moves away from 2*2 ^s it rises. Therefore, observing an 

 object of known index n shows that as 1 is increased from zero, the 

 lint order at BB' rises if /? < /;'. 



Let us now assume that the index n of the known object is higher 

 than n'. Starting again from the value .1 = 0, the arrangement is 

 shown in Fig. 7.32. Let us increase 1 by actuating the microscope 

 control as mentioned before. The wave {1\) most shifted to the right 

 drops, or, which amounts to the same thing, the wave most shifted 



