360 Mr. R. T. Glazebrook on 



a — j8 to « + j3. The same is true for positions of N on the 

 other side of AX. 



Hence, in the first case, the average displacement normal to 

 the plane is 



2p | I 'cos dd- f ""^cos doie\ 



\l— sin(a-/3) + l-sin(a + /3)J- 



2-7T sin a v 



TTsma 



= MJ__ C0S/3 } 



7r i sin a J 



= — {coseca— cos/3J (11) 



And in the second case it is 



_2p_ 



27rsin 



^— r^cos0d0=— 4- isin(a+j8)-sin(a-|8)} 

 -sinaj^^ 7rsina* v r/ v ^ /J 



= -^cotasin/3 (12) 



IT 



The first is clearly least when a is ~- ; the second decreases as 

 a increases, but has no minimum ; for after a time we should 



IT 



reach a point at which « + /3 became equal to ~, and then the 



limits would require changing : for this value, of course, the 

 two integrals are the same. 



Thus the average displacement normal to the plane OAX 

 is least when OX is at right angles to the optic axis, and hence 

 the average error in the position of the plane of polarization 

 is least also. The average displacement just calculated is of 

 course that for a given value of /3. If we require the average 

 for any value of /3 between and /3 1? we must multiply our 

 expressions (11) and (12) by d(i, and, integrating from to 

 f&i, divide the result by ft. 



To show the difference in this respect between the new prism 

 and Nicol's, let us calculate the displacement normal to the 

 plane AOX in the two cases, supposing the value of /3 is 5°. 



In NicoPs prism, a = 63° 30'; so that a + /3 is less than ^, and 



the second formula (12) must be taken. The ratio of the two 

 displacements is therefore 



«j f 3Q/ f P =cot63°30'cqtft 



1— cos/3 , 2 



