376 Mr. 0. J. Lodge on some Problems connected 



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the same size, shape, and strength, and situated at an equal 

 distance on the other side of the line so as to be opposite to the 

 first pole. 



Again, in order to make a given straight line into an equi- 

 potential line, the blocks must be arranged skew-symmetrically 

 with respect to the line — that is to say, in pairs one on each 

 side of it and at equal distances from it, but the sign of one of 

 the constituents of each pair must be opposite to that of the 

 other. This fact is expressed thus : — The image of a pole in a 

 straight equipotential line is a pole of the same size, shape, and 

 numerical strength, but of opposite sign, and situated at an 

 equal distance on the other side of the line. 



Hence, if we have given a sheet of tinfoil bounded by a 

 straight edge (i. e. by a straight flow-line) and with any distri- 

 bution of poles in it whatever, and we wish to find the distri- 

 bution of poles which would make that straight line a flow- 

 line in the unlimited sheet, we have only to regard the line as 

 a plane mirror whose silvered face is turned towards the sheet, 

 and the optical images of the several poles will give the distri- 

 bution required. The distribution of poles will be obtained in 

 the same way if the boundary is a straight band of copper (i. e. 

 an equipotential line) ; but in this case the sign of every image 

 must be reversed. 



§ 4. Now take a sheet bounded by two straight flow-lines 

 intersecting one another at any angle 6 : every pole in such a 

 sheet will be reflected backwards and forwards between the 

 two edges according to optical laws ; and hence it should have 

 a finite series of images lying on a circle and agreeing with 

 those in the kaleidoscope, the reflection coming to an end as 

 soon as a couple of images fall in the space which is behind 

 both mirrors. But here arises a difficulty. When 6 is a sub- 

 multiple of 7r, the last couple of images coincide and the whole 

 set is symmetrical with respect to both the straight lines, as is 

 required. But if 6 is not a submultiple of ir, all that happens 

 in the case of the kaleidoscope is that the last two images sepa- 

 rate, the apparent number of images being simply increased 

 by 1. But now the images are not arranged symmetrically 

 with respect to both the straight lines, and hence they cannot 

 both be flow-lines. In order to obtain symmetry, it is neces- 

 sary to continue the reflecting process as if the back of the 

 mirrors were silvered as well as the front ; and in that case, of 

 course, the number of images becomes infinite. But then some 

 of these infinite images will fall inside the angle 6 (that is, will 

 come into the sheet itself), which will be inconsistent with the 

 given distribution of poles in the sheet. Hence it appears to 

 be impossible for the uniform infinite sheet to contain two 



