OF ARTS AND SCIENCES. 305 



and one point of the slit, we see that they will strike the lens at au 

 angle of incidence about equal to i, will traverse it in a plane which 

 we will call the plane of refraction inclined to the first plane i — r, 

 and finally emerge in a plane parallel to the first. The plane of re- 

 fraction will intersect the lens along two circles whose distance apart 

 at the centre will be greater than the thickness of the lens in the ratio 

 of cos r to 1 ; hence their radii R' will be less than the radius of 

 curvature R of the surfaces of the glass in the same ratio, or R' = R 

 cos r. Again, the apparent index of refraction n' will be difFcient, and 



1 " 2(/i — 1) T 1 2(n' — 1) , ., ^n — lR' 



since j=-^-j^ and - = ^^. , we have /' =f~^^ - jr 



z=zf cos r - — -. It therefore only remains to determine n', the apparent 



index of refraction. As the problem is one in spherical trigonometry, 

 suppose a sphere described around the centre of the lens and projected 

 in Fig. 2, the eye lying in the axis of the lens prolonged. Let 

 CA = i, the angle through which the lens has 

 been turned, and CE = r, the corresponding 

 anjile of refraction. Then if the surface of 

 the glass is vertical, as at the centre of the 

 lens, the incident ray will he AC and the re- 

 fracted ray CB. Next suppose the surface slightly inclined by the 

 amount CD=: RC^v, as is the case for the upper and lower parts 

 of the lens. AB=.i' will now be the angle of incidence; and, to 

 construct the refracted ray, we have first the condition that sin ^' = n 

 sin 7-', and secondly that the incident and refracted ray shall lie in the 

 same plane with the normal BCD. To construct it, pass a plane 

 through the normal ^(7 and the incident ray AC, which will intersect 

 the sphere along the great circle AB and FD ; on this, lay oflf BE' 

 = r' such that n sin r'=:sin i\ but, as v is infinitesimal, i' will be sen- 

 sibly equal to ^, and r' to r. Now in the right-angled spherical triangle 

 FGB, sin CD = sin DF sin CFD, or sin v = v = sin i X F, or F=z 



-i-. ; and in the triangle FEE', sin EE' = sin FE' sin FEE', or 

 sin i ' = ' ' 



EE =2 siu (i — r) F, or, substituting the value of i^ just found, EE' 



— V ^—^ — r-^. Calling i" and r" the angles of incidence and re- 



fraction of the ray with regard to the section of the lens made by the 

 plane of i-efraction, then i" will not equal BC, but will be the an- 

 gle which, when projected on the plane of the section of the lens, will 



be BG or i" cos r=.BC .' . i" =z . Again EE' is the angle 



cos r = ° 



VOL. X. (n. 8. II.) 



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