Curvature and Refractive Index. 37 



As the light, in its passage to and from the apparent centre 

 ;\, is twice refracted by 2 and once reflected by 1, it would 

 seem at first sight that the value of /j might be obtained by 

 combining in the usual way the expression for twice the focal 

 length of an equiconvex lens with surfaces having the same 

 curvature as 2 and the radius of 1 ; thus 



A 2F + n l - 



But this operation, depending on a false assumption, leads to 

 an erroneous result. It makes /i = — x — , instead of 



TJTD /i ll^ + Rg 



The error arises in this way: — When a double 



fl — I^ + jjlB.2 



convex lens is employed, either to bring or to hurry light to 

 a focus, the bending-powers of the two surfaces depend on the 

 angles they make with the ray in the lens. Now, if one of 

 these angles is great, the other must be small; so that, as a 

 combination, they have the same focus-shortening power, how- 

 ever the light falls on them. But when a ray passing from 

 and returning to the apparent centre / strikes the front sur- 

 face, that surface makes an angle with the ray in the lens 

 which is greater than the mean in the ordinary way ; there- 

 fore the surface produces a greater diverting effect; and hence 

 the distance / is less than it would be if the supposition made 

 were correct. 



All that has been shown at present is only true when the 

 thickness of the lens is inappreciable. When this is not the 

 case, rays, whether from the principal focus F or from an 

 apparent centre /, will not cut the two surfaces at points 

 equidistant from the axis. First, consider the case of an 

 equiconvex lens. Let fig. 5 represent a portion of a thick 

 equiconvex lens. As before, since the central rays are 

 sufficient to give an image, arcs, sines, and tangents may be 

 considered identical. On one surface take any point p. 

 Through it draw a radius pR, and the line pbc parallel to the 

 axis. Now a ray of light parallel to the axis, meeting the 

 surface in the point p with an angle of incidence equal to 6, 



a 



will be refracted so that the angle dpe is equal to - ; therefore 



ii— 1 P 



the angle bpd—&- . Therefore the line pd continued will 



meet the axis in a point a such that 



^ « 



But this ray is diverted at the point d } and bent down so as to 



