214 Dr Searle, Experiments illustrating 
Distance from lens of pin and coincident image formed by light 
reflected at BK =p=51:90 em. Hence P= 1/p=0:019268 cm.™. 
Distance from lens of pin and coincident image formed by light 
reflected at AK =q = 52°38 cm. Hence Y=1/q=0-019091 cm... Hence, 
by (15), §8, 
Ff 0:009818 
— - = = ———————ux—_— = 5 9 5 
Le pea =e ao) ONT DS wee 
This value agrees closely with that obtained from the secondary image. 
Also 2(P+Q)=0:07672cm.+ and £+¥,=0°07691cm.~. By 
(16), § 8, these two quantities should have the same value. 
§11. Secondary focal lengths for a system of two thin lenses 
in contact. Let AKB, CLD (Fig. 5) be two thin lenses in contact. 
Let the radii of the faces AK, BK, CL, DL be a, b, c, d, the radu 
being counted positive when the surfaces are convex, as in Fig. 5. 
Let the planes KM, LN through the edges of the lenses cut the 
axis in M, N, and let MK =NL=h. Let S be a luminous object 
point and let 7, be one of its secondary images formed by rays 
which have suffered two reflexions in their passage through the 
system. Let SM=u, NT,=»,. Let the refractive index of the 
lens AKB be p and let that of the lens CLD be wp’. 
The position of T, is found by making the optical length of 
the path from S to 7, for the ray which starts along SK equal to 
that for the ray which starts along SA. 
Since the lenses are “thin” and since the angle KSA is 
“small”, the ray between K and L is never inclined to the axis at 
more than a “small” angle, and consequently the optical equation 
will be, to the accuracy required, the same as if the path of the 
ray were actually SKLT,. 
If X, Y, be points on SK, TLL such that SX =SA and 
T,Y,=T,D, the optical length of the path from X to Y, is to be 
equal to that from A to D. 
