210 Dy Searle, Experiments illustrating 4 
Since the distance of M from the centre of curvature of AK is 
(a? — h2)?, we have 
AM =a ({1- J/1—h2/a%}, 
and similarly for BM. Expanding as far as h?, we find 
AM=th?/a, BM=th?/b. 
Hence AB=th? Q/a-+ 1/0) \oeeeeeee eee (2). — | 
Now, as far as /?, 
SK =Vw+h=ut $l/u, TK =Vv +h? =v04+4h/0, 
SX =SM-AM=u-3hJa, TY =TM-—BM=v0-4}h]b. — | 
Hence KK =SK-SX=J( jus Von (3), 
KV =TK—TY =i Ajo 21) (4). | 
) 
Multiplying the optical equation (1) by 2/h? and substituting from | 
(2), (3) and (4), we have 
(L/w +1/a) + (1/v + 1/6) = » (A/a + 1/0), : 
or é Ljutdjo=(e2— 1) A/a + 1/0) See eee (5). 
Hence, the focal length is given by : 
Uf=(e—1)je+1)) (6), 
and the “ power” /’ by 
Fl =1)(Q/e + 1/) eee (7). 
§7. Secondary image with thin lens. Some of the light from 
S (Fig. 2) which falls on B& will be reflected and will strike AK. 
Here it will be again reflected and then, on refraction at BK, it 
will pass out of the lens, forming a secondary image of S at 7. 
Let UT, = 2p. 
If Y, he on 7,K and if T,Y,= T,B, the optical equation is 
XK KY, = 3a B 2. eee (8), 
since the light which moves along the axis has traversed the 
thickness of the lens three times. This equation only differs from 
the optical equation (1) corresponding to the primary image by 
having 3 in place of w and v, in place of v. Hence the method 
of § 6 gives at once 
1/u + 1/o,= Be — 1) (a + 1/b)ee eee (9). 
Hence, if f, be the focal length and F, the power of the lens for 
the secondary image, 
cates ee einer eyes il’ Tl 
2 Sea (5+ |= ek: 10), 
and ho veeiesd ee (11) 
