Posted by matt on Wednesday, May 30, 2012 at 6:52pm.
For the first divergent lens
F1= - 8 cm,
object distance u = 4 cm.
Apply
1/F1 =1/v+1/u,
plug the value of F1 and u, and
solve for v
-1/8 = 1/4 + 1/v,
the image distance for first lens (to the left ofthe first lens)
v = -2.67cm.
Distance between two lences
d= 15 cm.
Image formed by the first lens is object to 2nd lens,
then object distance for 2nd lens is
u1 =15+v =17.67 cm (here take only magnitude,
because we have taken sign conventions)
Focal length of 2nd lens
F2 = - 8cm
apply
1/F2 =1/v1 +1/u1
plug values of F1 and u1 we get v1(that is image distance to the left of 2nd lens)
v1 =- 5.5 cm
-represants towards left of second lens.
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