Question
Two identical diverging lenses are separated by 15 cm. The focal length of each lens is -8.0 cm. An object is located 4.0 cm to the left of the lens that is on the left. Determine the final image distance relative to the lens on the right.
Answers
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.
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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