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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.

  • physics -

    For the first divergent lens
    F1= - 8 cm,
    object distance u = 4 cm.
    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
    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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