1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Focal Length & Position of an object.

  1. Aug 5, 2009 #1
    1. The problem statement, all variables and given/known data

    A 2.0-cm-diameter spider is 3.0 m from a wall.

    (a) Determine the focal length of a lens that will make a half-size image of the spider on the wall.

    (b) Determine the position (measured from the wall) of a lens that will make a half-size image of the spider on the wall.

    2. Relevant equations

    thin-lens equation
    1/f= 1/s + 1/s'

    3. The attempt at a solution

    (a) Used s and s' in the thin-lens equation to find the focal length:
    1/2.0cm + 1/0.3cm = 3.83
    f= 1/3.83 = .261cm

    (b) .261/2 = .1305
  2. jcsd
  3. Aug 5, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    For part (a) (Worry about part (b) later)

    An additional equation you need is the magnification. It will allow you to use the fact that the image is half the size of the object. What is that equation? Also, you use 2.0 cm for s. You can't do this. Quantity s is the the distance of the spider from the lens. Quantity 2.0 cm is the size of the spider's of the image on the wall.

    Also, what is the 0.3 cm in your attempt represent? If it is a mistakenly substituted 3.0 m, then it should not be there. The 3.0 m given in the problem is the distance from the object to its image, i.e. s + s'.

    Can you put it together?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook