Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Estimating image distance in lens and mirror

  1. Jul 9, 2010 #1
    I'm having difficulty understanding how to estimate the location of an image in a lens/mirror.

    I tried understanding it by going off the books question/answer section:
    "A spherical concave lens has a focal length of 16 cm, and an object is placed 8 cm from the lens." It says the answer is 6.0 cm.

    What formula is being used, I cannot figure out how they got 6 cm? All I see in the book for formulas is f=R/2 and R=2f.


    I'm confused :(
     
  2. jcsd
  3. Jul 9, 2010 #2

    jtbell

    User Avatar

    Staff: Mentor

    For a lens, it's usually called the "thin lens equation", sometimes with "Gaussian" in front of it. Your textbook surely has it. Try the index at the end of the book. :smile:

    The two formulas you named, give the focal length of a mirror with radius R. There's a similar (but a bit more complicated) formula for the focal length of a lens, which is usually called the "lensmaker's equation." But you don't need that for this problem.

    Oh, and either you left out a minus sign, or the book has a typographical error, or it's being deliberately sneaky. A concave lens has a negative focal length, not positive.
     
  4. Jul 9, 2010 #3
    OK, I believe I have it after running a search for it...it's actually not in the book :(


    Is the equation:

    (1/Object Distance)+(1/Focal Length) = (1/image location)
    (1/45) + (1/36) = .05; then (1/.05) = 5.999

    This is it? I'm just making sure :)
     
  5. Jul 9, 2010 #4

    jtbell

    User Avatar

    Staff: Mentor

    It's actually usually written as

    [tex]\frac{1}{o} + \frac {1}{i} = \frac {1}{f}[/tex]

    where o, i and f are object distance, image distance and focal length. Some books use s and s', or p and q, instead of o and i.

    Why are you using 45 and 36 for object distance and focal length, when your original problem had 8 and 16? (actually -16 because it has to be negative as I said before)
     
  6. Jul 10, 2010 #5
    I have no idea where 45 and 16 came from, sorry it was a long night :blushing:

    I input (1/8o) + (1/-16i) = .0625f, then go (1/.0625f) and come out with 16.

    Either I'm missing something big, or I haven't drank enough coffee this morning. I even plugged the numbers into this website (http://www.photonics.byu.edu/Thin_Lens_Calc.phtml) and it's not coming up with 6 cm for the image location.


    Sorry, I'm not a math person but I'm trying to grasp this :confused:
     
  7. Jul 10, 2010 #6
    OK

    I have figured it out, the book is wrong and it was verified by the professor :)

    I inputted: (1/16f) + (1/8o) = .1875, then (1/.1875)= 5.33.


    They did not use a negative sign.


    Thank you! =)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook