1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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!

Intensity of a laser through a converging lens

  1. Apr 3, 2009 #1
    1. The problem statement, all variables and given/known data

    A laser beam passes through a converging lens with a focal length of 19cm. At what distance past the lens has the laser beam's intensity increased by a factor of 6?

    2. Relevant equations

    [tex]I=\frac{c\epsilon_0 E_0^2}{2} [/tex]??

    3. The attempt at a solution

    To be honest, I'm not even sure how to start on this question. I think I just need a hint to get started. Any help would be greatly appreciated.
  2. jcsd
  3. Apr 3, 2009 #2


    User Avatar
    Homework Helper

    Intensity is inversely proportional to square of the distance. Assume that at focus the intensity is maximum, say Io.
    What will be the intensity (I1) at the lens position?
    What will be the intensity (I2) at a distance x from the focus?
    What is the relation between I1 and I2?
  4. Apr 3, 2009 #3
    Thanks for the quick reply.

    I'm now trying to understand the formula given in my text [itex]I=\frac{P}{A}[/itex]

    So I'm thinking that [tex]I_1=\frac{P}{\pi r^2}[/tex]?

    If so then [tex]I_2=\frac{P}{\pi r^2}=6I_1=\frac{6P}{\pi 0.17^2}[/tex].
    so r=0.0694m. Thus the distance from from the lense is 10.1cm. I entered this into my assignment online and was informed I was incorrect. Any idea as to where I went wrong?
  5. Apr 3, 2009 #4


    User Avatar
    Homework Helper

    Check your calculations. From where did get 0.17?
    The equations should be I1 = P/pi*0.19^2
    And I2 = P/pi(0.19- x )^2 where x is the distance from the lens. Put I2 = 6I1.Solve for x.
  6. Apr 3, 2009 #5
    Thanks. Somehow .17 turned into .19. I've got it now. Really appreciate the help.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook