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A couple question on illumination.

  1. Mar 20, 2010 #1
    Q1:

    Two lamps illuminate a screen equally. The first lamp has an intensity of 12.5cd and is 3.0m from the screen. The second lamp is 9.0m from the screen. What is its intensity?

    I know I need to use P=4*pi*(I), but there is no variable included for distance (d). Another formula given is: Eill=P/(4*pi*d^2). How should I go about solving this problem now?
    ----------------------------------------------------------------------------------------------
    Q2:

    A 15cd point source lamp provide equal illuminations on a wall. if the 45cd lamp is 12m away from the wall, how far from the wall is the 15cd lamp.

    Again, there is no variable for distance in the equation: P=4*pi*I.
     
  2. jcsd
  3. Mar 20, 2010 #2

    rl.bhat

    User Avatar
    Homework Helper

    Illumination on a screen is inversely proportional to the square of the distance and directly proportional to the luminous intensity of the lamp.
    So L1/d1^2 = L2/d2^2.
    You can use this relation both the problems.
     
  4. Mar 20, 2010 #3
    .......
     
    Last edited: Mar 20, 2010
  5. Mar 20, 2010 #4
    Instructor's answers:
    #1: 110 cd
    #2: 6.9 m
    ------------------------
    My answers and work:
    #1:

    12.5 cd xxxxxx cd
    -------- = --------
    3.0m^2 9.0m^2

    My answer: 112.5cd OR 113.0 cd.
    ---------------------------------------------------
    My answers and work:
    #2:

    45 cd 15 cd
    ------ = ------
    12 m xxxx m

    My Answer: 4.00 m
    ---------------------------------------------------

    As you can see, I'm just slightly off from what my instructor's answers were.

    So, are my answers correct?
     
  6. Mar 20, 2010 #5
    Bump. Am I correct?
     
  7. Mar 20, 2010 #6
    Bump. All I need is clarification...
     
  8. Mar 20, 2010 #7

    rl.bhat

    User Avatar
    Homework Helper

    First answer is correct.
    In the second problem
    45/12^2 = 15/d^2.
    Solve for d.
     
  9. Mar 20, 2010 #8
    Thank you, sir. :cool:
     
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