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!

Calculate the image height

  1. Oct 15, 2012 #1
    1. The problem statement, all variables and given/known data

    A converging lens with a focal length of 30cm and a diverging lens with a focal length of -64cm are 139cm apart. A 4.0cm tall object is 50cm in front of the converging lens.

    distance between image and diverging lens has been calculated to be -32cm, and it is right.


    2. Relevant equations

    M = -(s'/s) = (hi/ho)

    3. The attempt at a solution

    -(-32cm/50cm) = (hi/4.0cm)
    hi = 2.56cm

    I already got the mastering physics problem wrong. Used all my guesses. But the answer is 3.0cm and I have no clue how?
     
  2. jcsd
  3. Oct 16, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The magnifications multiply, and it is not the same as the image distance from the second lens divided by the object distance from the first lens.

    ehild
     
  4. Oct 16, 2012 #3
    I also calculated it and it came out to be 2.5 cm .... from what i see there is no role of concave lens. I took it as a gimmik becouse the image would never reach the concave lens . Its a gimmik.
     
  5. Oct 16, 2012 #4
    Here ...
     

    Attached Files:

  6. Oct 16, 2012 #5

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You did something wrong.
    The first image is at distance of 75 cm after the converging lens. The magnification is N1=-di/do=-75/50. The image height is 6 cm and it is inverted. The first image is object for the diverging lens.
    The object distance from the diverging lens is 139-75=64 cm. The second image distance is -32, as you got. N2=-di/do=32/64. The image height is half of the first image.


    ehild
     
  7. Oct 17, 2012 #6
    Image distancecis coming -150/8 cm

    By the formula 1/v - 1/u = 1/f

    V = image distance
    U = object distance
    F = focal length
     
  8. Oct 17, 2012 #7
    Wait i was taking focal length as negative :P . I started this chappter two days ago ... but the ray is not reaching the concave lens so why consider it ?
     
    Last edited: Oct 17, 2012
  9. Oct 17, 2012 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The ray reaches the second lens if it is big enough. Or draw those rays which reach it :)
    The focal length of the converging lense is positive, that of the diverging lens is negative.
    The second image is virtual: It forms in front of the second lens. You have to draw the rays backwards to get the intersection.

    ehild
     

    Attached Files:

  10. Oct 17, 2012 #9
    Your ray diagram seems tovbe wrong because according to convention rays are drawn left to right :)
     
  11. Oct 17, 2012 #10

    ehild

    User Avatar
    Homework Helper
    Gold Member

    He-he. In my country,they can go in any direction, (the arrows show the direction); moreover we use the lens formula 1/t+1/k =1/f and define magnification as N=k/t and it is positive when the image is inverted.:biggrin:

    Have you understood that second image?

    ehild
     
  12. Oct 17, 2012 #11
    What is t and k ?
     
  13. Oct 17, 2012 #12

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The same as object distance and image distance.

    ehild
     
  14. Oct 17, 2012 #13
    No formula for lens is 1/image distance - 1/ object distance = 1/f
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Calculate the image height
  1. Image Height (Replies: 2)

  2. Image Height Question (Replies: 1)

Loading...