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Thin lens equation and image formation

  1. Apr 21, 2014 #1
    1. The problem statement, all variables and given/known data
    An object 4.29 cm high is projected onto a screen using a converging lens with a focal length of 32 cm. The image on the screen is 51.7 cm. Where are the lens and the object located relative to the screen?

    ho=4.29 cm
    hi=51.7 cm
    ƒ=32 cm



    2. Relevant equations
    1. m=(hi)/(ho)

    2. m=ƒ/(ƒ-do)

    3. m=-di/do


    3. The attempt at a solution
    I used equation number 1 to calculate the magnification since I was given the heights of the image and the object. I got 12.0513.

    m=(51.7cm)/(4.29cm)=12.0513



    Next, I used equation 2 to solve for do. I got 29.345 cm

    m=ƒ/(ƒ-do)

    (ƒ-do)=ƒ/m

    ƒ-ƒ/m=do

    32cm-(32cm/12.0513)=do=29.345 cm

    Since do is measured as the distance from the surface of the lens/mirror, the object is at a distance of 29.345 cm relative to the lens.



    Now that I have values for do and m, I can use another version of the magnification equation to solve for di. I got -353.645 cm

    m=-(di/do)

    -mdo=di

    -(12.0513)(29.345cm)=di=-353.645 cm ←which would be relative to the lens, on the opposite side.


    The question asks first for the distance between the lens and the screen that formed the image. Here is where I am not sure if I am thinking about it correctly.

    An image would have a height of hi at a distance di from the lens. Since the di I calculated was -353.645 cm, the image was 353.645 cm away from the lens on the right side(taking the left side as the location of the object).

    So the lens is 353.645 cm from the screen


    The next question asks for the distance of the object from the screen. That would be the length of the distance between the screen and the lens (353.645 cm), plus the length of distance from the lens to the object (29.345 cm).

    So the object is 382.99 cm from the screen.

    However they are both wrong. The biggest problem with these problems are that mistakes are typically due to incorrect usage of sign conventions, or from misinterpreting the question asked. Right away I see that 382.99 cm is a pretty big number compared to the other values for image height, object height, focal length, etc. Maybe I just haven't done enough problems to judge whether that is average or not.

    Any help would be appreciated.

    P.S. I know the answers are wrong because the homework they are on require me to type in the answer in a blank field which I then "submit" by pressing a "Check my work" button. I get 3 attempts for this question, this is my first attempt and it was wrong.

    Thanks for the help.
     
  2. jcsd
  3. Apr 21, 2014 #2
    What about significant figures? I would consider wrong anything but 350 cm from the screen for the first and 380 from the screen for the second.
    I would not assume that 32 cm is an exact value, therefore, as you used it in a division, you should know that there are only 2 significant figures in the answer.
    This may not be the cause of the wrong answer, however.
    Overall, your solution seems correct to me.
     
  4. Apr 21, 2014 #3
    I always forget to account for sig. fig's in the final answer. I tried using 3 sig. fig's (354 cm and 383 cm) as well as only 2 sig. fig's (350 cm and 380 cm) but they were both still wrong. Although I get 3 attempts, I can refresh the page and have a few more attempts, so I was able to check these whilst not wasting any of my attempts.
     
  5. Apr 21, 2014 #4
    Are you sure you got correct data?
    I mean,
    ho=4.29 cm
    hi=51.7 cm
    ƒ=32 cm
    for sure?
     
  6. Apr 21, 2014 #5

    adjacent

    User Avatar
    Gold Member

    Yes.
    If you have got a negative answer for the displacement of the image,then it's a virtual image which means that it can't be projected onto a screen.
     
  7. Apr 21, 2014 #6
    I am sure that those values are my given values.

    So if I got a negative value for the displacement of the image, does that mean that when using the equation

    m=-di/do

    should do be taken as negative? to give a positive value for di?

    I don't see how that will change my answer though. Even if the displacement is negative, when calculating images from a converging mirror or lens, it is common to use the concave side (the side with the radius of curvature and focal point) as the positive side, and the opposite side of the lens as negative, so if the image is visible on the opposite side of the lens (because light gets through) wouldn't it cross over to what I have labeled as the negative side of the coordinate axis (if I had set the origin at the center of the lens itself(with negligible thickness))?

    I hope I'm not just confusing myself, or others.
     
  8. Apr 21, 2014 #7

    adjacent

    User Avatar
    Gold Member

    I think I have confused you by saying that.Sorry.
    But think about this:The focal length of the lens is 32 cm.The distance from the object to the mirror is 29.345 cm. That means the image should be virtual.
     
  9. Apr 21, 2014 #8
    What do you mean?

    If there is a projection, I am right to assume that hi should be < 0 or ho < 0 and hi > 0, right?

    No. As I remember, when dealing with lenses, do should ALWAYS be positive as it does not make sense to talk about virtual objects.
    If hi < 0, then m (or A) is < 0 and both p (do) and p' (or di) can be positive.
    In lenses, if I am not mistaken, if the object is in one side and the image on the other, both d (or p) values should be > 0. I may be mistaken.
     
  10. Apr 21, 2014 #9
    If hi is < 0, do becomes > 32 cm, that is why I am inclined to say that the provided data is sloppy.
     
  11. Apr 21, 2014 #10
    I understand (after drawing diagrams) where the image(whether virtual or real, inverted or upright) should appear for mirrors, and I am only slightly less confident of the position for lenses.

    In a problem that superimposes/asks me to superimpose a coordinate system, then I would have to make sure and keep track of the sides which the image forms. In this problem, however, I am just asked for the distance from an object(screen) that is the farthest thing on one side, so I can take the magnitudes of the distance and ignore the sign conventions....right?
     
  12. Apr 21, 2014 #11
    and the question was copied and pasted from the online homework, so the question stated is typed verbatim.
     
  13. Apr 21, 2014 #12
    Yeah, only the magnitude.
    As it does not make sense to me to talk about virtual objects, it does not make sense to talk about negative distances.

    What I am inclined to think is that the problem itself may be wrong. Not your typing.
    I would try to do it with 2 significant figures and using hi = - 51.7 cm.
     
  14. Apr 21, 2014 #13
    mafagafo with the save!

    Yeap, that gave me the correct answers of

    do=34.6553 cm

    di=417.644 cm

    So the distance of the lens from the screen is ≈ 418 cm

    and the distance of the object is 417.644 cm + 34.6553 cm ≈ 452 cm

    Thanks!
     
  15. Apr 21, 2014 #14
    Glad for it.

    Two problems here.
    No need for "≈" when you are eliminating non-significant figures.
    I suggested you to use 2. Maybe 3 will be accepted by the web script, I, however, would not give 100% if there were 3 significant figures.

    You are the most welcome.
     
  16. Apr 21, 2014 #15
    I should not have used the approximation symbol, that was my mistake in expressing the values.

    I do need to do a better job of using significant figures correctly. Thanks for mentioning it.
     
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