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Lenses Question -- image is to be the same size as the object

  1. Mar 18, 2015 #1
    1. The problem statement, all variables and given/known data
    At what distance from an object must a 55mm camera lens be placed if the image is to be the same size as the object?

    2. Relevant equations
    1/u + 1/v = 1/f
    m = v/u
    do/di = ho/hi


    3. The attempt at a solution
    My thinking is that (m) needs to equal 0 therefore v/u needs to equal 0...and that's where i'm up to
    or is it just 55mm

    1/55 + 1/v = 1/55
    0.01 + i/v = 1/55
    1/v = 0.01 - 0.01 = 0
    1/0 = 1/55...

    Any help would be appreciated
     
    Last edited by a moderator: Mar 18, 2015
  2. jcsd
  3. Mar 18, 2015 #2

    BvU

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    Hello Reptic, welcome to PF :smile: !

    What are all those variables in your equations ? in particular the last one ? Do you need that one ?

    Can you say something about v and u if you know m = 1 ? So what does that give when you use the first equation ?
     
  4. Mar 18, 2015 #3
    Hi thanks for your reply!

    I wasn't sure about the last one, thought it might be relevant as it says that object height and image height is the same as image and object distance...i guess not

    If m is 1 then would that mean that u and v need to equal 1 to find the answer...i don't know how to write that into the first
    equation

    is there a specific equation i can either use or rewrite?

    Im no closer to an answer unfortunatly
     
    Last edited: Mar 18, 2015
  5. Mar 18, 2015 #4

    BvU

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    object and image distance... hadn't you chosen different variable names for those in the first equation ? That way you risk confusion !

    And what do you know about ho/hi ?

    No, u and v do not have to be equal to one. They only have to be equal to each other, so that v/u = 1. In other "words": v = u.

    That means that in the first equation you can replace one of the two by the other, e.g. replace u by v. What do you get when you do that ?
     
  6. Mar 18, 2015 #5
    Im afraid i am now completely lost:

    Very little, in fact nothing

    I don't see how interchanging the values of v or u makes a change if they're the same. Then even if i do i don't know how to put them into an equation that gets me towards the correct result.
     
  7. Mar 18, 2015 #6

    BvU

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    If the image is the same size as the object, and you divide the image height by the object height, what do you expect you will get ?

    No, interchanging is not very useful, I agree. But it is very useful if you replace one by the other and write down what you get. Why don't you just do it, e.g. replace u by v. What do you get when you do that ?

    PS. Do you by any chance have an old reflex camera with a detachable 55 mm lens lying around ? (In fact any lens with a known focal length will do). Place the lens in front of you, a candle on one side and with a piece of paper find where the image is on the other side. Play around until flame and image of flame are about the same size, and at that configuration measure v and u !​
     
    Last edited: Mar 18, 2015
  8. Mar 19, 2015 #7
    1/f = 1/v + 1/u

    1/55 = 0.5 x 1/55 + 0.5 x 1/55

    0.0181 = 0.009 + 0.009

    0.009 = Approx 111 mm??
     
  9. Mar 19, 2015 #8

    BvU

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    What you do is not what we normally call solving.

    Solving means substituting variables by expressions from the equations and juggling the symbols until you end up with a usable expression (in symbols) for whatever you want to find.
    Then you check the dimension(s). Only then you substitute numbers.
    Next: estimate order of magnitude (and approximate value, if possible) and finally you employ a calculator.​

    In this excercise the dimension check is almost unnecessary and the calculator isn't needed.

    If you replace u by v in
    1/f = 1/v + 1/u​
    what do you get ?
     
  10. Mar 19, 2015 #9
  11. Mar 19, 2015 #10

    BvU

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    Almost. You get 1/f = 1/v + 1/v = 2/v. Now how can you rewrite 1/f = 2/v in such a way that you get something like v = ... in terms of f ?
     
  12. Mar 19, 2015 #11
    v = 2 x 1/55?
     
  13. Mar 19, 2015 #12

    BvU

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    No way:
    1. it now has the wrong dimensions: length on the left and 1/length on the right
    2. there is no f in sight. Still too early to put in numbers!
    If you have 1/f = 2/v you multiply left and right by f (this is allowed, because f is not equal to zero). That gives you 1 = 2f/v . (Still ok dimensionally !)
    Repeat the trick: multiply left and right by v (also allowed: v = 0 can be readily excluded) and tataaa: an exact expression for v ! No need for "approx" and no need for a calculator (I hope :wink:)

    I have somewhat violated the PF rules, but I'll take the punishment in stride if this helps you understand how to deal with this kind of exercises in a physicist's way :smile:
     
  14. Mar 19, 2015 #13
    hahaha put it like that and nothing seems easier so the answer is just 2f or 110

    so to express it in the answer i would say something like

    1/f = 1/v + 1/u
    ∴ 1/f = 1/v + 1/v = 2/v
    1/f = 2/v
    1 = 2f/v
    v = 2f
     
  15. Mar 19, 2015 #14

    BvU

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    Well done ! And it's a good exercise to draw a diagram showing the rays in this special configuration (m = 1)

    They show it about 2/3 of the way down in this link
     
  16. Mar 19, 2015 #15
    Thank you so much for that. I look forward to using that in the future...
     
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