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Lens question

  1. May 5, 2016 #1
    1. The problem statement, all variables and given/known data
    Screen Shot 2016-05-05 at 11.40.31 PM.png

    2. Relevant equations
    1/do+1/di=1/f

    3. The attempt at a solution
    I tried finding the distance of image at 27m and then at 30.5m and taking the difference but that didn't work.
     
  2. jcsd
  3. May 5, 2016 #2

    jtbell

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    Does that mean you tried to find the average speed of the image? I think their talking about "initial speed" indicates that they want instantaneous speed.
     
  4. May 6, 2016 #3

    ehild

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    Your method gives the average velocity in the first second. It is a good approximation, if you calculate the image distances with enough significant digits (five at least). Show your work.
     
  5. May 6, 2016 #4
    How would I perform the calculation with instantaneous speed?
     
  6. May 6, 2016 #5
    I tried doing this but unfortunately this didn't give me the correct answer. Is there a more accurate way of doing this?
     
  7. May 6, 2016 #6

    jtbell

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    How do you get instantaneous speed (or more precisely, velocity) from position?
     
  8. May 6, 2016 #7
    derivative?
     
  9. May 6, 2016 #8

    jtbell

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    Yup.
     
  10. May 6, 2016 #9
    Ok cool, I'm still a little confused how i would apply it to this problem however.
     
  11. May 6, 2016 #10

    ehild

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    What did you get? Yes, taking the derivative of di would be more accurate, but not much different.
     
  12. May 6, 2016 #11

    jtbell

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    You know the relationship between ##d_0## and ##d_i##: $$\frac 1 {d_o} + \frac 1 {d_i} = \frac 1 f.$$ Take the derivative with respect to t, of both sides of this equation, and you'll have a relationship between ##\frac {dd_o}{dt}## and ##\frac {dd_i}{dt}##.
     
  13. May 6, 2016 #12

    jtbell

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    With Δt = 1 s I get about a 13% difference in the final answer, using the full precision of my calculator.

    I agree, if Dan shows us his working, we can tell him if he at least calculated that approximation correctly.
     
  14. May 6, 2016 #13

    ehild

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    Yes, it is true.
     
  15. May 6, 2016 #14

    jtbell

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    Good luck... if I want to get up for work in the morning I need to go to bed now. :oldwink:
     
  16. May 6, 2016 #15
    I am getting .00011 if i do it using the approximation with average velocity. It marks this as incorrect.
     
  17. May 6, 2016 #16

    ehild

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    OK, is the image moving towards the lens, or away from it?

    The method with differentiation must be more accurate. Differentiate the equation ##\frac {1} {d_o} + \frac {1} {d_i} = \frac {1} {f} ## with respect to time and solve it for di'.
     
  18. May 6, 2016 #17

    jtbell

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    Yes, that's what I got. Another way to get more accuracy would be to use a smaller time interval. In principle, if you make it small enough, the answer will get close enough to the exact answer to make your software happy. But you have to be very careful to avoid roundoff errors.
     
  19. May 6, 2016 #18
    Great i ended up doing this and got the right answer. Thanks!
     
  20. May 6, 2016 #19

    ehild

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    If you differentiate the equation ##\frac {1} {d_o} + \frac {1} {d_i} = \frac {1} {f}## you get ##-\frac {1} {d_o^2} d_o' -\frac {1} {d_i^2} d_i'=0##. Solve for di'.
     
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