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Velocity of an image in optics

  1. May 14, 2013 #1
    1. The problem statement, all variables and given/known data
    An object 20 cm. to the left of a converging lens with focal point 0.3 cm is moved to the left at a constant 5 cm/s. How fast is the image moving?


    2. Relevant equations
    v = vo + at
    1/f = 1/u + 1/v

    3. The attempt at a solution
    Since there is no acceleration, I got 5 m/s as the velocity but I don't know how to treat a non-constant value of u. The units don't match up in the lens equation, so what should I do?
     
  2. jcsd
  3. May 14, 2013 #2

    NascentOxygen

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    Can you express u in terms of v and f? Then differentiate both sides wrt time.
     
  4. May 14, 2013 #3
    Can I just differentiate:
    [itex]
    \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\
    0 = -\frac{1}{u^2} \cdot \frac{du}{dt} - \frac{1}{v^2} \cdot \frac{dv}{dt} \\
    0 = -\frac{1}{20^2} \cdot 5 - \frac{1}{0.305^2} \cdot \frac{dv}{dt} \\
    [/itex]
    Having solved for v using the regular 1/f = 1/u + 1/v = 1/0.3 = 1/20 + 1/v, then solve for [itex]\frac{dv}{dt}[/itex]?

    I get dv/dt = -0.001. If this is correct, does this mean the image is moving away from the lens to the right, since the velocity towards the left is a positive value?
     
  5. May 14, 2013 #4

    NascentOxygen

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    That's what I had in mind.

    To check your answer, use this piece of knowledge: as the object moves nearer to infinity, its image moves ever closer to f. (Remember, a converging lens focuses the sun's rays right at f.)
     
  6. May 14, 2013 #5
    My solution is incorrect then because the sign has opposite direction to the object which means it's moving away. What did I do wrong?
     
  7. May 15, 2013 #6
    The object distance is a function of time:

    u(t) = 0.05 t + 0.02

    so I don't think one can just substitute values for

    u2 and v2

    which means you are evaluating the differential equation at t = 0.

    It might be easier to use Newton's formula

    x1 x2 = f2
     
  8. May 16, 2013 #7

    NascentOxygen

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    You should check this in a physics book: but the formula 1/f = 1/u + 1/v suggests to me that for u and v, the positive direction is measured outwards from the lens. So your negative speed indicates motion towards the lens. That is what I expected, it's moving closer to the focus.
     
  9. May 16, 2013 #8
    My professor says the answer is wrong because it doesn't include focal length which affects the velocity.
     
  10. May 17, 2013 #9
    1/f = 1/u + 1/v

    = (v + u)/uv

    uv = fv + fu

    uv - fv = fu

    v(u - f) = fu

    v = fu/(u - f)
     
  11. May 17, 2013 #10
    Solving for v and then differentiating the equation is mathematically the exact same thing as differentiating the equation then solving for dv/dt..
     
  12. May 17, 2013 #11

    haruspex

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    That's a puzzling statement. You used the focal length to compute v, so it is 'included'. Symbolically, you have deduced ##\frac{dv}{dt} = -\left(\frac vu\right)^2\frac{du}{dt} = -\left(\frac f{u-f}\right)^2\frac{du}{dt}##
     
  13. May 17, 2013 #12
    Right, and he insists that the physics is incorrect even though I showed him that (v/u) = (f/(u-f)) he says that it's physically impossible.
     
  14. May 18, 2013 #13

    NascentOxygen

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    You could check your answer by approximating the calculus. For example, determine v and u when the object is in a particular position, then take a tiny time interval, say 0.001 secs, and determine the new v and u after that elapsed time (using plenty of significant figures on your calculator). Knowing ∆u and ∆t, you can calculate a good approximation to its velocity. No one could argue with that approach!

    BTW, I'm confident the analytical method is correct. But if your professor insists that it be answered differently (even if he is wrong) you should do it the way he demands. He is the one who determines your grade.
     
  15. May 18, 2013 #14

    rude man

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    The formula haruspex came up with is correct except I would substitute u = (-20cm - 5cm/s*t) to make dv/dt a function of constants and time only.

    This assumes by "focal point" was intended "focal length". Is that what it's called in the UK or Australia or ???
     
  16. May 18, 2013 #15

    haruspex

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    No, you're right, it should be focal length. Focal point is a place, not a distance.
     
  17. May 18, 2013 #16

    rude man

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    So I suppose they could have meant v = 0.3cm instead of f = 0.3cm? Wouldn't make much of a difference I suppose.

    PS does everyone use u for object and v for image distance nowadays? I always had p and q. u and v should be reserved for velocity ... in fact the OP used v for both image distance and velocity, just to cite one example ...
     
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