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Simple lens question (camera)

  1. May 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Was wondering if someone could help...

    2. Relevant equations
    1/f = 1/u + 1/v
    m = v/u

    3. The attempt at a solution
    Using:
    1/f = 1/u + 1/v

    1/50 = 1/8000 + 1/(v)
    v = 50.3 mm


    Then to calculate the linear magnification of the image, using m = v/u with same v and u values from above? (both in metres)

    i.e.

    m = 0.0503 / 8 = 0.0063

    Though as the image gets flipped upside down it's probably:
    m = -0.0503 / 8 = -0.0063


    Thank you :smile:
     
  2. jcsd
  3. May 22, 2010 #2

    Redbelly98

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    Looks good. In fact the correct equation for magnification does have the minus sign:

    m = -v/u
     
  4. May 22, 2010 #3
    Thanks very much Redbelly98. I hadn't realised that!
     
  5. May 24, 2010 #4
    Is that definitely correct?

    I looked the equation up in my CGP Physics Revision Guide and it states the equation as:

    m = v/u
     
  6. May 24, 2010 #5
    Yes, it is m=-v/u.
     
  7. May 24, 2010 #6
    So substituting with the numbers from the initial post:
    m = -v / u
    m = -0.0503 / 8 = -0.0063

    So the answer for m is definitely a -'ve answer and not +'ve ?
     
  8. May 24, 2010 #7

    Redbelly98

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    Yes. The image is inverted, and m is negative.
     
  9. May 25, 2010 #8
    I had thought, from what I learned in class, that the -'ve sign indicated that the image was virtual.

    I hadn't realised it was an indication of whether the image was inverted.

    Which is why I thought the m would be +'ve since the camera image is real.

    I haven't been studying diagrams for long, so it would appear that I have not quite got the concept correct yet.
     
  10. May 25, 2010 #9

    Redbelly98

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    The fundamental definition of magnification is the ratio of the heights, himage/hobject, with an inverted image treated as a negative height. Many physics textbooks use a geometric argument, using similar triangles, to show that this is equivalent to -v/u. The -v/u formula is generally more useful, since it's more common to give information about the distance from the lens rather than the height.
     
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