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Thread: Questions relating to magnification and focal points

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    Questions relating to magnification and focal points

    Hi there,

    I have a few questions for you all!

    1. An inverted image half the size of an object is required. If the object must be placed 120mm from the lens, what is the power of the lens and where would the image form?

    2. Calculate the power of the lens required to produce an upright virtual image which is 0.8x the size of the object, if the object distance is to be 100mm.

    3. What is the size and position of the image produced when an object 36mm high is placed 25mm outside the focal length of a +8.00D lens?

  2. #2
    ATO Member HarryChiling's Avatar
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    Lensgrinder wrote a great piece on Ray Tracing and put it up on the ATO site, great piece of work. You'll not only find the answers but you'll understadn them too.

    http://onlineopticianry.com/wordpress/?p=108

    Also I believe that Darryl Meister has an article on Ray Tracing on his site:

    http://www.opticampus.com/cecourse.php?url=ray_tracing/
    1st* HTML5 Tracer Software
    1st Mac Compatible Tracer Software
    1st Linux Compatible Tracer Software

    *Dave at OptiVision has a web based tracer integration package that's awesome.

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    Thanks loads Harry.
    These websites are so useful!
    I wouldn't know about them if it wasn't for optiboard.
    I've got question 3.
    I am just working on 1 and 2 as we speak.

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    ATO Member HarryChiling's Avatar
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    If you need to bounce answers off of me just let me know.
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    Hi Harry
    I am still having loads of problems with question 1 and 2.
    Could you nudge me in a direction?

    I've done 11 questions so far on the topic and nothing has troubled me until these. I think I am missing something?

    Whitney

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    ATO Member HarryChiling's Avatar
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    1. An inverted image half the size of an object is required. If the object must be placed 120mm from the lens, what is the power of the lens and where would the image form?
    inverted image half the size of the object, inverted tells me that the object size went from the positive to the negative, in order to do that one of the factors in our magnification ratio must be negative while the other is positive.

    Magnification = Image Size / Object Size = -1/2 = -0.5x

    Magnification of the size of the object is a ratio equal to the distance from the lens or lateral magnification.

    Lateral Magnification = Magnification = Image Distance / Object Distance = Image Size / Object Size

    What do we have

    Magnification = -0.5x
    Lateral Mag = -0.5x
    Object Distance = 120mm

    Mag = Image Distance / Object Distance
    Image Distance = Mag * Object Distance
    Image Distance = -0.5x * 120mm
    Image Distance = -60mm

    So now we have our image and our object distances, so lets find out the power of the lens.

    ( 1 / Image Distance ) = ( 1 / Focal Distance ) + ( 1 / Object Distance )

    These powers are expressed in Diopters so the distances are expressed in Meters.

    (1/-0.06) = (1/f) + (1/0.12)
    -16.67 = (1/f) + 8.33
    1/f = -25D

    2. Calculate the power of the lens required to produce an upright virtual image which is 0.8x the size of the object, if the object distance is to be 100mm.
    Upright Virtual Image

    Magnification = 0.8x
    Object Distance = 100mm

    That means that our

    Image Size = (+)
    Image Distance = (-)
    Object Size = (?)
    Object Distance = (+)

    Mag = Image Distance (- Virtual) / Object Distance
    -0.8x = Image Distance (-) / 100mm
    Image Distance (-) = -0.8x * 100mm = -80mm

    ( 1 / Image Distance ) = ( 1 / Focal Distance ) + ( 1 / Object Distance )

    (1/-0.08) = (1/f) + (1/0.1)
    -12.50 = (1/f) + 10
    1/f = -22.50

    Hope all that is right, double check it for errors and or run it through a ray trace program. I also made some assumptions that the object distances given were positive like they were writen and since none of the clues pointed otherwise i assumed that it was what it was.
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    Thanks a bunch Harry.
    You pointed exactly what was wrong with the way I have been taught.
    It hadn't been spelt out to me that:
    Mag = Image Distance / Object Distance

    I have also been struggling with rearranging.

    Thanks so much, I think I have these sort of question now.

    Smart, smart man huh!?

    Whitney

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