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Light - Lens help

  1. Mar 4, 2004 #1
    Can anyone help me on these 2 questions?

    1.An object is placed 30.0cm from a convex lens of f(focal length)=10.0cm. Find the position and the magnification of an image.

    2. An object is placed 20.0cm from a convex lens of f=30.0cm. Find the position and the magnifying factor of an image.

    thanx :wink:
  2. jcsd
  3. Mar 4, 2004 #2


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    It's not clear to me what kind of "help" one can give on a problem like this. I feel sure your text has a formula that you can just plug the numbers into.

    I don't happen to have an optics textbook handy or know the formulas offhand. Here's my rather idiosycratic way of solving a problem like these (may not help you much!):

    Draw a picture: A vertical line segment representing the lens, dots at the height of the middle of the "lens" and distance "f" on either side representing the focal points (in your first problem, f would be 10 cm.), and an upright arrow at distance "A" from the "lens" on the left side representing the object (in your first problem A would be 30 cm.)

    Draw a horizontal line from the object to the top of the "lens". Draw a line from the top of the lens through the "focus" on the other side. A light ray from the object parallel to the line of foci (and perpendicular to the lens) will pass throught the focus (that's what's meant by "focus". All light rays perpendcular to the lens pass through the focus.)

    Draw a second line from the same point on the object through the center point of the lens. Because the lens is symmetric, a light beam through the center will just continue through the lens unchanged.

    The whole point of that is that the image (of that point on the object) will be where those two lines intersect (actually, it is an image because ALL lines from that point through the lens converge there).

    Now that I know the image will be where those two lines intersect, I think about setting up equations for those two lines.

    Set up a coordinate system with (0,0) at the center of the lens. The focus at distance f, on the right, will have coordinates (f,0). The object is at distance A from the lens (and on the left) so any point on it will have x-coordinate -A. I didn't say how high the object was so just call the y-coordinate of my "point on the object" h. The horizontal line from that point to the lens hits the lens at (0,h) and must pass through the focus (f,0). The slope of that line is (h-0)/(0-f)= -h/f and goes through (0,h) so the equation of the line is y= -(h/f)x+ h.

    The other line from that same point on the figure, (-A,h), passes through the center of the lens: (0,0). The slope of that line is
    (h-0)/(-A-0)= -h/A so the equation of the line is y= -(h/A)x.

    I'm looking for the point where y= -(h/A)x and y= -(h/f)x+ h. That means -(h/A)x= -(h/f)x+ h and we can immediately divide through by h. This is the same as (1/f- 1/A)x= 1 or ((A-f)/Af)x= 1.
    x= Af/(A-f) so the image appears that distance to the right of the lens.

    Now, find y: y= -(h/A)x so y= -(h/A)(Af/(A-f))= -(f/(A-f))h. The negative is because the image is reversed. The "f/(A-f)" means that the image will be f/(A-f) times the size of the original object: the "magnification" is f/(A-f).

    Wow! I am impressed that I was able to do that but if I had to do a problem like this on a test, I would surely want to have memorized a formula! In fact, you now have formulas for those- just plug the numbers you are given in each problem for "A" and "f" into those formulas.
    Last edited: Mar 4, 2004
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