WEBVTT
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This is College Physics Answers
with Shaun Dychko.
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We have a thin lens with a focal
length of 100 millimeters
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and our object is placed at 103
millimeters from the lens.
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And the question is
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where will the image be?
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Because that's where one would
place the projection screen
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and it's going to be an inverted
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and magnified image.
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So in other words the question
is what is *di*?
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So we use the thin lens equation and
we're going to solve it for *di*
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so we'll subtract one over
*d naught* from both sides.
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And we have one over image
distance is one over
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focal length minus one
over object distance
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and then we'll take the reciprocal
of both sides
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so the image distance then is
one over the focal length
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minus one over the object distance
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and then take that difference to
the power negative one.
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So that's one over 100 millimeters
minus one over 103 millimeters
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all to the negative one
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which gives 3433
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and make sure you notice that
since I plugged in millimeters
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into the equation here our answer
is also in millimeters
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and we'll multiply that by one meter
for every thousand millimeters
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and get 3.43 meters.
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So that's the distance that the screen
should be from the lens
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on the other side
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compared to the object.
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Now we want to figure out what the
dimensions of the image will be
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given the known dimensions of the object.
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So magnification which is the
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image height divided by
the object height
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we know is the negative of
the image distance
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divided by the object distance
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and so we'll find the image height
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by multiplying both sides
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by the object height that we know
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and then we solve for *hi*
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and it’s object height times
negative image distance
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divided by object distance.
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And we know this image distance
and object distance.
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So that's negative 3.433 meters
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divided by 103 times ten to
the minus three meters.
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This is the object distance.
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And this gives a height in the
image of negative 33.33
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multiplied by whatever the
object dimension is.
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So we’ve found the magnification
in other words
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and then we use that to figure out
the length and width of the image.
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Now we don't really care about
the negative sign here,
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the fact that it's inverted,
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so it does take the absolute value
of that magnification.
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So the length of the image is
going to be 33.33 times
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24 millimeters.
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And that's going to work out
to this many millimeters
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which we multiply by one meter
for every thousand millimeters
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giving length that 0.8 meters.
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And then the width will be 33.33
times 36 millimeters
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giving an image width of 1.20 meters.