WEBVTT
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This is College Physics
Answers with Shaun Dychko.
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We're told that the letters on a page are
both three and a half millimeters high.
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That is the object height.
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We also know what the object distance is.
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The book is held 30 centimeters
in front of the eye.
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And so, here's the lens of the eye
and then there's the book here.
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That's the object distance
with 30 centimeters.
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And, we're told also that
the lens retina distance
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should be considered two centimeters
for all the problems in this chapter.
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And so, we have these three pieces
of information to work with.
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Now, we have to find out
what the image height is.
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And so, what height will this three and a
half millimeter letter on the page have
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when it's on the retina.
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So, we know that magnification is image
height divided by object height,
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and that equals also negative of image
distance divided by object distance.
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And, we have three of these things
and we need to solve for *Hi*.
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So, we multiply both sides by *Ho*.
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So, the image height then is
negative of image distance
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times object height divided
by object distance.
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So, that's negative of two centimeters
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written as times ten to
the minus two meters
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times three and a half times
ten to the minus three meters
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divided by 30 times ten
to the minus two meters.
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And, this gives this many meters,
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which we'll convert in to millimeters
and it's negative 233 millimeters.
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So, this negative sign means that the
letter is flipped on the retina,
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but then your brain is smart enough
to flip it back right side up again.
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Now, part B is asking compare
the size of the print
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to the size of the rods
and cones in the fovea
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and the fovea is the
center part of your eye.
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And, we'll see how much
resolution could be possible.
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So, now the possible
resolution will depend on
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how the area of a cone compares with the
size of the image that's on the retina.
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So, we need to figure out
the area of one cone.
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Now, the information we
are given in section 26.3
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is that there are six million
cones in the fovea
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and the fovea has the size
of about five millimeters.
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So, we'll find the area of the cone
is going to be the area of the fovea
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divided by the number of
cones that are there.
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And, we can also say that the area of a cone
is pi times its diameter squared over four.
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And then, we'l figure out the
diameter of the cone down here
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and find that it's two micrometers.
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So, the area of the fovea then is
pi times the fovea's radius squared
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divided by the number of cones.
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And, we don't know what the radius
is but we know the diameter,
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so we'll substitute that in, diameter
over two, in place of radius.
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This works out to pi times
diameter of the fovea squared
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over four times the number of cones.
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And, the area of a cone also is pi times
diameter of a cone squared over four.
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So, we can equate this with this because
they're both equal to area of the cone.
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And, we did that here.
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And then, we can solve for *Dc squared* by
multiplying both sides by four over pi.
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And then, take the square
root of both sides.
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So, the diameter of a cone is
the diameter of the fovea,
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which we know is five millimeters,
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and divided by the square
root of the number of cones.
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So, that's the square root of six million,
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which is 0.002 millimeters,
which is two micrometers.
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So, two micrometers is much smaller
than the 233 micrometer image size.
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This 0.223 millimeters is the
same as 223 micrometers.
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And so, because the cones
are smaller than the image,
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that means you will
have a good resolution.