in this video we're looking at a lens combination we have two converging lenses
one with a focal length of 15 centimeters on the left one with a focal length of 20 centimeters
on the right and these lenses are separated by 70 centimeters we've placed an object 20
centimeters from the first lens and notice that i labeled that as d01 because the image produced by
the first lens is going to be used as the object for the second lens and that would be do 2. and
what we're after in this slide is we're trying to make as accurate array diagram as we possibly can
and we're going to be able to approximate the final image location and the final magnification
relative to the original object on the next slide we'll run through the lens equation and do it
quantitatively so we start by taking a ray headed parallel to the principal axis off of this object
that's going to refract through the focus of the first lens then we take a ray headed through
the center of the first lens and that ray is unrefracted and finally we take the principal ray
that passes through the near side focus and that ray is refracted parallel to the principal axis
those rays all converge at a single point and that's where the first image of the arrowhead is
going to form so there's a couple different things that can happen with the lens combination and this
is the simplest one the image from the first lens landed on the left side of the second lens
and rays are diverging from that arrowhead in all directions so i'm allowed to use the first
image just as a simple object for the second lens the second case is when the first image lands on
the wrong side of the second lens that's called the virtual object and it makes the ray diagram
really tricky and i'll post a link to an example when i have it done so thankfully we have this
simple case here and i treat my first image as the object for the second lens i'm going
to take a ray parallel off of that and that'll be refracted through the focus on the opposite side
i take a ray passing through the center and that's unrefracted and then i'll take a ray
coming from the direction of the near side focus and unfortunately we've run out of
grid space here and we went past the edge of our lens but we know the lens is still at this
horizontal value and that ray coming from the near side focus is refracted parallel
to the principal axis so on the right side of the second lens i see three diverging rays in
other words we have a virtual image here so we're going to trace those rays back to a common origin
and that's where you'll see the object if you look through this lens combination so
there's the first one traced back the second one traced back and the third one traced back and so
the head of the arrow in this virtual image forms at the intersection of those three rays
and we're asked to find the location of the final image relative to lens two
so it looks like it landed almost exactly 20 centimeters to the left of lens two and we'll
check that out quantitatively on the next slide and just a little clarification that first one is
lens one on the left lens two on the right and then our magnification we can approximate that
using the grid spacing my original object had a height of about one and a half spaces
my magnification is going to be the final height divided by the initial height and my final height
looks like it has a little more than nine spaces maybe nine and a half and the image is inverted
so the magnification should have a minus sign on it so we get a magnification of about negative 6.3
and we'll check that out quantitatively in a moment next we're going to handle this whole
thing quantitatively with the thin lens equation and the magnification equation
so the magnification at each stage is given by negative image distance over object distance
and the strategy is just to use the image from the first lens as the object for the second lens
and i'm going to index these as one and two for the first image formation and the second image
formation so for our first lens we had an object distance the distance between the object and the
lens of 20 centimeters and that lens had a focal length of 15 centimeters and we plug into the thin
lens equation and 1 over d i is going to be 1 over f minus 1 over d o so 1 over 15 minus 1 over 20.
we use a calculator to get that 1 over 15 minus 1 over 20 and then we take 1 over the answer to get
the reciprocal and that gives an image distance i really should be good about indexing these things
of di1 equals 60 centimeters the magnification at this stage
is negative di1 over d01 and that gives me negative 60
over 20 or negative 3 which means the size of the image is 3 times the size of the object
and the image is inverted now we take the image from our first lens and we
run it through the second lens as the object and our lens separation was 70 centimeters
but our image was 60 centimeters to the right of the first lens that makes it 10 centimeters to the
left of the second lens so do2 is 10 centimeters the focal length of that lens is 20 centimeters
and i run everything through the thin lens equation so di2 is what we're after and
that would be one over f 2 minus 1 over do 2 that's 1 over 20 minus 1 over 10.
i calculate 1 over 20 minus 1 over 10 in my calculator take 1 over the answer and i've got di2
this gives me an image distance of negative 20 centimeters in other words 20 centimeters
to the left of lens 2. and that agrees exactly with what we got in our ray diagram
next i'll get the magnification at this second stage so that's negative di2
over d02 that's negative negative 20 in other words a positive magnification
in other words the orientation doesn't flip at this stage the image stays upside down
and my object distance was 10 centimeters so i get a magnification of plus 2 at this stage that
just means the size of the object was blown up by a factor of 2 and the orientation was not flipped
finally to get the total magnification you just multiply these magnification factors at each stage
so we blew it up by a factor of 3 and flipped it that's a magnification of
negative 3 and then we blew it up again by a factor of 2 so that's a total magnification
of 6 but with a minus sign because the final image is inverted compared to the original object
from our ray diagram we had predicted 6.3 for the magnification and that's an acceptable difference
a ray diagram should be used as a check on your work not forgetting the actual analytical solution
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