in this problem we're given two diverging lenses and each of these lenses has a focal length of 12
centimeters the lenses are also separated by 12 centimeters and that means that two of the foci
are landing on the lenses so if i label all the foci with which lens they belong to
here's lens number one on the left and there's one of its foci and here's the other focus
landing right on lens two and then the foci that belong to lens two well one of them lands on
lens one and the other one's to the right of lens two and we're placing an object
six centimeters from lens one we can see on the grid that that's three spaces that means each
of these spaces represents two centimeters and our job here is to make a complete ray diagram
using three principal rays at each step so we're going to process the image through lens one and
then we're going to use that image as the object for lens two and get a final image out of this
after we approximate the image distance and magnification using the ray diagram we're going
to back it up with the algebraic calculations using the thin lens and magnification equations
so we start with the principal rays as we process this thing through lens one
and i start with a ray coming horizontally off the head of that object and that's going to be
refracted away from the principal axis so that it comes from the direction of the focus f1
the next ray that i'll choose is the one that passes through the center of lens one
and that's unrefracted and then finally i'm going to choose the ray that aims at the far side focus
for lens one and that ray is going to be refracted horizontal to the principal axis so we're going to
use a dashed line to remind us of the direction that ray was originally aimed but the light ray
doesn't actually go that way so that's why it's a dashed line and the refracted ray coming out
horizontally now on the right side of this lens we see three diverging rays and so what we do
is trace back and find a common origin for these three rays and we already have the traces for two
of these the one on top and the one on bottom but the one in the middle doesn't have a trace
back to the direction that it originates when we observe those diverging rays and all three
of those should intersect at the same point that's a nice way of checking our work and
making sure that these principal arrays all make sense and there's our first image
now because this lands to the left of lens number two we can go ahead and just process that first
image as if it is the object for lens number two there's another case you have to be careful of
where if the first image lands on the opposite side of the second lens you can't treat that
first image as the object for the second lens and i'll post a link to the video where this case is
covered it's called the virtual object case well fortunately we don't have to deal with that here
okay because we're allowed to use the first image as the second object we can go ahead and
just start processing the principal rays from the first image through the second lens and the first
thing i do is take a ray coming horizontally off of that that's going to be refracted away
from the principal axis from the direction of f2 that's one of the foci for the second lens
then the easiest one of course is the ray that passes through the center
of lens two that's unrefracted and then finally we have the ray that's aiming at the opposite focus
and again that one is refracted parallel to the principal axis so again we see three diverging
rays to the right of lens number two and we want to trace those back to a common origin
i can already see where this is supposed to be because the intersection of two of those
is right here and then i just trace back the third ray to verify that it lands in the same
spot i'm tracing backwards from that third ray all three of those do intersect at a common point
so i'm confident i've got the right spot and my final image ends up right here and now we
can get a geometric approximation for the final image distance relative to lens two and the total
magnification of the image so remember that every grid space on this grid is two centimeters and i
count one two three and a little bit it looks like slightly less than seven centimeters for
this image distance i'm just gonna call it like 6.8 i'm going to call that di2 because it was the
image formed after processing through the second lens that counts as a negative because it's to the
left of lens two so i'm going to call it negative 6.8 centimeters then my total magnification i like
to use a capital m for that you would find that by taking the final height of this second image and
dividing by the original height so the original object had a height of four spaces on this grid
the final image is slightly over one space so i could call that about a 1.2 spaces for the final
image so 1.2 over 4 for our total magnification and this gives us approximately 0.3 the fact that
it's positive means the image is upright and then 0.3 means it's only 30 percent the original height
next we want to get the algebraic solutions so this means finding the image location using
the thin lens equation and then finding the magnification at each stage using the
magnification equation so we have to split into two parts here there's the processing of the first
image and the processing of the second image so when i write down the thin lens equation for the
first part i'm going to write 1 over d01 that's the object distance for the first lens part of the
process plus 1 over di1 is equal to 1 over f1 well this means 1 over d i1 is 1 over f1 and f1 is 12
centimeters but it's a diverging lens and we count that as negative so i have a negative 1 over 12
and then minus 1 over do1 well the object distance there the object was 6 centimeters from lens 1. so
i have minus 1 over 6. so i combine those numbers on the right in my calculator then i take the
reciprocal of the result to isolate di1 and when i run the numbers on this i get negative four
centimeters i can also find the magnification for the first stage and i'm going to call that little
m1 and that's the negative of the image distance which is negative four so this magnification is
going to come out positive and then divided by the original object distance which was six so end
up with a positive four over six or two thirds now we have to process that first image as the
object for the second lens and the first thing we have to deal with here is that the object distance
is not that four centimeter value it's the distance between that first image and the second
lens and that's 12 centimeters that's the lens separation plus the additional four centimeters
so that's 16. so d02 is going to be 16 centimeters then i plug into the thin lens equation and i have
1 over do2 plus 1 over di2 that'll be the final image distance relative to the second lens
is equal to one over f this means one over d i two is going to be again negative one over twelve
because the focal length of the lens is negative twelve centimeters and then minus a one over
the object distance which we just figured out was 16 centimeters run the numbers on this one and we
find that di2 is negative 6.86 centimeters so that's the real answer and we can go back and
check with our geometric approximation to make sure it's pretty close our approximation was
negative 6.8 centimeters relative to lens number two we got negative 6.86 so this is pretty close
now the magnification for the second stage is again a negative image distance so there's
double negative there 6.86 over the original object distance which was 16. this gives me a
magnification of 0.43 now the final thing we need to do for this problem is find the total
magnification and compare to what we got from the ray diagram and the magnifications just multiply
at each stage so i have the total magnification should be m1 times m2 m1 we wrote as an exact
fraction that was two-thirds m2 we wrote as 0.43 and when i run the numbers on this i get 0.287
which is also a close match for the approximation that we made using the ray diagram so everything
looks good and we're done with the problem if you enjoyed this video or at least found it useful
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