in this video we're looking at a lens combination with a diverging lens on the left a converging
lens on the right and we have an object sitting five centimeters to the left of that diverging
lens now just to look at the specs on the problem real quick the diverging lens has a focal length
of negative 20 centimeters and the minus sign there is to use in the thin lens equation
the two foci for this lens have a distance of 20 centimeters from the diverging lens
and given that the object distance is 5 centimeters from that diverging lens we
can see on the grid that every two spaces counts for five centimeters so i have 5 10
15 and 20 and i see f1 is the distance of 20 centimeters to the left of the diverging lens
i can do the same thing finding f1 to the right of the diverging lens 510 15 and 20. then our
converging lens is 15 centimeters away from this diverging lens that's six spaces two four six
away from the diverging lens and if i start at that converging lens and i count spaces to the
left and right i'll find my focal length at 30 centimeters away so again every two spaces is 5
centimeters 5 10 15 20 25 30 there's f2 on the left 5 10 15 20 25 30 there's f2 on the right
okay now that the introduction is out of the way what's the point of the problem we're going to do
a complete ray diagram using all three principal rays at each stage to find the final image formed
by the combination of these two lenses we're going to try to be as accurate as we possibly can and
use our ray diagram to make approximations for the final image distance relative to the second lens
and the final magnification of the image in the second part of the video we're
going to actually algebraically determine that final image distance and total magnification
and hopefully we see pretty good agreement between the two approaches so let's get
started with the ray diagram the first thing we do is take a ray parallel to the principal axis
off of that object and that's going to refract away from the principal axis as if it's coming
from f1 next we can take the ray emanating from the head of the object and passing through the
center of that first lanes and that's unrefracted and finally we can take a ray that's headed at f1
on the opposite side of the lens and that one is refracted so that it's parallel to the principal
axis now on the right side of the diverging lens i see three diverging rays and if we trace these
back to a common origin that's where the first image forms two of them are already traced back
and i can see that they intersect right here and i can verify that everything is correct by
tracing that middle ray back and seeing that it crosses in the same place as the other two rays
traced back so this is where the first image forms and now because this first image lands
to the left of the second lens we can treat this as a real object that will produce an image when
it interacts with that second lens if your first image lands on the wrong side of the second lens
that's the virtual object case and i'll post a link to an example where i treat that case
the one we're looking at is simpler i have a real second object i can treat it just like an ordinary
object interacting with that converging lens for the second stage so we start to take rays off of
this first image and the first one i'm going to look at is one that comes off parallel to the
principal axis and that's going to be refracted through the focus for the second lens which was f2
way out here then i look at a ray that comes off at the right angle to pass through the center
of the second lens and that is unrefracted and finally i look at a ray that's coming at the
second lens from the direction of f2 on the near side and that ray is refracted parallel to the
principal axis now i have a final set of three diverging rays to the right of the second lens
and i need to trace all these back to a common origin when i get to that common origin that's
where the final image is going to form so i'm going to go way back here and i can see that
these three rays converge somewhere around here and this is where our final image is going to form
now we can get into our geometric approximations for the final image distance and the magnification
that final image distance we're going to call that di2 and that's the distance between the
final image and lens number two now if i start counting spaces again every two spaces is five
centimeters so i have 5 10 15 20 25 30 35 40 45 and almost 50. i'll just call it 49. now with the
sign conventions for lens problems i'm going to call that a negative 49 meaning it landed
on the same side of that second lens my total magnification the notation for that is a capital m
and i would take the size of the final image and divide by the size of the original object well the
original object is a little bit obscured now but it looked like it was about one and a half spaces
on the grid and then my final image is a little bigger than three spaces i'm going to call it
about 3.3 and this gives me a total magnification of about 2.2 now these geometric approximations
are very sensitive to how perfectly you make the ray diagram but we're hoping this agrees
reasonably with what we get using the thin lens and magnification equations so let's get into
the algebraic approach to the problem first i want the final image distance so i'm going to
label the original object distance between the original object in the first lens as d01 and
remember that was five centimeters and the focal length of that first lens that diverging lens
was negative 20 centimeters and we plug into the thin lens equation and we solve for one over di1
that's the image distance for the image produced by the first lens and that gives me 1 over f1 so
a negative 1 over 20 minus 1 over do 1 so minus 1 over 5. so i smash the two numbers on the right
in my calculator and then i take the reciprocal of the result to get di1 this gives me an image
distance of negative four centimeters let's go back to the ray diagram and see if this looks
correct so i'm back in my original ray diagram and this is the first image that i'm talking about
and given that every two spaces is five centimeters well this is a little less than
two spaces so it's reasonable to say that's about four centimeters so things look good so far now i
need to use that first image as the object for the second lens and i've got to figure out how
far is that first image from the second lens well the separation between lenses was 15 centimeters
and this image that i'm using as my second object that was four centimeters additional
to the left of lens one that gives me a total of 19 centimeters from lens two and the focal length
for that converging lens was 30 centimeters and i plug into the thin lens equation again so 1 over f
is 1 over 30 and then i have a minus 1 over do2 so minus 1 over 19. we smash the numbers on the
right hand side take the reciprocal of the result and i get a final image distance of negative 51.8
centimeters going back to the previous slide remember that we approximated that
image distance as negative 49 centimeters so we have a very small percent difference here
and we're happy with that now let's look at the magnification the magnification has to be broken
into two stages you have the magnification created by the processing of the image through each lens
and at each stage the magnification i'll just move my first one down here
across from my first image distance my magnification is going to be the negative
of the image distance over the object distance and that minus sign there is included to make it so
that if an image is upright it's a positive magnification and if the image is inverted
it's a negative magnification well my image distance for that first lens was negative four
my object distance was five and i get a positive four over five or positive point
eight my magnification at the second stage again is going to be the negative of the image distance
over the object distance remember the object distance was 19 there
and again i get a positive magnification and this time it comes out to 2.73
now my total magnification is just the product of the two magnifications and that's 0.8 times 2.73
and when i run the numbers here i get 2.18 positive for my total magnification positive
meaning the final image is upright and if i go back and check to see what we got in the ray
diagram we got about 2.2 from the ray diagram so that's also a very small percent error
so we're happy with that result as well if you enjoyed this video or at least found it useful
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