in this video we have a lens combination so they're labeled as lens number one
and lens number two the first lens has a 20 centimeter focal length and the lenses are 20
centimeters apart which puts the second focus of the first lens sitting right on the second lens
the second lens has a focal length of 15 centimeters this whole grid has been set
up to scale and every two spaces means five centimeters in part a we're asked to draw an
accurate ray diagram for how the image is processed by both lenses
and we should be able to get reasonable estimates of the image distance relative to the second lens
and the total magnification so we begin by using the principal rays for the first lens a ray coming
in horizontally to that first lens is refracted through the focus array coming in through the
center of that first lens is unrefracted an array coming from the direction of the near side focus
from the head of that object is refracted parallel to the principal axis now on the right hand side i
see that we have three diverging rays and we need to trace these back to a common origin
we see that all three rays have a common origin back here and that's where we're going to see
a virtual image of the head of the arrow form and now that virtual image becomes the object
for the second lens so we start taking principal rays and processing them through the second lens
now so i have a ray coming in parallel to the principal axis that's refracted through the focus
on the opposite side of that second lens i have a ray coming through the center of the second lens
that's unrefracted and then a ray coming through the near side focus of the second lens
that's refracted parallel to the principal axis
and i see all three of these rays converging to a point on the opposite side of the second lens
so there's my final image now array diagram isn't going to give you exactly the correct quantitative
results but it should allow you to estimate fairly accurately so the image distance relative to the
second lens i can just count spaces on my grid and estimate that and that's seven and a half
spaces again we're using a grid where every two spaces is five centimeters in other words
every space is two and a half centimeters so i get an approximate image distance of 18.8 centimeters
right of the second lens and then my approximate magnification as far as i can tell it's really
really close to one so our original our original object was about one and a half spaces high our
final object is about one and a half spaces high it looks like it might be slightly less than
that but it's really hard to tell so i'll say our magnification is about one maybe a little smaller
and just one final point on this slide the magnification should come with a minus sign
because i can see that my final image is inverted relative to the original object
now we get quantitative i want the image location relative to lens 2 and i want the total
magnification so really the best way to organize this is in terms of the processing of the image
through the first lens and then we'll do the processing through the second lens
so i write down what i know with that first lens and i was told that my original object distance
so i'll call that d01 15 centimeters from a lens with a focal length of 20 centimeters so i used
the thin lens equation and i solved for one over the image distance for that first lens
that's going to be 1 over the focal length for that lens minus 1 over the object distance and you
can do this all at once in a calculator you do 1 over 20 minus 1 over 15 and then take 1 over that
answer and when i run the numbers i get an image distance of negative 60 centimeters in other words
it's a virtual image on the left side of that first lens just like we saw in the ray diagram my
magnification at that step is going to be negative d i one over d d01 and when i plug in the negative
60 i end up with a positive result and that's 60 over 15 so it's a magnification of plus four
so the first image is four times taller than the original object which also agrees qualitatively
with what we saw in the raid diagram then i look at the processing through the second lens
in my object distance well i'm using my first image as the object for the second lens
and my first image was 60 centimeters to the left of the first lens in addition the lenses
are 20 centimeters apart and this means my object distance is 80 centimeters my focal length for
the second lens was 15 centimeters i plug into the thin lens equation solving for 1 over di2
that's 1 over the focal length which was 15 minus 1 over the object distance which was
80. i smash those numbers in my calculator and take 1 over the answer and i get di2
is 18.5 centimeters so that's the answer to part b and it's pretty close to the 18.8 centimeters that
we predicted from our ray diagram so that's good then i go on to get my magnification at the second
step again negative image distance over object distance so my image distance was positive 18.5
object distance positive 80. so i get a negative magnification which is good because
this is where we saw the image get inverted and i get a magnification of negative 0.23
at this step the only thing left to do is part c the total magnification and magnifications
are just a scale factor telling you how much bigger or smaller the image got at each step
and if you just multiply those as you go through a series of lenses you get the total magnification
so our total magnification is negative which is good we saw a final image that was inverted
and i kept some extra digits in my calculator so i get negative 0.925 from our ray diagram we knew
that the magnification was going to be very close to negative one and probably a little smaller so
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