in this problem we have a lens combination and we have an object sitting 25 centimeters in
front of the first lens which is a diverging lens with a focal length of negative 15 centimeters
next we have a converging lens with a focal length of 17 centimeters and that's 20 centimeters from
the diverging lens we're asked to draw a ray diagram including all three principal rays at
each step to find the image we're going to use the diagram to approximate the image distance
relative to the second lens and the total magnification of the image
finally we're going to use the thin lens equation and magnification equation
to find the image distance and magnification to three significant digits and hopefully
these answers compare favorably so we start by getting principal rays through the first lens
a ray coming in parallel to the principal axis will be refracted away from the direction of
the nearside focus second a ray coming through the center of the diverging lens is unrefracted
and third a ray aiming toward the focus on the opposite side of the lens will be refracted
parallel to the principal axis so we have three diverging rays on the right side of this diverging
lens and we trace all those back to a common origin point if you look through that single lens
this is where you would perceive the image to be the common origin of the diverging rays
two of these rays have already been traced back to the common origin point
there's only one left to do those line up pretty nicely and that's where the image
of the head of the arrow is going to form after the first lens next we're going to run this image
through the converging lens to find the final image formed by the combination i start with
a ray parallel to the principal axis and that's going to be refracted through the
focus of that converging lens then i can do a ray through the center of the converging lens
finally i take a ray through the near side focus of the converging lens so that's the 17 centimeter
focus right here and that's going to be refracted parallel to the principal axis now when the rays
come together at these shallow angles it's pretty hard to be precise with the ray diagram but
somewhere around here my rays are converging to form my final image next i'll get my approximate
image distance and magnification and the image distance here i mean relative to the second lens
so every two of these spaces is five centimeters and i just count the spaces and the final image is
a little before the 40 centimeter mark i'm going to call it 38 centimeters and our approximate
magnification can be found by comparing the height of the final image to the height of the original
objects the final image i'll say is about 1.3 spaces the original object was three spaces
there should be a minus sign on this because the final image is inverted so that's about negative
0.43 for the magnification next we want to check our approximations by using the thin lens equation
and magnification equation to get a precise answer for the final position and magnification
of the image so to do this we break it into two steps we do the refraction through the first lens
and then we use the image from the first refraction as the object for the second refraction
and our toolkit here is the thin lens equation to find the image distance at
each step and the magnification equation to find the magnification at each step then we
write down what we know at the first lens and our object distance was 25 centimeters
our focal length was negative 15 centimeters
so diverging lenses have negative focal length and we want to find the image distance out of this
so i take 1 over f minus 1 over do and this can be done in one step in a calculator just negative
1 15 minus 1 25 1 over that answer and i get negative 9.38 centimeters approximately for that
first image distance now these things should actually be indexed for our first refraction
and then when that image passes through the second lens those will all have a subscript of two so our
first magnification i'll call that m1 that's a positive 9.38 over 25 since the image distance
was negative and i get a magnification of 0.375 at that first step then we go into lens 2 so we're
just taking the image from the first lens remember that was an upright virtual image to the left of
the diverging lens and we're running that through the converging lens so to get the object distance
i need the distance to the diverging lens added to the distance between the lenses which was 20
centimeters that's a total of 29.38 and the focal length of that converging lens was 17 centimeters
i solved the thin lens equation for one over d i that's going to be one over f
minus one over d o and again i can do it all at once in my calculator and i get an image distance
of 40.3 centimeters that's to the right of the converging lens my ray diagram approximation
was 38 centimeters so that's not too bad my second magnification is negative di2 over do2
and i get negative 1.37 for this now the only thing left to do is to get the total magnification
of the image and the magnification at each step is just a scale factor for
what percentage you change the height of the object by so these magnifications simply multiply
and the total is given by m1 m2 first we shrank it by multiplying by 0.375
and then we flipped it upside down and made it grow a little bit
by multiplying by negative 1.37 the total effect of this is a magnification of negative 0.514
and our approximation from the ray diagram was about negative 0.43 so that's off by a bit
so we missed by about 15 in our ray diagram estimate and the issue there is that the
image was so tiny that our estimate of its final height had a huge uncertainty in it
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