in this video we're looking at a single diverging lens with a focal length of 12 centimeters and
actually i should call that negative 12 centimeters because in the thin lens equation
we're going to have to call it negative and we have an object sitting at a distance of
8 centimeters from this thing so that's inside the focal length and we're asked to produce a
ray diagram with all three principal rays to approximate the image location and the
magnification and then we're going to find the real answers by using the thin lens equation
and the magnification equation so let's get our principle raised into this thing
first a ray leaving the head of that object through the center of the lens is undeflected
next a ray leaving parallel to the axis for the problem will be bent away from the axis such that
it appears that it's coming from the nearside focus so that's what it looks like where i've
used a dashed line to indicate a place where there's not actually physically a light ray but
that's the direction it's coming from and then finally a ray headed toward the opposite focus
so again a dashed line where the light ray is not actually physically at but it was aiming towards
the opposite focus that's going to be refracted parallel to the axis now on the right hand side
of this lens i see three diverging rays so where the image is formed is where we perceive the
convergence of these three rays so there should be a common point that they're all coming from so
i only have to trace back on one of these because the others already have traces back in the diagram
and i see that all three of these diverging rays have a common point of origin and that's where
we're going to see the image form and that's an upright image and i can see that it's shorter
than the original object now i get into my approximations so what's the approximate image
distance in this diagram i can see that every grid space counts for two centimeters because the
original object was four spaces left of the lens that was an object distance of eight centimeters
the foci have a distance of 12 centimeters from the lens that was six spaces in the diagram
so where is our image formed it's a little more than two spaces so it looks like a little bit
less than five centimeters and because that's on the near side of the lens we call that negative
i'm going to go ahead and call this negative 4.8 then for the magnification my original height for
the object was two grid spaces my final height is a little more than one grid space i'll call
it maybe 1.2 so my magnification would be the factor this thing shrank by so 1.2 divided by 2
and that gives me 0.6 and that counts as positive because the image is upright
now to get the real answers i'll use the thin lens equation in the magnification equation so i
have one over d i plus one over d o is one over f remember f counts as negative for a diverging lens
so i have one over d i is equal to one over f so negative one over twelve minus one over d
o that was a one over eight so i take care of the right hand side in my calculator and then i take
the reciprocal of the result to find d i and i end up with an image distance of exactly negative 4.8
centimeters and you'll have to take my word for this or if you don't believe me watch my other
thin lens videos i don't cheat on these when i was approximating this i was actually just using
the ray diagram and as i was guessing what number to write down on that i just got lucky this time
and got exactly the right answer now we'll wrap things up with the magnification that's negative
d i over d o the minus sign is there just so that we can count an upright image as positive and an
inverted image as negative so when i plug in my image distance of negative 4.8 centimeters then
my object distance of 8 the double negative is going to make the answer come out positive and
my 4.8 over 8 comes out to exactly 0.6 and we're done if you enjoyed this video or at least found
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