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- [Voiceover] If I had been handed

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this problem on a physics test

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I probably would have freaked out.

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This looks really intimidating

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but it's actually not that bad.

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It's a classic example
of a two lens system

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and overall, before we
get lost in details here.

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The overall idea of how we're going

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to approach is this.

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We've got two lenses.

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This 1st lens is going to create

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an image of this object over here.

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It creates an image of this object.

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Then the 2nd lens is going to create

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an image of that image and so

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what we're basically going to do

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is we're going to use
the thin lens formula.

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We're first going to pretend like

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this 2nd lens doesn't exist.

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We're going to figure out, what image

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does this 1st lens create?

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Then, we'll pretend like the 1st lens

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doesn't exist and we'll treat that image

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that the 1st lens creates as if

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it's the object for the 2nd lens.

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We'll do another calculation, figure out

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where this 2nd lens creates an image

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of the thing it thinks is the object

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but it's actually the
image from the 1st one

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and so we're going to figure out

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first the object gets turned into an image

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from the 1st lens, then the
2nd lens turns that image

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into another image and that's what

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our eye is going to see.

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So that's the question we want to answer

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is what image would our eye see

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in this case if we had it over here

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looking through these two lenses?

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So let's do this.

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First thing we got to do is I'm going

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to pretend like the 2nd lens

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doesn't exist so it doesn't confuse me.

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We use the thin lens equation.

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Alright here we go.

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1 over f, so 1 over, okay here we go,

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12 centimeters, do I make
it positive or negative?

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It's on the left.

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Maybe I make it negative, nope.

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The only thing you look at is what kind

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of lens this is.

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This is a convex lens.

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I can tell by the way it's drawn.

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Since this is a convex
lens my focal length

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is going to be positive 12 centimeters.

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So that's what I plug in,

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Positive 12 centimeters equals, alright,

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object distance so 1
over the object distance

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that's 24 right?

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No, all distances are measured from

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the center of the lens to the thing

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you're looking at and so
for an object distance

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I'd have to go from the center of the lens

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all the way to the object.

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Don't just plug in any
number they're giving you.

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You've got to plug in that distance

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and that's going to be 24 plus 12.

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It's actually 36.

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So I'd have 1 over and this is going

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to be 36 centimeters.

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Positive or negative?

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Well that's going to be positive here.

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We've got two lenses
but for this first case

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it's on the left hand side.

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Everything's cool, it's on the opposite

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side as my eye.

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It's going to be a positive 36.

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I'll show you in a minute how this

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could possibly be negative in a second.

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Plus 1 over the image distance.

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I'll just label that with a di.

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Alright so now we just got to solve.

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If you solve this you going to go

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1 over 12 centimeters
minus 1 over 36 centimeters

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equals 1 over di what you're going to

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end up getting is 1 over 18 centimeters

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on the left hand side equals 1 over di

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but that's 1 over di, if you solve that

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for di you'll get that
di equals 18 centimeters

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positive 18 centimeters.

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And so what does that mean?

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Where is this?

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Well, alright, di,
positive image distance.

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Remember, positive means on the same

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side as your eye for a lens.

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So I'm going to have
an image that's formed

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over on this side of the lens

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and 18 means 18 centimeters from where?

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From the center of the lens.

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So if I draw this.

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I'm going to go 18 from here.

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That's going to be past the focal point

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because I know the focal point is only 12

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so I'm going to be somewhere around here.

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I'm going to label that
and that's going to be

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18 centimeters and that's where

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the first images, so that's where

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the image is going to be that

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this 1st lens creates.

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It's going to be right there.

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Okay so now what do I do?

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I told you, here's what we do.

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We pretend, we'll bring the 2nd lens back.

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We know the image that the 1st lens

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created is right here.

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This 2nd lens is going to think

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that that's the object.

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It's going to create
an image of that image

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as if that were an object.

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So we do another thin lens formula

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but this time we treat this positive 18

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not at if it's the image, we treat it

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like it's the object.

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So this is going to be
well, it's not going

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to be equal to do because remember,

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you have to measure everything from where?

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From the center of the lens so for this

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second calculation, my object distance

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is going to be from the center of the lens

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all the way to, let me not use that color.

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Let me see, let's pick this one here.

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From the center of the lens all the way to

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where it's object is and it's object

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is over here which is not 18.

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That's why it gave you this
distance between lenses.

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The distance between
lenses is 33 centimeters.

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So if the distance from
this 1st lens is 18

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well the difference has to
be equal to this length.

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So 33 minus 18 is going to be 15, whoops.

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15 centimeters will be where the "object"

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for this 2nd lens is going to be.

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So as we turn this positive 18 into

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object distance of
positive 15 centimeters.

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Why positive?

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Well it's still on this left hand side

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opposite side as your eye.

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That's where the objects
are supposed to be.

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Now here, this is where I told you

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there's one case where you can get

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negative object distances if our 1st lens

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would have created an image of this object

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way over here on the
wrong side of the lens.

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In other words, if this
1st lens created an image

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that was closer to our eye than

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the 2nd lens was, well
that's just all messed up.

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That's going to be a
negative object distance.

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Your not supposed to hold an object

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between your eye and the lens.

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So we'd count that, you'd do

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your calculation just the same.

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I mean everything would work out right.

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It's not like the whole
thing's going to break.

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You'd get an image.

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It's just when you do your calculation

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for the second case.

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If your image ended up on the wrong side

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of the lens for the first image

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that was created you'd have to treat it

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as a negative object distance.

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That didn't happen here.

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Our image is still formed on this side.

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We're going to pretend like
this 1st lens didn't exist.

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We got an object over here, we're going to

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treat it like our object.

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Everything's fine.

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Eye looking through
the lens at our object.

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So we still treat this as a positive 15.

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So let's do our calculation.

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We'll have for this lens it's going

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to be 1 over.

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Alright, focal length.

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What do we look at for focal length?

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We look at what type of lens it is.

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This lens is a diverging lens.

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Diverging lenses always get contributed,

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always have a negative focal length

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that they contribute into this equation.

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So a negative 10 centimeters.

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That's the focal length of the 2nd lens,

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equals 1 over our object
distance we just found.

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It's not really an object but that's okay.

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It's an image but this
lens doesn't know that.

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It thinks it's an object
so 1 over positive 15

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because it's still on the side that

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that object was supposed to be on,

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plus 1 over our image.

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This is going to be a second image.

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This is the second image that this

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lens system is going to create

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and so I'll just put di.

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Alright, so we do the math, alright

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1 over negative 10 centimeters

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minus 1 over 15 centimeters
equals 1 over di,

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you solve that on the left hand side.

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You flip it over, you're going end up

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getting the di is, once
you do that inversion

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negative 6 centimeters.

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That's what our second
image is going to be.

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So what does this mean?

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Negative 6 centimeters.

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This is always from the lens and negative.

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Remember negative distance means

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on the left hand side or the opposite

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side as your eye.

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So I'm going to come up here

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and I'm going to draw a line that goes

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from the center of my lens, 6 centimeters

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to the left somewhere around there.

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So that's going to be 6
centimeters right here.

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Whoops, I'll label that.

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That's going to be 6 centimeters

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because that's our image distance

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and that means our final
image is right there.

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Our eye is going to see an image

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of something our arrow right at that spot

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right there that's where our image

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is going to be and just
to recap what happened.

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So this 1st lens created an image

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of the object over at this point here,

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this 15 centimeter point and then

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the 2nd lens created
an image of that image.

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It treated it like it was an object

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and it created an image
at this white point

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right here that's where we're going

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to see the image but we don't know

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what it's going to look like.

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These calculations, this thin lens formula

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only shows us where the
image is going to be.

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If we wanted to figure out how big it was

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we have to use the magnification formula

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so let's do that.

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Over here, magnification
equals negative di.

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Remember it's negative di over do.

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Alright, so what was
di for this first case?

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We're going to do it one step at a time.

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For this first case,
the di was positive 18.

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So I'm going to put a positive 18 up here.

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I'm going to leave off the units

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because they're going to cancel.

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Do was not 24 it was 36.

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So what do I get?

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I get negative 18 over 36
that's 1/2, negative 1/2.

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What does that mean?

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That means this 1st lens, the image

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that it created is going to be an image

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that's 1/2 as big as my object

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and it's going to be
negative magnification.

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That means it's upside
down so I'm going to

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draw this first image.

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I'm going to do it in orange

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because that's what I
labeled the image distance.

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It's going to be upside down

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and half as big as the object.

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So if the object was that big.

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My image is going to be upside down

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and about half as big.

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00:10:28,978 --> 00:10:32,322
So this was my image one.

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00:10:32,322 --> 00:10:34,185
I'll label it image one.

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00:10:34,185 --> 00:10:37,624
That's what I'm treating
as my second object.

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00:10:37,624 --> 00:10:42,137
So that's image one that's object two

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00:10:42,137 --> 00:10:43,488
essentially because that's what this

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00:10:43,488 --> 00:10:45,610
2nd lens is going to think is the object.

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00:10:45,610 --> 00:10:47,898
What magnification does it provide?

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00:10:47,898 --> 00:10:49,354
Well let's do that one.

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00:10:49,354 --> 00:10:54,354
Magnification equals negative di over do

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00:10:54,740 --> 00:10:57,469
so negative of my image distance for

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00:10:57,469 --> 00:11:00,237
this 2nd lens was, it was negative 6

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00:11:00,237 --> 00:11:01,155
remember right here?

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00:11:01,155 --> 00:11:02,426
That's what we solved.

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00:11:02,426 --> 00:11:05,739
The 2nd lens created an
image distance negative 6.

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00:11:05,739 --> 00:11:06,954
I'll leave off the units
because they cancel

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00:11:06,954 --> 00:11:08,546
and my object distance for

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00:11:08,546 --> 00:11:12,242
the 2nd lens was 15 centimeters.

283
00:11:12,242 --> 00:11:13,960
That's what this was right here.

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00:11:13,960 --> 00:11:16,498
The distance from the lens to where

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00:11:16,498 --> 00:11:21,498
it's "object" was and that was positive 15

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00:11:22,799 --> 00:11:24,527
and what do I get?

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00:11:24,527 --> 00:11:27,888
If you take negative 6 over 15

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00:11:27,888 --> 00:11:31,556
you end up getting positive 2/5.

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00:11:31,556 --> 00:11:34,610
So this is the second magnification.

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00:11:34,610 --> 00:11:37,199
The magnification of the
1st lens was negative 1/2.

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00:11:37,199 --> 00:11:40,529
The magnification of the
2nd lens was positive 2/5.

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00:11:40,529 --> 00:11:42,445
That means at this point over here,

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00:11:42,445 --> 00:11:43,791
I'll make it white because that's where

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00:11:43,791 --> 00:11:45,647
I labeled this image distance.

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00:11:45,647 --> 00:11:47,087
I'm going to get an image that's

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00:11:47,087 --> 00:11:50,191
2/5 as big as what?

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00:11:50,191 --> 00:11:53,199
As the 1st image.

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00:11:53,199 --> 00:11:56,837
Now this is a magnification of the image

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00:11:56,837 --> 00:11:59,151
because that's what
this 2nd lens is doing.

300
00:11:59,151 --> 00:12:01,472
So I don't look back
at my object over here.

301
00:12:01,472 --> 00:12:03,023
I'm done with this guy for now.

302
00:12:03,023 --> 00:12:04,862
Now I'm just looking at what this 2nd lens

303
00:12:04,862 --> 00:12:06,943
thinks is the object.

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00:12:06,943 --> 00:12:10,239
So 2/5 as big as this object and positive

305
00:12:10,239 --> 00:12:12,942
means right side up but in this case

306
00:12:12,942 --> 00:12:15,743
this really just means not inverted.

307
00:12:15,743 --> 00:12:18,607
So this started off our object for

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00:12:18,607 --> 00:12:21,063
the 2nd lens started off upside down.

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00:12:21,063 --> 00:12:22,823
This positive means that you maintain

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00:12:22,823 --> 00:12:25,369
the direction, you don't invert it.

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00:12:25,507 --> 00:12:26,909
So I leave this thing upside down.

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00:12:26,909 --> 00:12:27,800
That's what it means

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00:12:27,800 --> 00:12:29,442
I leave it upside down.

314
00:12:29,442 --> 00:12:30,804
If it would have started right side up

315
00:12:30,804 --> 00:12:32,513
I would have left it right side up

316
00:12:32,513 --> 00:12:33,993
but in this case, I leave it upside down

317
00:12:33,993 --> 00:12:35,778
and I make it 2/5 as big.

318
00:12:35,778 --> 00:12:39,766
2/5 is a little under 1/2 and so I'm going

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00:12:39,766 --> 00:12:41,345
to draw this about like that.

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00:12:41,345 --> 00:12:43,856
That's where my image
would be, right there.

321
00:12:43,856 --> 00:12:46,391
That's what I would see.

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00:12:46,391 --> 00:12:49,655
I would see a very little image

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00:12:49,655 --> 00:12:51,975
that's upside down and it would be

324
00:12:51,975 --> 00:12:55,049
right around there and
so that's an example

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00:12:55,049 --> 00:12:57,673
of where you can do a two lens system.

326
00:12:57,673 --> 00:12:59,408
You treat each lens separately

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00:12:59,408 --> 00:13:01,676
and you use these formulas accordingly.

328
00:13:01,676 --> 00:13:02,873
The last thing I'll say is,

329
00:13:02,873 --> 00:13:04,775
if you wanted to know why we had

330
00:13:04,775 --> 00:13:06,807
two magnifications here, is there some way

331
00:13:06,807 --> 00:13:10,044
to figure out the total magnification?

332
00:13:10,044 --> 00:13:11,488
Some slick way of doing it?

333
00:13:11,488 --> 00:13:13,063
Yeah and it's really easy it turns out.

334
00:13:13,063 --> 00:13:15,380
You can just take your
total magnification.

335
00:13:15,380 --> 00:13:17,800
If all I wanted to know
was if I had this object

336
00:13:17,800 --> 00:13:20,719
what my total magnification
was going to be.

337
00:13:20,719 --> 00:13:23,231
What could I multiply
this object's height by,

338
00:13:23,231 --> 00:13:25,048
in other words, what factor could

339
00:13:25,048 --> 00:13:27,375
I multiply it by to get both

340
00:13:27,375 --> 00:13:30,787
the orientation and height
of this final image?

341
00:13:30,787 --> 00:13:32,707
Well you just take the magnification

342
00:13:32,707 --> 00:13:34,837
of the 1st lens times the magnification

343
00:13:34,837 --> 00:13:35,779
of the 2nd lens.

344
00:13:35,779 --> 00:13:37,331
If I had more I'd just multiply them all.

345
00:13:37,331 --> 00:13:39,091
That would give me the factor.

346
00:13:39,091 --> 00:13:41,619
That's the overall
magnification that I can

347
00:13:41,619 --> 00:13:43,922
multiply this object's height by

348
00:13:43,922 --> 00:13:46,195
to get the height of the final image

349
00:13:46,195 --> 00:13:48,259
and the orientation of it.

350
00:13:48,259 --> 00:13:50,699
It doesn't give you the position.

351
00:13:50,699 --> 00:13:52,763
This just gives you the
overall magnification,

352
00:13:52,763 --> 00:13:54,331
That could be handy if your making

353
00:13:54,331 --> 00:13:56,458
a microscope or something but it doesn't

354
00:13:56,458 --> 00:13:57,645
tell you the position.

355
00:13:57,645 --> 00:14:00,492
For the position you've got to use

356
00:14:00,492 --> 00:14:02,188
these thin lens formulas.

357
00:14:02,188 --> 00:00:00,000
So that's how you treat
a multiple lens problem.