00:03
All right, so you have quite a optics problem here.
00:08
So there's a lot of steps to it.
00:11
I'll go through it.
00:13
Hopefully it makes sense when we're all finished because there's a lot of steps to get this one done.
00:20
There was a handful of things given in the problem, so i tried to draw that out to help myself figure out what do we have to do to solve this problem.
00:30
We were given one of the two focal points.
00:33
Focal point for lens one was negative 12 .3 because it was a diverging lens.
00:38
F2 is going to be a positive focal point because it's a converging lens.
00:44
And then we were given the object's height in distance from the first lens.
00:51
Now, i'm assuming that that's in front of both lens two and lens one, just because of the way it's worked at some of the questions that are being asked, but it doesn't specify that just because it's in front of lens one, it doesn't mean that it's in front of lens two because it doesn't say lens two is behind lens one and the problem.
01:14
So this might end up being completely wrong because i might have guessed to put the lens two in the wrong spot.
01:21
But i think i got it right because if i want a view screen behind lens two, i'm looking at it for that last question, i think they all have to be in line like this.
01:35
Well, the overall magnification was given as well.
01:38
We want an image forming on the view screen that is twice as big as the original object and upside down.
01:44
So that was all that was given.
01:46
So we've got a lot of work to get to all the numbers we need.
01:51
So i'm just going to go through the steps of how to solve all of these numbers that we need to get to the final answer.
02:00
So step one is to figure out what is the image formed.
02:05
From the first lens.
02:06
So i'm going to call these the f1, d -o -1, and d -i -1, so that we keep separate, which image and object are we talking about here? if we put that into the lens equation and solve for d -i -1, you get negative 7 .8 centimeters, meaning that it's a virtual image in front of the lens.
02:33
That's normally what's going to happen with a diverging lens, we're going to get a virtual image, and it's in front of the lens.
02:43
That's important.
02:44
If we put that behind the lens, we get all different numbers.
02:48
So we've got negative 7 .8, which basically means that it's forming 47 .8 in front of lens 2.
02:57
That's what's happening here with this one.
03:02
So step number two, we need to figure out, well, what's the magnification of this first lens? lens one.
03:09
So i made an m1 for magnification of lens one.
03:14
If i find the magnification of lens one, that'll tell me what the magnification of lens number two is so that the total magnification is negative two.
03:23
So i did the work.
03:24
I got negative di over d .o.
03:27
Remember, i'm using the one, di one, d .o.
03:31
I get a positive 0 .37.
03:34
So the image is upright and smaller by less than half.
03:42
So far so good.
03:43
It should be making sense, hopefully.
03:46
That's to the point where we could start looking at the second lenses information.
03:51
We've got everything figured out for lens one.
03:54
Where's the image for me? how far is it from lens two? what was the magnification that we got for lens one so that we can figure out lens two's magnification? all right.
04:07
So step three.
04:09
Step three, we have to figure out what's the magnification of the lens.
04:12
Number two in relationship to the lens number one's magnification so that they put the total magnification on the view screen is negative two.
04:26
Well, basically, if i have two lenses and i multiply their magnifications together, i'll get their total magnification.
04:35
So if i had two met, two lenses that each magnified at 10, 10 times 10 is 100.
04:41
So we want it to be negative two.
04:43
And our first lens has magnification of 0 .37.
04:48
So if we do that, that math, we've got a magnification for lens 2 of negative 5 .5.
04:58
So that means it's an inverted, and it's going to invert the image, which is what it asked us to do, make it upside down for the total magnification.
05:07
And if we take 5 .5 times 0 .37, we'll get to.
05:13
So that's the next step.
05:14
We've got the magnification of lens 2, which is going to allow us to figure out where do we put the actual view screen now.
05:24
So i'm going to take that new magnification for lens 2, and i'm going to put it into the magnification equation once again, this time with my d .i .2 over d .o .2...