00:01
Hi there, so for this problem, we are told that an object is in front of a converging lens with a focal length that is equal to 0 .21 meters.
00:14
The magnification of the lens is given and that magnification is 2 .7.
00:22
So, for part a of this problem, we are asked about relative to the lens in what direction should the object we move so that the magnification changes to a value that is the same as before but negative.
00:42
So since we have a converging lens, we know that if the object is position at a position that is graded at the focal length, then the object is going to form in the other side of the lens and then is going to be with a negative magnification.
01:09
However, in this case we have initially that the minification is positive.
01:16
So that means that the object position is inside the focal length.
01:22
So to change the magnification to a negative value, we need to move the object away from the lens.
01:38
So that the magnification changes from a positive value to a negative value.
01:49
Now, for power b of this problem, we are asked about through what distance should the object be moved.
02:01
So what we are going to do is to obtain the object distance in each case, when the magnification is 2 .7 and when the magnification is minus, as 2 .7 and the difference between those two values will give us the distance that we should move the object.
02:26
So first when we have 2 .7, so first with this positive value, well we know that the magnification is defined as minus the image distance divided by the object distance.
02:43
So with that we can write in this case that the image distance is equal to minus the magnification times the object distance so that will be that the image distance is minus 2 .7 times the object distance and for that we use the following equation that the inverse of the focal length is equal to the inverse of the object distance plus the inverse of the image distance so now we substituting here what is the the condition in here, so that will be this right here.
03:27
So we can take out the object distance.
03:32
So there will be one minus 1 divided by 2 .7.
03:40
So remember that this is the inverse of the focal length.
03:46
So simplifying this value in here, let me just calculate that value.
03:53
That is going to be approximately, this value in here is going to be 0 .629.
04:10
Now to obtain the object distance because that's the value that we need to obtain.
04:20
So what we are left is with this...