00:01
In this problem, we're asked how far in front of the lens should the object be placed, so that the size is reduced by a factor of 1 .9.
00:09
So we know that magnification is equal to the height of the object divided by, or the height of the image divided by the height of the object.
00:18
So for our first image, we have height of object 1 as equal to h0 times negative i1 divided by 0.
00:26
1.
00:28
So we can plug this into the lens equation.
00:31
1 over f is equal to 1 over object distance plus 1 over image distance.
00:36
So 1 over i1 is equal to 1 over f minus 1 over object distance.
00:42
That's equal to 1 negative 12 cm minus 1 over 18 and we get i1 of negative 7 .2 cms.
00:51
Now we can do the same way similarly for the second image.
00:55
So image 2 is equal to h0 times negative i2 divided by o2...