00:01
In this problem, we are given that there is a diverging lens.
00:04
And the focal length of this diverging lens that is given as minus 12 centimeters.
00:11
And this is because the focus lies to the left through which the light rays appear to pass through after it comes out of the object.
00:19
And we take all the distance measured to the left as negative.
00:23
And all the distance that we measure to the right, we take it as positive.
00:27
And we have an object and that is kept at a distance.
00:30
Of 18 centimeters in front of this lens.
00:34
So this is minus because the object is kept on the left of this lens as stated here.
00:41
And we are required to determine the distance from the lens where the object should be placed so that the image that is formed is reduced by a factor of.
00:54
So here we first use this lens formula and we compute the image distance.
00:59
And when we use the value of f and u and put it here in this equation we get one by v minus 1 by u that's minus 18 equal to 1 by f that's minus 12 and solving this gets us 1 by v as 1 by 18 this is of course minus and from this we subtract minus 1 by 12 and this implies that we will be equal to 18 times 12 divided by minus 18 minus 12 so we multiply 18 with 12 and we divide this with minus 18 minus 12 so we get the image distance as minus 7 .2 centimeters and now we use this expression to get the magnification in this case and that turns out to be we by you which is minus 7 .2 divided by minus 18 and when we do that we get the magnification as 0 .4.
02:02
And as we are given that, the size of image is reduced by a factor of...