00:01
Now in this question, we look at basically two lenses.
00:04
One is convulsion lens.
00:05
There are these divergent lens.
00:07
So in the case of, let's look at the first question, which is about the convex lens, it's convergent lens.
00:13
So basically you have, so suppose we have a convergent lens, right, like this, and you have a prince of axis of this, of this lens, and you place an object, which, you place an object, to 17 .9 centimeters in front of convex lens.
00:34
So somewhere here.
00:36
So there's an object.
00:37
And it forms an image on the far side of the lens with magnification, which is managed.
00:43
So it means that it must be on, actually, on this facet, on this side, it must be somewhere here, right? that's an image.
00:49
So let's call this d -i, okay? and this is d -o, right? and then you know that d .i over d .o.
01:00
Must be the magnification that is minus 3 .3 according to the question, right? so, right, that must be this.
01:09
So you can work out, since d .o is given, which is 17 .17 .9 centimeter.
01:16
So you can work out the i, right? d .i is simply given by minus 3 .3 times d .o.
01:21
That's 17 .9 centimeter.
01:23
And from this, you can work out the focal lens.
01:27
And the focal lens will be given by if.
01:29
So i would write if 1 of a equals 1 over d .i plus 1 of d .o, right? you plug the values for d and d .o and you find the focal lens...