00:01
So let's first of all analyze the contact lens.
00:07
So the contact lens is able to help her see things which are 25 centimeters away, although her far point, a near point is as far as 79 centimeters away.
00:24
But since the image form will be a virtual image, so the i would be negative 79 centimeters.
00:30
The lens helps her to see things which are actually at 25 centimeters by forming an image at 79 centimeters.
00:39
So we have 1 over f is equal to 1 over do plus 1 over d .i, which is equal to 1 over 25 plus 1 over negative 79.
00:55
So solving this for f gives me f is equal to 36.
01:00
6 centimeters.
01:02
So we're going to use this value of f for solving part a.
01:06
So in part a, we are told that a poster is placed some unknown distance away from the i, but the image is formed at a distance of 217 centimeters again since it's a virtual image.
01:23
So the i will be equal to negative and f we just figured was equal to 36 .6 centimeters so we have one over d o plus one over di i is equal to one over f one over d o is equal to one over f minus one over d i which will be equal to one over 36 .6 minus one over negative to 17 so solving this for do gives do is equal to 31 .3 centimeters.
02:02
So the poster is 31 .3 centimeters from her i...