00:01
In this problem, a child has a near point at 10 cm, and she also has a fire point, which is at 125 centimeter.
00:10
Also, it is given that each eye is 2 centimeter from retina.
00:16
So to solve the problem a of this, when the child is able to see the object at the far point without using the corrective lens, she will have the objective, distance at 125 centimeter.
00:36
So her object distance is going to be 125 centimeter, and that's going to be 1 .25 meter.
00:49
And remember, in this case, she's going to be able to focus the image on retina, which is at 2 centimeter from the lens, right? so that means q is going to be equal to 2 centimeter which is 0 .02 meter so now what we need to do here is to find the power for the far point for the far point which is also one over focal length for the far point is basically one over p plus one over q and p is here going to be p max and this is going to be one over one over 1 .25 plus 1 over 0 .02.
01:46
All right.
01:49
So from here, so this is 0 here, by the way.
01:54
So 1 over 0 .02, from here, the power we get is going to be positive 50 .8 diopters.
02:10
So this is going to be power for the far point.
02:14
Now to find the power for the near point, we're going to use.
02:19
The near point as the object distance...