00:01
So in this problem, the near point of an eye is 75 cm.
00:05
So for part a, if we want to find the corrective lens to be able to see at normal near point of 25 centimeter, then when the object distance is at normal near point, which is 25 centimeter, or 0 .25 meter, then the image distance has to be at negative 75 centimeter, negative 75 centimeter, okay, which is negative 0 .75 meter.
00:53
Now, the image, the reason why you see this negative here is because the image formed is going to be virtual and upright, right? so from here, we can actually find the power of the list, that corrective list.
01:06
Is going to be 1 over f equals 1 over p plus 1 over q.
01:13
This is going to give me q plus p over q .p over q .p.
01:20
And this is gonna be negative 0 .75 plus 0 .25 over negative 0 .75 times 0 .25.
01:41
And this would be positive 2 .67 diopters.
01:57
So in part b, the person who made the lens messed it up.
02:06
Okay? and the lens had different focal length instead...
