00:01
In this problem, a patient has a near point at 45 centimeter and far point at 85 centimeter.
00:10
So to solve the part a in this problem, the answer to whether a single corrective lens can solve his problem is yes, it can.
00:24
Since the patient needs corrective vision for both near vision and far vision, he can use bifocal lens.
00:32
So the answer to that is yes.
00:40
Or alternatively, he can also use two pairs of glasses, one for reading and one for distance viewing.
00:49
So for part b, to correct the near vision, if the patient is looking at the object at the normal near point, then the object distance would be p equals 25 centimeter.
01:06
Let me write down in meters, 0 .25 meter.
01:13
Then a virtual upright image should form at his actual near point, right? so q in that case would be the actual near point.
01:24
Negative value, 45 centimeter.
01:28
I'm going to write this down in meters, 0 .45 meter.
01:32
Negative 0 .45 meter.
01:37
Now from here, if you want to calculate the power, we can just use the lens equation.
01:45
So the lens equation is 1 over f.
01:51
The power equals 1 over f in meters equals 1 over p plus 1 over q.
02:00
And this would be 1 over 0 .25 minus 1 over 0 .45...