The near point of defective eye is 0.75,m whereas it is 0.25m for normal eye. So the lens used in spectacles should form a virtual image at 0.75m when object is placed at o.25m.
Hence object distance u= -0.25m and image distance v = -0..75m
So applying formula of lens
1/v-1/u=1/f
We get
-1/0.75-1/(-0.25)=1/f
=>4/3+4=1/f
=>8/3=1/f
=>f=0.375m
+ve value of focal length means, the lens used in the spectacles is convex one.
So power of the lens = 1/f=1/0.375=2.66D
The eye here is long-sighted .
