The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? (Assuming that the near point of a normal eye is 25 cm)
+3 D
If the object is placed at 25 cm (N′) , then the image has to be formed at 1 m (N) (because N is the far point for the hypermetropic eye) , which act as the object for eye lens, which forms the final image at the retina.
Let v be the image distance, u be the object distance and f be the focal length.
Using lens formula we have,
1v−1u=1f
Now, applying cartesian convention,
v=−1 m, u=−25 cm=−14 m
Therefore, −11−−41=1f
1f=+3 m
Thus power is +3 D