1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Draw a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1m. What is the power of the lens required to correct this defect? Assume that near the point of the human eye is 25cm.

Open in App
Solution

## A person suffering from hypermetropia can see distinct objects clearly but faces difficulty in seeing nearby objects clearly. It happens because the eye lens focuses the incoming divergent rays beyond the retina. This defect of vision is corrected by using a convex lens. A convex lens of suitable power converge the incoming light in such a way that the image is formed on the retina, as shown in the following figure: The convex lens actually creates a virtual image of a nearby object (N’ in the figure) at the near point of vision (N) of the person suffering from hypermetropia. The given person will be able to clearly see the object kept at 25 cm (near the point of the normal eye) if the image of the object is formed at his near point, which is given as 1m. Object distance, u = - 25 cm Image distance, v= - 1 m = - 100 cm Using the lens formula 1v - 1u = 1f 1−100 - 1−25 = 1f 1f = 125 - 1100 1f = 4−1100 f = 1003 cm, P = 1f = 10.33m =+3.0D

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Power of Accommodation and Defects
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program