Let the point of least distance at which the eye lens forms a clear image on the retina for a person with hypermetropia be ‘H’. See figure 6(b).

If an object is at H or beyond H, the eye can form its image on retina (see figures 6(b) and 6(c)). If the object is between H and point of least distance of distinct vision (L) then it cannot form an image. See figure 6(a)

point of minimum distance at which the eye lens can form an image on the retina is called near point (d). The people with defect of hypermetropia cannot see objects placed between near point (H) and point of least distance of distinct vision (L).

• How can you correct this defect?

Eye lens can form a clear image on the retina when any object is placed beyond near point. To correct the defect of hypermetropia, we need to use a lens which forms an image of an object beyond near point, when the object is between near point (H) and least distance of distinct vision (L).

This is possible only when a double convex lens is used.

• How can you decide the focal length of the convex lens to be used?

To find the focal length of lens, let us consider that the object is at point of least distance of distinct vision (L). Then the defect of vision, hypermetropia, is corrected when the image of the object at L is formed at the near point (H) by using a bi-convex lens as shown in figure 6(d).

This image acts like an object for the eye lens. Hence final image due to eye is formed at retina (see figure 6(d))

Here object distance (u) = -25 cm

Image distance (v) = distance of near point = -d

Let ‘f’ be the focal length of bi-convex lens

Using lens formula, 1/f = 1/v – 1/u

1/f = 1/ -d – 1/(-25)

1/ f = -1/d +1/25

1/ f = (d – 25)/25d

f= 25d / (d – 25) (f is measured in centimeters)

we know that if d > 25cm, then ‘f’ becomes +ve i.e., we need to use biconvex lens to correct defect of hypermetropia.