he near point of a hypermetropic eye is at 75 cm from the eye. What is the power of the lens required to enable him to read clearly a book held at 25 cm from the eye?

To calculate the power of the lens required, we need to use the lens formula:

1/f = 1/v - 1/u

Where: f = focal length of the lens v = image distance (near point for the hypermetropic eye) u = object distance (normal near point)

In this case, the near point of the hypermetropic eye is 75 cm (v), and we want to enable him to read a book held at 25 cm (normal near point, u).

However, in optics, the convention is to use distances as positive if they are on the side of the light source (object side) and negative if they are on the opposite side (image side). So, we will use -75 cm for v and -25 cm for u.

1/f = 1/(-75) - 1/(-25) 1/f = -0.0133 + 0.04 1/f = 0.0267

So, f = 1 / 0.0267 = 37.5 cm

The power P of a lens is the reciprocal of the focal length (in meters), so we convert the focal length to meters:

f = 37.5 cm = 0.375 m

P = 1/f = 1/0.375 = 2.67 diopters

So, the power of the lens required is +2.67 D. The positive sign indicates that a converging lens is needed, which is used for correcting hypermetropia.