he far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem

To correct myopia (nearsightedness), a diverging lens is needed. The far point of a myopic person is the maximum distance up to which the person can see clearly. Beyond this point, objects appear blurry.

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

1/f = 1/v - 1/u

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

In this case, the far point of the myopic eye is 80 cm (v), and the normal far point is at infinity (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 -80 cm for v.

1/f = 1/(-80) - 1/∞ 1/f = -0.0125

So, f = 1 / -0.0125 = -80 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 = -80 cm = -0.8 m

P = 1/f = 1/-0.8 = -1.25 diopters

So, the power of the lens required is -1.25 D. The negative sign indicates that a diverging lens is needed, which is used for correcting myopia.