The 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 the vision of a myopic (nearsighted) person, we need to find the power and nature of the lens required to bring the far point to infinity. The far point of a myopic person is the maximum distance at which they can see clearly.

Given:

. The far point of the myopic person is 80 cm in front of the eye, which means they can see objects clearly at a distance of 80 cm.
To correct this problem, we need to find the lens power (P) required. The lens power formula is:

P= 1/f

. P is the power of the lens in diopters (D).
. f is the focal length of the lens in meters (m).

To bring the far point to infinity (i.e., to correct the vision), we need to calculate the focal length required to achieve this. The focal length should be such that the image of distant objects is formed at infinity.

We can calculate the focal length as follows:

f = 1/d_i

Where d_i is the image distance, which should be at infinity.

Therefore:
f = 1/∞ = 0 m
So, to bring the far point to infinity, the focal length of the lens should be 0 meters.

Now, we can calculate the lens power using the lens power formula:

P = 1/f = 1/0m

However, since we cannot have a lens with a focal length of zero, the lens required to correct the myopic person’s vision is a concave (diverging) lens with a focal length of 0 meters. In practice, this would be considered an extremely weak lens with a power close to zero.

So, the nature of the lens required is a concave (diverging) lens, and the power of the lens is approximately 0D. This extremely weak lens helps to bring the far point to infinity, correcting the nearsightedness.