The far point of a myopic person is 80 cm infront of the eye. What is the power of the lens required to enable him to see very distance objects clearly?

Explanation:

To correct myopia (nearsightedness), a diverging lens is used to form a virtual image at the individual's far point, allowing them to see distant objects clearly.

Given:

• Far point (v) = 80 cm = 0.8 m
• Object distance (u) = -∞ (since the object is at infinity)

Using the lens formula:

1/f = 1/v - 1/u

Substituting the known values:

1/f = 1/0.8 - 1/∞ = 1/0.8 = 1.25 D

Therefore, the focal length (f) of the required lens is 0.8 m.

Calculating the power (P) of the lens:

P = 1/f = 1/0.8 = 1.25 D

Since a diverging lens is needed to correct myopia, the power is negative:

P = -1.25 D

Final Answer

The power of the lens required to enable the person to see distant objects clearly is –1.25 D.