Question

A farsighted eye is corrected by placing a converging lens in front of the eye. The lens will create a virtual image that is located at the near point (the closest an object can be and still be in focus) of the viewer when the object is held at a comfortable distance (usually taken to be 25 cm). 1) If a person has a near point of 63 cm, what power reading glasses should be prescribed to treat this hyperopia? (Express your answer to two significant figures.)

Answer 1

To correct hyperopia, the power of the lens needed can be calculated using the lens equation. By plugging in the object distance and the image distance, we can determine the focal length of the lens. Then, using the formula Power = 1/f, we can calculate the power of the lens needed to treat hyperopia.

Step-by-step explanation:

To determine the power reading glasses needed to treat hyperopia (farsightedness), we can use the lens equation:

1/f = 1/do + 1/di

Where:

• f is the focal length of the lens
• do is the object distance (the comfortable distance at which the object is held, usually taken to be 25 cm)
• di is the image distance (the near point of the viewer, in this case, 63 cm)

We need to solve for the focal length f and convert it to power using the formula Power = 1/f. Rearranging the lens equation, we can solve for f:

1/f = 1/do + 1/di

1/f = 1/0.25 + 1/0.63

1/f = 4 + 1.59

1/f = 5.59

f = 1/5.59

f ≈ 0.179 m (to three significant figures)

Now, we can calculate the power of the lens using the formula Power = 1/f:

Power = 1/0.179

Power ≈ 5.59 diopters (to two significant figures)

Answer 2

To correct hyperopia (farsightedness), reading glasses with a power of 0.18 D should be prescribed based on the person's near point of 63 cm.

Step-by-step explanation:

To calculate the power of the reading glasses needed to correct farsightedness (hyperopia), we can use the lens formula:

1/f = 1/v - 1/u

Where f is the focal length of the lens, v is the image distance, and u is the object distance. In this case, the object distance (u) is 25 cm and the image distance (v) is the near point of 63 cm. Rearranging the lens formula and substituting the values:

f = 1 / ((1/v) - (1/u))

Plugging in the values, we get:

f = 1 / ((1/0.63) - (1/0.25)) = 5.55 cm

The power of the lens is the reciprocal of the focal length:

Power = 1 / f = 1 / 5.55 cm = 0.18 D

Therefore, a power of 0.18 D reading glasses should be prescribed to correct this hyperopia.