A person with eyesight problems may wear either diverging or converging contact lenses.

Question

A short-sighted eye cannot focus on distant objects because the power of the eye is too great.
One student with short sight cannot focus on objects further than 1.5 m without wearing her contact lenses.
To view distant objects, it is determined that the combined power of her eye and her contact lens should be 41.7 D.
Determine the power and type of lens needed to correct her vision. Assume the equations for thin lens apply to both lenses.

Distance from eye lens to retina = 2.4 cm

Answer 1

First, we can use the thin lens equation to find the power of the eye:

1/f_eye = 1/do + 1/di

where f_eye is the focal length of the eye, do is the distance from the object to the eye, and di is the distance from the image to the eye. Since the eye cannot focus on objects further than 1.5 m, we can assume that the image formed by the eye is at infinity, so di is effectively infinite. Thus:

1/f_eye = 1/do

1/f_eye = 1/150 cm

f_eye = 150 cm = 1.5 m

The power of the eye is then:

P_eye = 1/f_eye = 1/1.5 = 0.67 D

Next, we can use the fact that the combined power of the eye and contact lens should be 41.7 D to set up another equation using the thin lens equation:

1/f_combined = 1/f_eye + 1/f_contact

where f_combined is the combined focal length of the eye and contact lens, and f_contact is the focal length of the contact lens. We can solve for f_contact:

1/f_contact = 1/f_combined - 1/f_eye

1/f_contact = 1/41.7 - 1/1.5

1/f_contact = 0.0239

f_contact = 41.7 cm

The power of the contact lens is then:

P_contact = 1/f_contact = 1/0.417 = 2.40 D

Since a short-sighted eye has too much power, we need a diverging lens (i.e. a negative lens) to correct the vision. The power of the diverging lens can be found using the thin lens equation again:

1/f_diverging = 1/f_combined - 1/f_eye

1/f_diverging = 1/41.7 - 1/1.5

1/f_diverging = 0.0239

f_diverging = -41.7 cm

The power of the diverging lens is then:

P_diverging = 1/f_diverging = -2.40 D

Therefore, the student needs a diverging contact lens with a power of 2.40 D to correct her vision.