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Your 68-year-old patient went to see her optometrist

yesterday. Her best-corrected visual acuity is 20/70. She
notes that this seems to get worse at times. What can
you advise her to do to maximize her visual function?
a. Improve contrast
b. Improve illumination
c. Reduce glare
d. Increase stereopsis
e. a, b, and c

User Peelman
by
7.5k points

1 Answer

0 votes

Final answer:

A young woman with normal distant vision and an ability to accommodate 10% more than normal power can see the closest object clearly at approximately 18.2 cm from her eyes.

Step-by-step explanation:

Understanding Visual Acuity and Accommodation

For a person with normal distant vision, the average power of the eye for focusing on distant objects is approximately 50 diopters (D). When a young woman with normal distant vision has a 10.0% ability to accommodate, it means she can increase her eye's power by 10% over the normal power for focusing on distant objects. This increased power allows her to focus on objects closer than what is possible with the unaccommodated eye.

The formula for calculating the closest distance at which she can see objects clearly, given her accommodation ability, is:

Power of accommodation = Normal power for distant vision × % ability to accommodate

Using this information:

New power = 50.0 D + (10.0% of 50.0 D)

New power = 50.0 D + 5.0 D

New power = 55.0 D

To find the closest distance (d) at which she can see clearly, we use the formula:

d = 1 / Power (in meters)

Therefore:

d = 1 / 55.0 D

d = 0.0182 meters or 18.2 centimeters (approximately)

Thus, the closest object that the young woman can see clearly is at about 18.2 cm from her eyes.

User Dpdearing
by
7.6k points
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