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It has become common to replace the cataract-clouded lens of the eye with an internal lens. This intraocular lens can be chosen so that the person has perfect distant vision. Will the person be able to read without glasses? If the person was nearsighted, is the power of the intraocular lens greater or less than the removed lens?

A person with an internal lens will need glasses to read because their muscles cannot distort the lens as they do with biological lenses, so they cannot focus on near objects. To correct nearsightedness, the power of the intraocular lens must be less than that of the removed lens.

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If the cornea is to be reshaped (this can be done surgically or with contact lenses) to correct myopia, should its curvature be made greater or smaller? Explain.

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Problems

Unless otherwise stated, the lens-to-retina distance is 2.00 cm.

What is the power of the eye when viewing an object 50.0 cm away?

P = 52.0 D

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Calculate the power of the eye when viewing an object 3.00 m away.

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The print in many books averages 3.50 mm in height. How high is the image of the print on the retina when the book is held 30.0 cm from the eye?

h i h o = d i d o h i = h o ( d i d o ) = ( 3.50 mm ) ( 2.00 cm 30.0 cm ) = −0.233 mm

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Suppose a certain person’s visual acuity is such that he can see objects clearly that form an image 4.00 μm high on his retina. What is the maximum distance at which he can read the 75.0-cm-high letters on the side of an airplane?

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People who do very detailed work close up, such as jewelers, often can see objects clearly at much closer distance than the normal 25 cm. (a) What is the power of the eyes of a woman who can see an object clearly at a distance of only 8.00 cm? (b) What is the image size of a 1.00-mm object, such as lettering inside a ring, held at this distance? (c) What would the size of the image be if the object were held at the normal 25.0 cm distance?

a. P = + 62.5 D ;
b. h i h o = d i d o h i = −0.250 mm ;
c. h i = −0.0800 mm

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What is the far point of a person whose eyes have a relaxed power of 50.5 D?

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What is the near point of a person whose eyes have an accommodated power of 53.5 D?

P = 1 d o + 1 d i d o = 28.6 cm

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(a) A laser reshaping the cornea of a myopic patient reduces the power of his eye by 9.00 D, with a ± 5.0 % uncertainty in the final correction. What is the range of diopters for eyeglass lenses that this person might need after this procedure? (b) Was the person nearsighted or farsighted before the procedure? How do you know?

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The power for normal close vision is 54.0 D. In a vision-correction procedure, the power of a patient’s eye is increased by 3.00 D. Assuming that this produces normal close vision, what was the patient’s near point before the procedure?

Originally, the close vision was 51.0 D. Therefore, P = 1 d o + 1 d i d o = 1.00 m

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For normal distant vision, the eye has a power of 50.0 D. What was the previous far point of a patient who had laser vision correction that reduced the power of her eye by 7.00 D, producing normal distant vision?

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The power for normal distant vision is 50.0 D. A severely myopic patient has a far point of 5.00 cm. By how many diopters should the power of his eye be reduced in laser vision correction to obtain normal distant vision for him?

originally, P = 70.0 D ; because the power for normal distant vision is 50.0 D, the power should be decreased by 20.0 D

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A student’s eyes, while reading the blackboard, have a power of 51.0 D. How far is the board from his eyes?

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The power of a physician’s eyes is 53.0 D while examining a patient. How far from her eyes is the object that is being examined?

P = 1 d o + 1 d i d o = 0.333 m

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The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?

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The far point of a myopic administrator is 50.0 cm. (a) What is the relaxed power of his eyes? (b) If he has the normal 8.00% ability to accommodate, what is the closest object he can see clearly?

a. P = 52.0 D ;
b. P = 56.16 D 1 d o + 1 d i = P d o = 16.2 cm

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A very myopic man has a far point of 20.0 cm. What power contact lens (when on the eye) will correct his distant vision?

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Repeat the previous problem for eyeglasses held 1.50 cm from the eyes.

We need d i = −18.5 cm when d o = , so
P = 5.41 D

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A myopic person sees that her contact lens prescription is –4.00 D. What is her far point?

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Repeat the previous problem for glasses that are 1.75 cm from the eyes.

Let x = far point
P = 1 ( x 0.0175 m ) + 1 x P + ( 0.0175 m ) P = 1 x = 26.8 cm

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The contact lens prescription for a mildly farsighted person is 0.750 D, and the person has a near point of 29.0 cm. What is the power of the tear layer between the cornea and the lens if the correction is ideal, taking the tear layer into account?

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Practice Key Terms 6

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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