A patient can\'t see objects closer than 48 cm and wishes to clearly see objects

Question

A patient can't see objects closer than 48 cm and wishes to clearly see objects that are 19.9 cm from his eye.

p = cm q = cm f = cm P = diopters

A patient can't see objects closer than 48 cm and wishes to clearly see objects that are 19.9 cm from his eye. Is the patient nearsighted or farsighted? If the eye-lens distance is 2.02 cm, what is the minimum object distance p from the lens? What image position with respect to the lens will allow the patient to see the object? Is the image real or virtual? Is the image distance q positive or negative? Calculate the required focal length. Find the power of the lens in diopters. If a contact lens is to be prescribed instead, find p, q, and f, and the power of the lens.

Explanation / Answer

1) he can't see the objecy closer than 48 cm so he is far sighthness.

2) p = 19.9 -2.02 = 17.88 cm

3) q = 48 - 2.02 = 45.98 cm

4) image ang object are on same side and distance is negative so virtual.

5) 1/f = 1/p + 1/q

q= 45.98 cm

p = 17.88 cm

so f = 29.25 cm

6) power = 100 / f = 3.41 diaopter

7) in case of contact lens

1/f = 1/19.9 + 1/(-48)

f = 33.99 cm

power = 100/ 33.99

= 2.94 diopter

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