A patient can\'t see objects closer than 33.4 cm and wishes to clearly see objec

Question

A patient can't see objects closer than 33.4 cm and wishes to clearly see objects that are 20.0 cm from his eye.

(a) Is the patient nearsighted or farsighted?

nearsightedfarsighted    

(b) If the eye–lens distance is 1.99 cm, what is the minimum object distance p from the lens? (Give your answer to at least three significant digits.)
cm

(c) What image position with respect to the lens will allow the patient to see the object? (Give your answer to at least three significant digits.)
cm

(d) Is the image real or virtual? Is the image distance q positive or negative? (Select all that apply.)

real imagevirtual imagedistance is positivedistance is negative

(e) Calculate the required focal length.
cm

(f) Find the power of the lens in diopters.
diopters

(g) If a contact lens is to be prescribed instead, find p, q, and f, and the power of the lens.

p = cm      (Give your answer to at least three significant digits.) q = cm      (Give your answer to at least three significant digits.) f = cm P = diopters

Explanation / Answer

farsighted

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(b)

object distance p = 20-1.99 = 18.01 cm

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(c)

image distance q = -(33.4- 1.99) = -31.41 cm

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(d)

virtual

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(e)

1/p + 1/q = 1/f

1/18.01 - 1/31.41 = 1/f

f = 42.2 cm

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(f)

power = 1/f(in meters) = 1/0.422 = + 2.37 D

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(g)

p = 20 cm

q = 33.4 cm

1/p + 1/q = 1/f

1/20 - 1/33.4 = 1/f

f = 49.8 cm

p = 1/0.498 = 2 diopters

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