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KIM [24]
3 years ago
10

A physicist's right eye is presbyopic (i.e., farsighted). This eye can see clearly only beyond a distance of 97 cm, which makes

it difficult for the physicist to read books and journals. Find the focal length and power of a lens that will correct this presbyopia for a reading distance of 25 cm, when worn 2 cm in front of the eye
Physics
1 answer:
Ivan3 years ago
5 0

Answer:

f = 19,877 cm   and  P = 5D

Explanation:

This is a lens focal length exercise, which must be solved with the optical constructor equation

        1 / f = 1 / p + 1 / q

where f is the focal length, p is the distance to the object and q is the distance to the image.

In this case the object is placed p = 25 cm from the eye, to be able to see it clearly the image must be at q = 97 cm from the eye

let's calculate

        1 / f = 1/97 + 1/25

        1 / f = 0.05

        f = 19,877 cm

the power of a lens is defined by the inverse of the focal length in meters

         P = 1 / f

         P = 1 / 19,877 10-2

         P = 5D

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Area = 3.801 * 10^(-4)    meter squared

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Stress = 75000 / (3.801 * 10^-4)

Stress = 197316495 Pa     OR       0.197 GPa

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200 = 0.19732 / Strain

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0.0009866 = Elongation of rod / 200

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(b) Poisson ratio = lateral strain / longitudinal strain

0.3 = lateral strain / 0.0009866

Lateral strain = 0.000296

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0.000296 = Change in diameter / 22

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4 0
3 years ago
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8 0
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2 years ago
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