A 40-year-old optometry patient focuses on a 6.50 cm -tall photograph at his nea

Question

A 40-year-old optometry patient focuses on a 6.50 cm -tall photograph at his near point. (See the table.) We can model his eye as a sphere 2.50 cm in diameter, with a thin lens at the front and the retina at the rear.(Figure 1)

Part A:

What is the effective focal length of his eye when he focuses on the photo?

Part B:

What is the power of his eye in diopters when he focuses on the photo?

Part C:

How tall is the image of the photo on his retina?

Part D:

If he views the photograph from a distance of 2.40 m , how tall is its image on his retina?

Explanation / Answer

Solution:

a) Use the thin lens equation

1/do + 1/di = 1/f

as do = 22cm[From the table] and di =2.50 cm,

so , 1/f = 1/22+1/2.50 => f = 2.24 cm.

b) The Power in diopters is = 1/f when f is in meters.

so, power = 1/0.0224 = 44.6 diopters

c) And Magnification = -di/do = - 2.5/22 = -0.11

so, the height of the image is 6.5 cm x 0.11 = 0.74 cm = 7.4 mm

d) Using M =-di/do and the image is always cast on the retina, so i =2.5

if do =240 cm, then

m = - 2.5/240 = 0.0142

and the image 0.011*6.5cm = 0.0677cm = 0.677mm tall.

I hope you understood the problem, If yes rate me!! or else comment for a better solution.

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