A 30-year-old optometry patient focuses on a 6.50 cm -tall photograph at his nea

Question

A 30-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? f =_____ cm

Part B What is the power of his eye in diopters when he focuses on the photo? D = ____diopters

Part C How tall is the image of the photo on his retina? |h| =_____ mm

Part D Is the image mentioned in part C erect or inverted? Real or virtual? Is the image mentioned in part C erect or inverted? Real or virtual?

a.The image is erect and virtual.

b. The image is inverted and real.

c.The image is inverted and virtual.

d.The image is erect and real.

Part E If he views the photograph from a distance of 2.10 m , how tall is its image on his retina? h''=_____mm

Explanation / Answer

a)

do = object distance = 14 cm

di = image distance = 2.50 cm

f = focal length of eye

using the lens equation

1/f = 1/di + 1/do

1/f = 1/2.50 + 1/14

f = 2.12 cm

b)

Power is given as

P = 100 /f = 100/2.12 = 47.2 D

c)

ho = height of photograph = 6.50 cm

hi = image height

using the formula

hi/ho = - di/do

hi/6.50 = - 2.50/14

hi = - 1.16 cm

d)

image is inverted since magnification is negative . and the image is real too

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