A 50-year-old optometry patient focuses on a 6.50cm -tall photograph at his near

Question

A 50-year-old optometry patient focuses on a 6.50cm -tall photograph at his near point. (See the table.) We can model his eye as a sphere 2.50cm in diameter, with a thin lens at the front and the retina at the rear. (Figure 1)

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

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

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

If he views the photograph from a distance of 2.50m , how tall is its image on his retina in mm?

Explanation / Answer

a)

We use the equation:

1/u + 1/v = 1/f

So, 1/40 + 1/2.5 = 1/f

So, f = 2.35 cm <-------answer (effective focal length)

b)

Power of eye = 1/f = 42.6 D

c)

Now, image height, h = object height*(v/u) = 6.5*(2.5/40) = 0.406 cm = 4.06 mm

d)

Now, image height, h = 6.5*(2.5/250) = 0.065 cm = 0.65 mm <------answer

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