Vision correction A human eye has a fixed distance deye = 2.40 cm between its le

Question

Vision correction
A human eye has a fixed distance deye = 2.40 cm between its lens and retina. The eye
muscles adjust focus by changing the focal length feye of the lens.
a) Calculate the range of feye for an eye with good vision, able to focus from 25.0 cm away
to infinity.
b) An imperfect eye can only focus on objects closer than 86.0 cm away. Is it nearsighted
or farsighted? What is the power in diopters of a corrective lens placed 2.00 cm in front of
the eye?
c) Consider an eye under water, which causes its focal length to become
about feye = 72.0 cm. What focal length corrective lens placed 2.00 cm in front of the eye
will bring objects at infinity into focus?
Hint: The image formed by the first lens becomes the object for the second lens.

Explanation / Answer

a. use 1/f = 1/u + 1/v

f = uv/(u+v)

f = 2.4 * 25/(2.4 + 25)

f = 2.189 Cm

--------------------------

b. again f = uv/u+v

f = 86 * 23/(86+23)

f = 0.1814 m

so Power P = 1/f

P = 1/0.1814

P = +5.5 Diopters

--------------------------------

f = 25* 2/(25+2)

f = 1.85 cm

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