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zzz [600]
3 years ago
15

The eye is actually a multiple-lens system, but we can approximate it with a single-lens system for most of our purposes. When t

he eye is focused on a distant object, the optical power of the equivalent single lens is 45.4 diopters. a. What is the effective focal length of the eye?Feff = ____________ cmb. How far in front of the retima is this "equivalent lens" located?di= ______________ cm
Physics
1 answer:
hram777 [196]3 years ago
5 0

Explanation:

Given that,

The optical power of the equivalent single lens is 45.4 diopters.

(a) The relationship between the focal length and the focal length is given by:

f=\dfrac{1}{P}

f=\dfrac{1}{45.4}

f = 0.022 m

or

f = 2.2 cm

(b) We need to find how far in front of the retina is this "equivalent lens" located. It is given by using lens formula as :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Here, u = infinity

\dfrac{1}{v}=\dfrac{1}{2.2}

v = 2.2 cm

So, at 2.2 cm in front of the retina is this "equivalent lens" located.

Hence, this is the required solution.

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