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bulgar [2K]
2 years ago
15

The near point of an eye is 110 cm. A corrective lens is to be used to allow this eye to focus clearly on objects 26.0 cm in fro

nt of it. What should be the focal length of this lens? What is the power of the needed corrective lens (in diopters)? Do not enter units.
Physics
1 answer:
irga5000 [103]2 years ago
6 0

Answer:

The focal length of the lens is 34.047 cm

The power of the needed corrective lens is 2.937 diopter.

Explanation:

Distance of the object from the lens,u = 26 cm

Distance of the image from the lens ,v= -110 cm

(Image is forming on the other side of the lens)

Since ,lens of the human eye is converging lens,convex lens.

Using a lens formula:

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\frac{1}{f}=\frac{1}{26 cm}+\frac{1}{-110 cm}

f = 34.047 cm = 0.3404 m

Power of the lens = P

P=\frac{1}{f}=\frac{1}{0.34047 m}=2.937 diopter

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Since the angle θ and the magnitude of the velocity is given, the vector representation can be written as follows:

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