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(a) What is meant by the power of accommodation of an eye?(b) A person with a myopic eye cannot see objects beyond 1.2 m directly. What should be the type of the corrective lens used? What would be its power?
(a) The term "power of accommodation" of the human eye refers to the ability of the eye to focus the distant objects as well as the nearby objects on the retina by changing the focal length (or converging power) of its lens.
(b) Concave lens (or, diverging lens) is used to correct the vision of a myopic eye person.
Myopia, also known as near-sightedness or short-sightedness, is a defect of vision in which a person can't see the distant object clearly (appears blurred), though can see the nearby objects clearly. The far point of a myopic eye is less than infinity.
Given:
The person with a myopic eye cannot see objects beyond 1.2 m directly, which means the far point of the person is 1.2 m in front of the eye (the person can see objects kept at infinity if the images of the objects are formed at the person's own far point of 1.2 m from the eye).
Object distance, , $u$ = $\infty$
Image distance, , $v$ = $-$1.2 m (the far point of the defective eye in front of the lens)
In order to find the power of the concave lens required, first, we have to calculate its focal length.
To find: Focal length, $f$, and Power of the lens, $P$.
Solution:
From the lens formula, we know that-
$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$
Substituting the given value, we get-
$\frac {1}{(-1.2)}-\frac {1}{\infty}=\frac {1}{f}$
$-\frac {1}{1.2}-0=\frac {1}{f}$ $(\because Any\ number\ divided\ by\ infinity\ is\ equal\ to\ zero)$
$-\frac {1}{1.2}=\frac {1}{f}$
$f=1.2m$
Thus, the focal length, $f$ of the lens is -1.2m.
Now,
We know that power of the lens is given as-
$P=\frac {1}{f}$
Substituting the value of $f$ we get-
$P=\frac {1}{1.2}$
$P=\frac {10}{12}$
$P=-0.83D$
Thus, the power, $P$ of the lens is -0.83D.
Hence, a concave lens of focal length 1.2m and power -0.83D is to be used.
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