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(a) What is short-sightedness? State the two causes of short-sightedness (or myopia). With the help of ray diagrams, show:(i) the eye-defect short-sightedness.(ii) correction of short-sightedness by using a lens.(b) A person having short-sight cannot see objects clearly beyond a distance of 1.5 m. What would be the nature and power of the corrective lens to restore proper vision?
(a) Short-sightedness, or myopia, is a defect of vision because of which a person cannot see distant objects clearly but has a normal nearby vision.
The two causes of short-sightedness (or myopia) are as follows:
1. The high converging power of the eye lens (because of its short focal length), which causes the formation of the images of objects in front of the retina, or
2. The eye ball is too long, which causes light to focus in front of the retina, instead of directly on the retina.
(i) The ray diagram showing the eye-defect short-sightedness.
(ii) The ray diagram showing the correction of short-sightedness by using a lens.
(b) A person suffering from short-sightedness can correct the defect by wearing spectacles containing concave lenses.
In order to find the power of the concave lens, we have to first calculate its focal length.
Given:
The far point of the short-sighted person is 1.5 m from the eye (the person can see the object kept at infinity if the image of the object is formed at the person's own far point of 1.5 m from the eye).
So,
Object distance, $u$ = $\infty$
Image distance, $v$ = $-$1.5 m (far point of the defective eye in front of the lens)
To find: Focal length, $f$ and power, $P$.
Solution:
From the lens formula we know that-
$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$
Substituting the given values we get-
$\frac {1}{(-1.5)}-\frac {1}{\infty}=\frac {1}{f}$
$-\frac {1}{1.5}-0=\frac {1}{f}$ $(\because Any\ number\ divided\ by\ infinity\ is\ equal\ to\ 0)$
$-\frac {1}{1.5}=\frac {1}{f}$
$f=-1.5m$
Thus, the focal length $f$ is -1.5m.
Now,
We know that power of the lens is given as-
$P=\frac {1}{f}$
Putting the value of $f$ we get-
$P=\frac {1}{-1.5}$
$P=-\frac {10}{15}$
$P=-0.667D\approx-0.67D$
Thus, the power of the concave lens required to rectify the defect is -0.67 dioptres.
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