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A student uses spectacles of focal length – 2·5 m. (a) Name the defect of vision he is suffering from.(b) Which lens is used for the correction of this defect?(c) List two main causes of developing this defect.(d) Compute the power of this lens.
A student uses spectacles of focal length $-2.5m$.
(a) The, defect of vision he is suffering from is myopia or short-sightedness because the focal length of the spectacles is negative.
(b) Concave lens (or, diverging lens) is used for the correction of this defect.
(c) The two main causes of developing this defect are:
1. Due to elongation of the eyeball.
2. Due to the high converging power of the eye lens.
(d) Given:
Focal length, $f$ = $2.5m$
To find: Power of the lens, $P$.
Solution:
We know that power of the lens is given by-
$P=\frac {1}{f}$
Putting the value of focal length in the formula we get-
$P=\frac {1}{-2.5}$
$P=-\frac {10}{25}$
$P=-0.4D$
Thus, power of this lens is $-$0.4D.
Explanation
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.
This defect occurs either due to the high converging power of the eye lens, (because of its short focal length). Or, due to the eye-ball being too long, which causes light to focus in front of the retina, instead of directly on the retina.
It is corrected by using spectacles containing concave lenses, which should be of such a focal length (or power) that it produces a virtual image of the distant object (lying at infinity) at the far point of the myopic eye.
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