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A doctor has prescribed a corrective lens of power, −1.5 D. Find the focal length of the lens. Is the prescribed lens perging or converging?
Given:
Power of the lens, $P$ = $-$1.5 D
To find: Focal length of the lens, $f$.
Solution:
Power of a lens is given by:
$P=\frac {1}{f}$
Substituting the given values we get-
$-1.5=\frac {1}{f}$
$f=-\frac {1}{1.5}$
$f=-\frac {10}{15}$
$f=-0.666\ m=-66.6\ cm$
Thus, the focal length of the lens $f$ is 66.6 cm, and the negative sign $(-)$ implies that the lens is diverging in nature. Hence, it is a concave lens.
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