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A perging lens has a focal length of 3 cm. Calculate the power.
Given:
Focal length, $f$ = $-$3 cm = $-$0.03 m $(\because lens\ is\ diverging\ in\ nature,\ focal\ length\ will\ be\ negative)$
To find: Power of a lens, $P$.
Solution:
We know that power of the lens is given by-
$Power\ (P)=\frac {1}{f}$
Substituting the given value we get-
$P=\frac {1}{-0.03}$
$P=-\frac {100}{3}$
$P=-33.33D$
Hence, the power of the lens is 33.33 D.
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