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A converging lens has a focal length of 50 cm. The power of this lens is :
(a) +0.2D (b) −2.0D (c) +2.0D (d) −0.2D
(c) +2.0 D
Explanation
Given:
Focal length of the lens, $f$ = $+$50 cm = $+$0.50 m $(\because lens\ is\ converging\ in\ nature,\ focal\ length\ will\ be\ positive)$
To find: Power of the lens, $P$.
Solution:
Power of a lens is given by-
$P=\frac {1}{f}$
Substituting the given values we get-
$P=\frac {1}{0.50}$
$P=\frac {100}{50}$
$P=+2D$
Thus, the power of the lens, $P$ is 2 D or 2.0 D.
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