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A convex lens of focal length 10 cm is placed in contact with a concave lens of focal length 20 cm. The focal length of this combination of lenses will be:(a) +10 cm (b) +20 cm (c) −10 cm (d) −20 cm
(a) +20 cm
Explanation
Given:
Focal length of the convex lens, $f_1$ = $+$10 cm = $+$0.1 m (power of converging lens is always taken positive)
Focal length of the concave lens, $f_2$ = $-$20 cm = $-$0.2 m (power of diverging lens is always taken negative)
To find: Combined focal length of the two lenses, $f$.
Solution:
To find the combined focal length of the two lenses, first, we have to find out the combined power of the two lenses.
We know that, power of the lens is given as-
$P=\frac {1}{f}$
Substituting the given value, we get-
$P_1=\frac {1}{0.1}=\frac {10}{1}=10m$
$P_2=\frac {1}{-0.2}=-\frac {10}{2}=-5m$
We know that power of the lens adds up when lenses come in contact. Therefore, the combined power of the lenses is given as-
$P=P_1+P_2$
Substituting the given values, we get-
$P=10D+(-5D)$
$P=10D-5D$
$P=+5D$
Thus, the combined power, $P$ of the lenses is +5D.
Now,
The combined focal length, $f$ is given as-
$f=\frac {1}{P}$
Substituting the value of $P$ in the expression, we get-
$f=\frac {1}{5}$
$f=+0.2m=+20cm$
Thus, the combined focal length, $f$ is +20cm.
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