# A convex lens of focal length 25 cm and a concave lens of focal length 10 cm are placed in close contact with one another.(a) What is the power of this combination?(b) What is the focal length of this combination?(c) Is this combination converging or perging?

Given:

Focal length of convex lens $f_1$ = $+$25 cm = $+$0.25 m   $(\because lens\ is\ convex,\ focal\ length\ will\ be\ positive)$

Focal length of concave lens $f_2$ = $-$10 cm = $-$0.1 m     $(\because lens\ is\ concave,\ focal\ length\ will\ be\ negative)$

To find: Power of the lens $P_1$ and $P_2$

Solution:

Power of the lens is given by-

$P=\frac {1}{f}$

Substituting the given value, we get-

$P_1=\frac {1}{f_1}=\frac {1}{0.25}=\frac {100}{25}=+4D$

$P_2=\frac {1}{f_2}=\frac {1}{-0.1}=-\frac {10}{1}=-10D$

(a) The power $(P)$ of the combination of lenses is given as-

$P=P_1+P_2$

$P=4+(-10)$

$P=4-10$

$P=-6D$

(b) The power of a lens and the focal length are related as:

$P=\frac {1}{f}$

So, the focal length $(f)$ of the combination of lenses is given as-

$f=\frac {1}{P}$

Putting the value of $P$ in the above expression we get-

$f=\frac {1}{-6}$

$f=\frac {1}{-6}$

$f=-0.166m=-16.6cm$

Thus, the focal length $(f)$ of the combination of lenses is -16.6 cm.

(c) The combination of the lenses has negative focal length, i.e, -16.6 cm, so it is a diverging or concave lens.

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Updated on: 10-Oct-2022

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