Question

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

Answer 1

(b) +20 cm
We know that:
Power (P) = $\frac{1}{\mathrm{Focal}\mathrm{length}}$
Therefore
Power of convex lens P₁= $\frac{1}{10\mathrm{cm}}=\frac{1}{0.1\mathrm{m}}=+10\mathrm{D}$

Power of concave lens P₂= $\frac{1}{-20\mathrm{cm}}=\frac{-1}{0.2\mathrm{m}}=-5\mathrm{D}$

We know that power of lens adds up when lenses comes in contact. Therefore, we have:
Net power (P) = P₁ + P₂
Net power (P) = 10 D $-$ 5 D = 5 D
Net focal length = $\frac{1}{\mathrm{Power}}=\frac{1}{5}=0.2\mathrm{m}=20\mathrm{cm}$