Question

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 diverging?

Answer 1

Focal length of convex lens f₁ = + 25 cm = + 0.25 m

Power of convex lens ${\mathit{\text{P}}}_{\mathit{1}}\mathit{\text{=}}\frac{\mathit{\text{1}}}{{\mathit{\text{f}}}_{\mathit{1}}}\text{=}\frac{\text{1}}{\text{0.25}}\text{=4 D}$
Focal length of concave lens f₂ = - 10 cm = - 0.10 m

Power of concave lens ${\mathit{\text{P}}}_{\mathit{2}}\mathit{\text{=}}\frac{\text{1}}{{\mathit{\text{f}}}_{\mathit{2}}}\text{=}\frac{\text{1}}{\text{-0.10}}\text{=-10 D}$

(a) The power of the combination of lenses is the algebraic sum of the powers of the individual lenses.
∴ Power of combination P = P₁ + P₂
⇒ P = 4 - 10 = - 6 D.

(b) Suppose, the focal length of the combination of the lenses is f.
The power of a lens and the focal length are related as:

$$\begin{gathered} \mathit{\text{P}}\text{=}\frac{\text{1}}{\mathit{\text{f}}} \\ \text{-6=}\frac{\text{1}}{\mathit{\text{f}}} \\ \mathit{\text{f}}\text{=}\frac{\text{1}}{\text{-6}}\text{=-0.167 m = - 16.7 cm} \end{gathered}$$

Therefore, the focal length of the combination of the lenses is - 16.7 cm.

(c) The focal length of the combination of the lenses is - 16.7 cm. Here, the negative sign shows that the combination of the two lenses acts like a concave lens. Therefore, this combination of lenses is diverging.