Question

The power of a combination of two lenses X and Y is 5 D. If the focal length of lens X be 15 cm :
(a) calculate the focal length of lens Y.
(b) state the nature of lens Y.

Answer 1

(a) The focal length of lens X, ie., f₁ = 15 cm = 0.15 m
Power of lens X, ${\mathit{\text{P}}}_{\mathit{1}}\text{=}\frac{\text{1}}{{\mathit{\text{f}}}_{\mathit{1}}}\text{=}\frac{\text{1}}{\text{0.15}}\text{=6.67 D}$
Let the power and the focal length of the lens Y be P₂ and f_2^, respectively.
The resultant power of the combination of two lenses is the algebraic sum of the powers of the individual lenses.
∴ Power of combination of lenses P = P₁ + P₂
$\Rightarrow$ 5 = 6.67 + P₂
​ $\Rightarrow$ P₂ = 5 - 6.67 = - 1.67 D
The power of the lens is related to the focal length of the lens as:

$$\begin{gathered} {\mathit{\text{P}}}_{\mathit{2}}\text{=}\frac{\text{1}}{{\mathit{\text{f}}}_{\mathit{2}}} \\ \Rightarrow \text{-1.67 =}\frac{\text{1}}{{\mathit{\text{f}}}_{\mathit{2}}} \\ \Rightarrow {\mathit{\text{f}}}_{\mathit{2}}\text{=}\frac{\text{1}}{\text{-1.67}}\text{=-0.60 m = -60 cm} \end{gathered}$$

The focal length of the lens Y is -60 cm.

(b) The focal length of the lens Y is negative. Therefore, the lens is concave and diverging in nature.