Question

An object 50 cm tall is placed on the principal axis of a convex lens. Its 20 cm tall image is formed on the screen placed at a distance of 10 cm from the lens. Calculate the focal length of the lens.

Answer 1

Height of object (h) = 50 cm
Height of image (h') = -20 cm (real and inverted)
Distance of image from the lens (v) = 10 cm
Distance of object from the lens (u) = ?
Focal length of the lens (f)= ?
We know, magnification (m) of the lens is given by:

$\mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{\mathrm{h}\text{'}}{\mathrm{h}}$

Thus, substituting the values of v, h and h', we get:

$$\begin{gathered} \frac{10}{\mathrm{u}}=\frac{-20}{50} \\ \frac{10}{\mathrm{u}}=\frac{-20}{50} \\ \mathrm{u}=\frac{-5}{2}\mathrm{x}10 \\ \therefore \mathrm{u}=-25\mathrm{cm}. \end{gathered}$$

Using the lens formula:

$\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$

$$\begin{gathered} \frac{1}{10}-\frac{1}{-25}=\frac{1}{\mathrm{f}} \\ \Rightarrow \frac{1}{10}+\frac{1}{25}=\frac{1}{\mathrm{f}} \\ \Rightarrow \frac{5+2}{50}=\frac{1}{\mathrm{f}} \\ \Rightarrow \frac{7}{50}=\frac{1}{\mathrm{f}} \\ \Rightarrow \mathrm{f}=\frac{50}{7} \\ \Rightarrow \mathrm{f}=7.14\mathrm{cm}. \end{gathered}$$