Question

The same size images are formed by a convex lens when the object is placed at $20cm$ or at $10cm$ from the lens. The focal length of a convex lens is ______ $cm$.

Answer 1

Step 1: Given Data

$$\begin{gathered} {\mathrm{u}}_{0_1}=-20\mathrm{cm} \\ {\mathrm{u}}_{0_2}=-10\mathrm{cm} \\ {\mathrm{m}}_1=-{\mathrm{m}}_2 \end{gathered}$$

Images formed by a convex lens when the object is placed at $20cm$is $u_{0_1}$

Images formed by a convex lens when the object is placed at $10cm$is $u_{0_2}$

Object distance in a convex lens is taken as negative due to the sign convention.

Step 2: Formulas Used

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

$$\begin{gathered} \mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}} \\ \Rightarrow \mathrm{m}=\frac{\mathrm{f}}{\mathrm{f}+\mathrm{u}} \end{gathered}$$

$m$ = magnification

$f$ is the focal length of the lens.

$u$ is the distance between the object and the optical center.

$v$ is the distance of the image formed from the optical center of the lens.

Step 3: Calculating the focal length

It is already given in the question that the magnification is the same in both cases.

Object distance in a convex lens is taken as negative due to the sign convention and a convex lens has a positive focal length, hence magnification at $10cm$ is taken as negative.

$$\begin{gathered} {\mathrm{m}}_1=-{\mathrm{m}}_2 \\ \frac{\mathrm{f}}{\mathrm{f}+{\mathrm{u}}_{{\mathrm{o}}_1}}=-\left(\frac{\mathrm{f}}{\mathrm{f}+{\mathrm{u}}_{{\mathrm{o}}_2}}\right) \\ \frac{\mathrm{f}}{\mathrm{f}-10}=-\left(\frac{\mathrm{f}}{\mathrm{f}-20}\right) \\ \mathrm{f}-20=-\mathrm{f}+10 \\ 2\mathrm{f}=30 \\ \Rightarrow \mathrm{f}=15\mathrm{cm} \end{gathered}$$

The focal length of the convex lens is $15\mathrm{cm}$.