Question

A concave lens forms the image of an object such that the distance between the object and image is $10cm$ and the magnification produced is $\frac{1}{4}$. The focal length of the lens will be

• A. $-6.2cm$
• B. $-4.4cm$
• C. $-8.6cm$
• D. $-10cm$

Answer 1

The correct option is B $-4.4cm$

$\begin{array}{cc}m& =\frac{1}{4}=\frac{v}{u}\\ \therefore & v=\frac{u}{4}\end{array}$

The image formed by concave lens is always virtual.

$\begin{array}{c}\text{distance between}\\ \text{object \&}\mathrm{imag}e=u-\frac{u}{4}=10cm\\ \Rightarrow \frac{3u}{4}=10cm\\ \Rightarrow u=\frac{40}{3}cm\end{array}$

$\begin{array}{cc}u& =\frac{40}{3}cm\text{(according to}{\mathrm{sig}}^{n}convertion)\\ v& =\frac{u}{4}=-\frac{10}{3}cm\\ & \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\left[\text{lens formula}\right]\\ \Rightarrow & \frac{1}{f}=\frac{-3}{10}-(-\frac{3}{40})\end{array}$

$\begin{array}{c}\Rightarrow \frac{1}{f}=\frac{3}{40}-\frac{3}{10}\\ \Rightarrow \frac{1}{f}=\frac{-9}{40}\\ \Rightarrow f=-\frac{40}{9}=-4.4cm\\ \text{:)}\text{focal length}=-4.4cm\end{array}$