Question

An object 3cm high is placed perpendicular to the principal axis of a concave lens of focal length 15cm. The image is formed at a distance of 10cm from the lens. Calculate the distance at which the object is placed and the size and nature of the image formed.

Answer 1

We know in concave lens image formed is always virtual,erect and diminished.
given f = -15cm
v = -10cm
use lens formula
$$\begin{gathered} \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \\ \frac{1}{-15}=\frac{1}{-10}-\frac{1}{\mathrm{u}} \\ \frac{1}{\mathrm{u}}=-\frac{1}{10}+\frac{1}{15} \\ \frac{1}{\mathrm{u}}=\frac{-3+2}{30}=-\frac{1}{30} \\ \Rightarrow \mathrm{u}=-30\mathrm{cm} \\ \mathrm{magnification}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{10}{30}=\frac{1}{3} \\ \Rightarrow \frac{\mathrm{Height}\mathrm{of}\mathrm{image}}{\mathrm{Height}\mathrm{of}\mathrm{object}}=\frac{1}{3} \\ \Rightarrow \frac{\mathrm{Height}\mathrm{of}\mathrm{image}}{3}=\frac{1}{3} \\ \Rightarrow \mathrm{Height}\mathrm{of}\mathrm{image}=1\mathrm{cm} \\ \mathrm{Image}\mathrm{is}\mathrm{virtual},\mathrm{errect}\mathrm{and}\mathrm{diminished} \end{gathered}$$