Question

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Fine the position and nature of the image.

Answer 1

Object distance u = -4 cm (left side of the lens)
Focal length f = -12 cm (left side of the lens)
Image distance v = ?
From the lens formula, we know that:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Substituting the values of v and u, we get:
$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}-\frac{1}{-4}=\frac{1}{-12} \\ \Rightarrow \frac{1}{v}=-\frac{1}{12}-\frac{1}{4} \\ \Rightarrow \frac{1}{v}=\frac{-3-1}{12} \\ \Rightarrow \frac{1}{v}=\frac{-4}{12} \\ \Rightarrow v=\frac{12}{-4} \\ \Rightarrow v=-3\mathrm{cm} \end{gathered}$$

Thus, the image is virtual and formed 3 cm away from the lens, on the left side.