Question

An object is 2 m from a lens which forms an erect image one-fourth (exactly) the size of the object. Determine the focal length of the lens. What type of lens is this?

Answer 1

Given:
Object distance (u) = $-$2 m
Image distance (v) = ?
Magnification (m) = 0.25 (one-fourth of the size of the image)
Focal length (f) = ?
Magnification (m) = $\frac{\mathrm{v}}{\mathrm{u}}$
$$\begin{gathered} 0.25=\frac{\mathrm{v}}{-2} \\ \mathrm{v}=-0.5\mathrm{m} \end{gathered}$$

Putting these values in lens formula, we get:
$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{-0.5}-\frac{1}{-2}=\frac{1}{f} \\ \Rightarrow \frac{1}{-0.5}+\frac{1}{2}=\frac{1}{f} \\ \Rightarrow \frac{1}{f}=\frac{1}{2}-\frac{1}{0.5} \\ \Rightarrow \frac{1}{f}=\frac{1}{2}-\frac{10}{5} \\ \Rightarrow \frac{1}{f}=\frac{-3}{2} \\ \Rightarrow f=-0.66\mathrm{m} \end{gathered}$$
Negative sign of focal length shows that lens is diverging in nature. Hence, it is a concave lens.