Question

A concave lens has focal length 15 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also find the magnification produced by the lens.

Answer 1

Focal length of concave lens f = - 15 cm
Image distance v = - 10 cm ( image formed by a concave lens is on the left side of the lens)
Using the lens formula:

$$\begin{gathered} \frac{\text{1}}{\mathit{\text{f}}}\mathit{\text{=}}\frac{\text{1}}{\mathit{\text{v}}}\mathit{\text{-}}\frac{\text{1}}{\mathit{\text{u}}} \\ \mathit{\Rightarrow }\frac{\text{1}}{\mathit{\text{-15}}}\mathit{\text{=}}\frac{\text{1}}{\mathit{\text{-10}}}\mathit{\text{-}}\frac{\text{1}}{\mathit{\text{u}}} \\ \Rightarrow \frac{\text{1}}{\mathit{\text{u}}}\text{=}\frac{\text{1}}{\text{15}}\text{-}\frac{\text{1}}{\text{10}} \\ \Rightarrow \frac{\text{1}}{\mathit{\text{u}}}\text{=}\frac{\text{2-3}}{\text{30}}\text{=-}\frac{\text{1}}{\text{30}} \end{gathered}$$

∴ u = - 30 cm.

The object is at a distance of 30 cm from the lens and on its left.
The magnification of the concave lens is given as:

$$\begin{gathered} \text{Magnification=}\frac{\text{image distance}}{\text{object distance}} \\ \text{⇒}\mathit{\text{m =}}\frac{\mathit{\text{v}}}{\mathit{\text{u}}}\text{=}\frac{\text{-10}}{\text{-30}}\text{=}\frac{\text{1}}{\text{3}} \end{gathered}$$

Since the magnification is less than 1, the image is smaller than the object.