Question

A concave lens has a focal length of 20 cm. At what distance from the lens a 5 cm tall object be placed so that it forms an image at 15 cm from the lens? Also calculate the size of the image formed.

Answer 1

Focal length of the concave lens, f = - 20 cm
Image distance from the concave lens, v = - 15 cm (image formed by a concave lens is on the left side of the lens)
Height of the object, h₁ = 5 cm
Height of the image = h₂
Using the lens formula:

$$\begin{gathered} \frac{\text{1}}{\mathit{\text{f}}}\mathit{\text{=}}\frac{\text{1}}{\mathit{\text{v}}}\mathit{\text{-}}\frac{\text{1}}{\mathit{\text{u}}} \\ \Rightarrow \frac{\text{1}}{\text{-20}}\text{=}\frac{\text{1}}{\text{-15}}\text{-}\frac{\text{1}}{\mathit{\text{u}}} \\ \Rightarrow \frac{\text{1}}{\mathit{\text{u}}}\text{=}\frac{\text{1}}{\text{20}}\text{-}\frac{\text{1}}{\text{15}} \\ \Rightarrow \frac{\text{1}}{\mathit{\text{u}}}\text{=}\frac{\text{3-4}}{\text{60}}\text{=-}\frac{\text{1}}{\text{60}} \end{gathered}$$

∴ u = - 60 cm.

Therefore, the object is at a distance of 60 cm from the concave lens and on the left side of it.
The magnification of the concave lens is given as:

$$\begin{gathered} \text{Magnification=}\frac{\text{image distance}}{\text{object distance}}\text{=}\frac{\text{height of image}}{\text{height of object}} \\ \Rightarrow \frac{\mathit{\text{v}}}{\mathit{\text{u}}}\mathit{\text{=}}\frac{{\mathit{\text{h}}}_{\mathit{2}}}{{\mathit{\text{h}}}_{\mathit{1}}} \\ \Rightarrow \frac{\text{-15}}{\text{-60}}\text{=}\frac{{\text{h}}_2}{5} \\ \Rightarrow {\text{h}}_2\text{=}\frac{5}{4}\text{=1.25 cm.} \end{gathered}$$

Therefore, the size of the image is 1.25 cm.