Question

A 2.0 cm tall object is placed 40 cm from a diverging lens of focal length 15 cm. Find the position and size of the image.

Answer 1

A concave lens is also known as a diverging lens.
Focal length of concave lens, f = $-$15 cm
Object distance from the lens, u = $-$40 cm
Height of the object, h₁ = 2.0 cm
Height of the image, h_{2 =} ?
Using the lens formula, we get:

$$\begin{gathered} \frac{\mathit{\text{1}}}{\mathit{\text{f}}}\mathit{\text{=}}\frac{\mathit{\text{1}}}{\mathit{\text{v}}}\mathit{\text{-}}\frac{\mathit{\text{1}}}{\mathit{\text{u}}} \\ \Rightarrow \frac{\text{1}}{\text{-15}}\text{=}\frac{\text{1}}{\text{v}}\text{-}\frac{\text{1}}{\text{-40}} \\ \Rightarrow \frac{1}{-15}=\frac{1}{v}+\frac{1}{40} \\ \Rightarrow \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{-15}}\text{-}\frac{\text{1}}{\text{40}} \\ \Rightarrow \frac{1}{v}\text{=}\frac{\text{-8-3}}{\text{120}} \\ \Rightarrow \frac{1}{v}=\text{-}\frac{\text{11}}{\text{120}} \end{gathered}$$

∴ v = - 10.90 cm

Therefore, the image is formed at a distance of 10.90 cm and to the left of the lens.
Magnification of the lens:

$$\begin{gathered} \text{Magnification=}\frac{\text{Image distance}}{\text{Object distance}}\text{=}\frac{\text{Height of the image}}{\text{Height of the object}} \\ \Rightarrow \frac{\mathit{\text{v}}}{\mathit{\text{u}}}\mathit{\text{=}}\frac{{\mathit{\text{h}}}_{\mathit{2}}}{{\mathit{\text{h}}}_{\mathit{1}}} \\ \mathit{\Rightarrow }\frac{\mathit{\text{-10.90}}}{\mathit{\text{-40}}}\mathit{\text{=}}\frac{{\mathit{\text{h}}}_{\mathit{2}}}{\mathit{2}} \\ \mathit{\Rightarrow }{\mathit{\text{h}}}_{\mathit{2}}\text{=+0.54 cm} \end{gathered}$$

The height of the image formed is 0.54 cm. Also, the positive sign of the height of the image shows that the image is erect.