Question

An object is placed 20 cm from (a) a converging lens, and (b) a diverging lens, of focal length 15 cm. Calculate the image position and magnification in each case.

Answer 1

(a) Focal length of the converging lens, i.e., convex lens f = + 15 cm
Object distance u = - 20 cm
Using the lens formula:

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

∴ v = +60 cm.

Therefore, the image formed is real. It is at a distance of 60 cm from the lens and to its right.

$$\begin{gathered} \text{Magnification=}\frac{\text{image distance}}{\text{object distance}} \\ \therefore \mathit{\text{m =}}\frac{\mathit{\text{v}}}{\mathit{\text{u}}}\mathit{\text{=}}\frac{\mathit{\text{60}}}{\mathit{\text{-20}}}\text{=-3.} \end{gathered}$$

The magnification is greater than 1. Therefore, the image is magnified.
The negative sign shows that the image is inverted.

(b) ​Focal length of the diverging lens, i.e., concave lens f = - 15 cm
Object distance u = - 20 cm

Using the lens formula:

$$\begin{gathered} \frac{\text{1}}{\mathit{\text{f}}}\mathit{\text{=}}\frac{\text{1}}{\mathit{\text{v}}}\mathit{\text{-}}\frac{\text{1}}{\mathit{\text{u}}} \\ \mathit{\Rightarrow }\frac{\text{1}}{\mathit{\text{-15}}}\mathit{\text{=}}\frac{\text{1}}{\mathit{\text{v}}}\mathit{\text{-}}\frac{\text{1}}{\mathit{-}\mathit{20}} \\ \Rightarrow \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{-15}}\text{-}\frac{\text{1}}{\text{20}} \\ \Rightarrow \frac{1}{v}\text{=}\frac{\text{-4-3}}{\text{60}}\text{=-}\frac{\text{7}}{\text{60}} \end{gathered}$$

∴ v = - 8.57 cm.

Therefore, the image formed is virtual. It is at a distance of 8.57 cm from the lens and to its left.

$$\begin{gathered} \text{Magnification=}\frac{\text{image distance}}{\text{object distance}} \\ \therefore \mathit{\text{m =}}\frac{\mathit{\text{v}}}{\mathit{\text{u}}}\text{=}\frac{-8.57}{\text{-20}}\text{=+0.42.} \end{gathered}$$

The magnification is less than 1. Therefore, the image is diminished. The positive sign indicates that the image is erect.