Question

An object is placed at a distance of 50 cm from a concave lens of a focal length of 20 cm. Find the nature and position of the image.

Answer 1

Step 1: Given:

The focal length of the lens, $f=-20cm$ (The focal length of the concave lens is negative)

object distance, $u=-50cm$ (distances measured left side of the lens should be taken negatively.)

Step 2: Finding the image distance using the lens formula:

We know that $\begin{array}{l}\frac{1}{f}=\frac{1}{v}–\frac{1}{u}\end{array}$

$$\begin{gathered} Bysubstitutingthegivenvalues,weget \\ \begin{array}{l}\Rightarrow \frac{1}{-20}=\frac{1}{v}–\frac{1}{-50}\\ \Rightarrow \frac{1}{-20}–\frac{1}{50}=\frac{1}{v}\\ \Rightarrow \frac{-5-2}{100}=\frac{1}{v}\\ \Rightarrow \frac{-1}{14.3}=\frac{1}{v}\\ therefore,v=-14.3cm\end{array} \end{gathered}$$

The position of the image is 14.3 cm in front of the lens towards the object.

The nature of the image is virtual and erect.