Question

An object is placed at a distance of 30 cm from a concave lens of a focal length of 30 cm.
i. Use the lens formula to determine the distance of the image from the lens.
ii. List four characteristics of the image (nature, position, size, erect/inverted) in this case.
iii. Draw a well-labeled diagram to justify your answer part (ii).

Answer 1

Step 1: Given data

Object distance, $u=-30cm$
Focal Length, $f=-30cm$

Step 2: Formula Used
Lens Formula, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

1. where f is the focal length of the concave lens,
2. v is the image distance from the optical center,
3. u is the object distance from the optical center.

Step 3: Calculating the image distance
Using the lens formula, we get:
$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \Rightarrow \frac{1}{-30}=\frac{1}{v}-\frac{1}{-30} \\ \Rightarrow \frac{1}{v}=\frac{1}{-30}+\frac{1}{-30} \\ \Rightarrow \frac{1}{v}=\frac{-2}{30} \\ \Rightarrow v=-15cm \end{gathered}$$

Step 4: Characteristics of the image formed in this case

1. Nature: As the image distance is negative, it is formed on the left-hand side, i.e., the image is virtual.
2. Position: The image distance calculated above is negative, indicating that the image is formed on the same side as that of the object, i.e., to the left of the lens.
3. Size: The image formed by the concave lens when the object is at focus is diminished.
4. The image is erect as all the virtual images are erect.

Step 5: Diagram of the image formation

As shown in the diagram below, the image is virtual, erect, diminished and on the same side as the object.