The below figure shows the formation of an image by a lens shown by a thick line.

Analyse the figure and answer the following questions.

(a) What is the type of lens used?

(b) What is the nature of the image?

(c) If the image is formed at a distance of 30 cm from the lens and the image is twice the size of the object, then where is the object placed?

(a) Convex lens

Explanation — The lens shown in the figure is a convex lens (converging lens) because the rays passing through the lens seems to be converged at a point on the same side of the lens, forming a virtual and magnified image.

(b) The image formed is virtual, erect and magnified.

(c) Given,

• Image distance (v) = -30 cm ∵ (image is on the same side of the object)
• Size of the image ($\text h$i) = 2 x Size of the object ($\text ho$)

Let, object distance be 'u'.

As, magnification of a lens is given by,

$\text {Magnification (m)} = \dfrac {\text h$i}{\text ho} = \dfrac {\text v}{\text u} \\[1em] \Rightarrow \dfrac {2\times \text ho}{\text ho} = \dfrac {\text v}{\text u} \\[1em] \Rightarrow \dfrac {2}{1} = \dfrac {\text v}{\text u} \\[1em] \Rightarrow \dfrac {\text u}{\text v} = \dfrac {1}{2} \\[1em] \Rightarrow \text u = \dfrac {1}{2} \times \text v \\[1em] \Rightarrow \text u = \dfrac {1}{2} \times -30 \\[1em] \Rightarrow \text u = -15 \text { cm}

This means the object is placed 15 cm in front of the lens.