Question

(a) Draw ray diagrams to show the formation of a three times magnified (i) real image (ii) virtual image, of an object kept in front of a converging lens. Mark the positions of object, F, 2F, O and position of image clearly in the diagram.
(b) An object of size 5 cm is kept at a distance of 25 cm from the optical centre of a converging lens of focal length 10 cm. Calculate the distance of the image from the lens and size of the image.

Answer 1

18 (a)

(i)


(ii)



18(b) Focal length of the lens, f = 10 cm

The distance of the object from the lens, u = 25 cm

Height of the object, h_o = 5 cm

Using lens formula, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$$\begin{gathered} \Rightarrow \frac{1}{10}=\frac{1}{v}-\frac{1}{25} \\ \Rightarrow \frac{1}{v}=\frac{1}{10}+\frac{1}{25}=\frac{7}{50} \\ \Rightarrow v=\frac{50}{7}\text{cm} \end{gathered}$$

Magnification,

$$\begin{gathered} m=\frac{v}{u}=\frac{h_{\text{i}}}{h_{\text{o}}} \\ \Rightarrow \frac{{\displaystyle 50}}{{\displaystyle 7\times 25}}=\frac{{\displaystyle h_{\text{i}}}}{{\displaystyle 5}} \\ \Rightarrow h_{\text{i}}=\frac{10}{7}\text{cm} \end{gathered}$$

Hence, the distance of the image from the lens is $\frac{50}{7}\mathrm{cm}$ and the size of the image is $\frac{10}{7}\mathrm{cm}$.