Question

An object 3 cm high is placed at a distance of 8 cm from a concave mirror which produces a virtual image 4.5 cm high:

(i) What is the focal length of the mirror?
(ii) What is the position of image?
(iii) Draw a ray-diagram to show the formation of image.

Answer 1

Distance of the object from the mirror 'u' = -8 cm

Height of the object 'h_o' = 3 cm
Height of the image 'h_i' = 4.5 cm
We have to find the focal length of the mirror 'f' and distance of the image 'v'.
Using the magnification formula, we get
$$\begin{gathered} m=\frac{h_i}{h_0}=-\frac{v}{u} \\ m=\frac{4.5}{3}=\frac{-v}{-8} \\ v=\frac{4.5\times 8}{3}=12\mathrm{cm} \end{gathered}$$

Therefore, the distance of the image 'v' is 12 cm behind the mirror.

Now, using the mirror formula, we get
$$\begin{gathered} \frac{1}{f}=\frac{1}{v}+\frac{1}{u}=\frac{1}{12}+\frac{1}{-8} \\ \frac{1}{f}=\frac{2}{24}-\frac{3}{24}=\frac{-1}{24} \\ f=-24\mathrm{cm} \end{gathered}$$

Thus, the focal length of the concave mirror 'f' is 24 cm.