Question

(a) An object is placed just outside the principal focus of concave mirror. Draw a ray diagram to show how the image is formed, and describe its size, position and nature.
(b) If the object is moved further away from the mirror, what changes are there in the position and size of the image?
(c) An object is 24 cm away from a concave mirror and its image is 16 cm from the mirror. Find the focal length and radius of curvature of the mirror, and the magnification of the image.

Answer 1

(a) Ray Diagram-

(b) The image will move towards the mirror, and its size will gradually decrease.
(c) Distance of object (u) = -24 cm
Distance of image (v) = -16 cm
We have to find the focal length of the mirror (f) and the magnification of the image (m).
Using the mirror formula, we get

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$$\begin{gathered} \frac{1}{f}=\frac{1}{-24}+\frac{1}{-16} \\ \mathrm{or}\frac{1}{f}=\frac{-2-3}{48} \\ or\frac{1}{f}=\frac{-5}{48} \\ orf=\frac{-48}{5} \\ f=-9.6\mathrm{cm} \\ Therefore,the\mathrm{focal}length\mathrm{of}the\mathrm{concave}\mathrm{mirror}\mathrm{is}9.6\mathrm{cm}. \\ R\mathrm{adius}\mathrm{of}\mathrm{curvature}\left(R\right)=2f \\ R=2\times -9.6 \\ \mathrm{R}=-19.2\mathrm{cm} \\ Therefore,the\mathrm{radius}\mathrm{of}\mathrm{curvature}\mathrm{of}\mathrm{the}\mathrm{mirror}\mathrm{is}19.2\mathrm{cm} \end{gathered}$$

Now, using the magnification formula, we get

$\mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}}$

$$\begin{gathered} \mathrm{m}=\frac{-16}{-24} \\ \mathrm{m}=0.66 \end{gathered}$$

Thus, the image is real, inverted and small in size.