An object 2 cm in size is placed 30 cm in front of a concave mirror of focal length 15 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? What will be the nature and the size of the image formed? Draw a ray diagram to show the formation of the image in this case.

We are provide with a concave mirror. 

Here,
Size of the object, h = + 2 cm
Object distance, u = - 30 cm
Focal length, f = -15 cm 

Now, using the mirror formula, 

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

we have therefore, 

$\therefore$                  $$\begin{gathered} \frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}} \\ = \frac{1}{-15}-\frac{1}{-30} \\ = -\frac{1}{30} \end{gathered}$$ 

i.e.,                 $\mathrm{v} = -30 \mathrm{cm}.$ 

Thus, the screen should be placed at 30 cm in front of the mirror so as to obtain the real image. 

Magnification, $\mathrm{m} = \frac{\mathrm{h}\text{'}}{\mathrm{h}} = -\frac{\mathrm{v}}{\mathrm{u}}$ 
Therefore, 

Image size,  h', 
                       $$\begin{gathered} =-\frac{\mathrm{vh}}{\mathrm{u}} \\ = -\frac{(-30) (+2)}{(-30)} \\ = - 2 \mathrm{cm} \end{gathered}$$ 

The image formed is real, inverted and is of the same size as the object.

The image formation is shown in the ray diagram given below.