An object of size 5 cm is placed at a distance of 25 cm from the pole of a concave mirror of radius of curvature 30 cm. Calculate the distance and size of the image so formed. What will be the nature of the image?

We are given a concave mirror. 
Here, 
Object size, h = + 5.0 cm
Object distance, u = - 25 cm
Radius of curvature,  R = - 30 cm     [R is -ve for a concave mirror] 

Therefore, 
Focal length,  $\mathrm{f} = \frac{\mathrm{R}}{2} = -15 \mathrm{cm}$ 

Now, using the mirror formula, 

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

we have,
                  $$\begin{gathered} \frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}} \\ = \frac{1}{-15} - \frac{1}{-25} \\ = \frac{-5+3}{75} \\ = \frac{-2}{75} \end{gathered}$$

i.e.,               $\mathrm{v} = -37.5 \mathrm{cm}.$

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

$\therefore$      Image size, $\mathrm{h}\text{'} = -\frac{v h}{\mathrm{u}}$ 

                                $$\begin{gathered} = -\frac{(-37.5 \mathrm{cm})\hspace{0.17em}(+5.0 \mathrm{cm})}{(-25 \mathrm{cm})} \\ = -7.5 \mathrm{cm}. \end{gathered}$$

As v is (-)ve, so a real, inverted image of height 7.5 cm is formed at a distance of 37.5 cm in front of the mirror.

We are given a concave mirror.

Here,
Object size, h = + 4.0 cm
Object distance, u = -25.0 cm
Focal length, f = - 15.0 cm
Image distance, v = ?
Image size, h' = ? 

Now, using the mirror formula,  

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

$\therefore$                $$\begin{gathered} \frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}} \\ = \frac{1}{-15}-\frac{1}{-25} \\ =-\frac{1}{15}+\frac{1}{25} \end{gathered}$$ 

                   $$\begin{gathered} \frac{1}{\mathrm{v}} = \frac{-5+3}{75} \\ = \frac{-2}{75} \\ \mathrm{v} = -37.5 \mathrm{cm} \end{gathered}$$

The screen should be placed at a distance of 37.5 cm on the object side of the mirror, to obtain a sharp image of the object. 

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

Image size,
                   $$\begin{gathered} \mathrm{h}\text{'} = -\frac{\mathrm{vh}}{\mathrm{u}} \\ = -\frac{(-37.5 \mathrm{cm}) (+4.0 \mathrm{cm})}{(-25 \mathrm{cm})} \\ = -6.0 \mathrm{cm}. \end{gathered}$$ 

The image formed is real, inverted (because h' is negative) and enlarged in size.

Here,   h = + 2 cm,   u = - 30 cm,   f = -15 cm
As               
        
or                                  
Thus, the screen should placed at 30 cm in front of the mirror so as to obtain the real image.
Magnification,  
Image size,  h' = 
The image is real, inverted and of the same size as the object. The image formation is shown in the ray diagram given below.

Given, a convex mirror.

We have, 

Radius of curvature, R = + 3.00 m          [R is +ve for a convex mirro]
Object distance, u = - 5.00 m 
Image distance, v = ?
Height of the image, h'  = ? 

Therefore,
Focal length,  $\mathrm{f} = \frac{\mathrm{R}}{2} = +\frac{3.00 \mathrm{m}}{2} = + 1.50 \mathrm{m}$ 

Using the mirror formula, 

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

$\therefore$                    $$\begin{gathered} \frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}} \\ = +\frac{1}{1.50}-\frac{1}{(-5.00)} \\ = \frac{1}{1.50}+\frac{1}{5.00} \end{gathered}$$

                            $= \frac{5.00+1.50}{7.50}$ 

                          $$\begin{gathered} \mathrm{v} = \frac{+7.50}{6.50} \\ = + 1.15 \mathrm{m} \end{gathered}$$ 

The image is formed at a distance of 1.15 m behind the mirror. 

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

                          $$\begin{gathered} = -\frac{1.15 \mathrm{m}}{-5.00 \mathrm{m}} \\ = + 0.23 \end{gathered}$$ 

The image is virtual, erect and smaller in size by a factor of 0.23.