Question

An object is placed 15 cm from (a) a converging mirror, and (b) a diverging mirror, of radius of curvature 20 cm. Calculate the image position and magnification in each case. 

Answer 1

 Case -1

It is a converging mirror, i.e. concave mirror.

Distance of the object 'u' = -15 cm

Radius of curvature of the mirror 'R' = -20 cm

Focal length of the mirror 'f' = ${\displaystyle \frac{R}{2}}$ = -10 cm

We have to find the position of the image 'v' and its magnification 'M'.

Using the mirror formula, we get 

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

${\displaystyle \frac{1}{-10}=\frac{1}{-15}+\frac{1}{v}}$ 

${\displaystyle \frac{1}{v}=\frac{1}{-10}+\frac{1}{15}}$ 

${\displaystyle v=-30cm}$  

the image wii be from at a distance of 30 cm in front of converging mirror 

magnification= ${\displaystyle \frac{-v}{u}}$ 

${\displaystyle m=-\frac{-30}{-15}}$ 

${\displaystyle m=-2}$ 

magnification =-2

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

Mirror is converging mirror i.e. convex mirror.

Distance of the object  u = -15 cm

Radius of curvature of the mirror  R = 20 cm

Focal length of the mirror  ${\displaystyle f= R_2 =10cm}$ 

We have to find the position of the image  v = ?

Magnification = ? 

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

⇒${\displaystyle \frac{1}{f}=\frac{1}{-15}+\frac{1}{v}}$  

⇒${\displaystyle \frac{1}{10}=\frac{1}{-15}+\frac{1}{v}}$ 

⇒${\displaystyle \frac{1}{v}=\frac{1}{10}+\frac{1}{15}}$ 

⇒${\displaystyle \frac{1}{v}=\frac{15}{150}+\frac{10}{150}}$

⇒${\displaystyle \frac{1}{v}=\frac{25}{250}}$ 

⇒${\displaystyle \frac{1}{v}=\frac{1}{6}}$ 

⇒v=6 cm  

Therefore, the imagewilll form 6 cm behind the mirror.Using the magnification formula, we get  ${\displaystyle m=\frac{-v}{u}}$ 
${\displaystyle m=\frac{-6}{-15}}$ 
m=0.4 
Thus, the image is virtual, erect and small in size.