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' = `R/2` = -10 cm

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

Using the mirror formula, we get 

`1/f=1/u+1/v` 

`1/-10=1/-15+1/v` 

`1/v=1/-10+1/15` 

`v=-30 cm`  

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

magnification= `(-v)/u` 

`m=-(-30)/-15` 

`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  `f =  R_2 = 10 cm` 

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

Magnification = ? 

`1/f=1/u+1/u` 

⇒`1/f=1/-15+1/v`  

⇒`1/10=1/-15+1/v` 

⇒`1/v=1/10+1/15` 

⇒`1/v=15/150+10/150`

⇒`1/v=25/250` 

⇒`1/v=1/6` 

⇒v=6 cm  

Therefore, the imagewilll form 6 cm behind the mirror.Using the magnification formula, we get  `m=(-v)/u` 
`m=(-6)/-15` 
m=0.4 
Thus, the image is virtual, erect and small in size.