Question

A  convex  mirror  with  a  radius  of  curvature  of  3m  is  used  as  rear  view  in  an
automobile.  If  a  bus  is  located  at  5m  from  the  mirror  find  the  position,  nature
and  size  of  the  image.

Answer 1

Given  , convex mirror

R = 3 cm

object distance ( u ) =  - 5 cm

f = 1.5 cm [ f = R / 2 ]
 
we know that , $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

so

$\frac{1}{1.5} = \frac{1}{v} - \frac{1}{5}$

so

 $\frac{1}{v} = \frac{1}{1.5} + \frac{1}{5}$

V = 1.15

the image is at a 1.15 cm distance from the mirror on the opposite side

Size of the image  = Magnification = $\frac{v}{u}$
            
                                                           = $\frac{1.15}{(-5)}$
                                              
                                                             = 0.23

Hence the size of the object is 0.23 times of the object

The Image is diminished , errect and Virtual

Thanks !!!!!!

Answer 2

Answer:

Given, R = 3cm (Convex Mirror)

f = 3/2 = 1.5m

u = -5m

v = ?

Mirror Formula:

= 1/f = 1/u + 1/v

= 1/1.5 = 1/-5 + 1/v

= 1/v = 1/1.5 + 1/5 = 5 + 1.5/7.5 = 6.5/7.5

∴ v = 7.5/6.5 = 1.15m

∴ The Image is Virtual seen at the back of the Mirror.

Magnification (m) = $h_{i}/h_{o}$= -v/u = -1.15/-5 = 0.23

∴ Image is Erect and Smaller in Size by a factor of 0.23.