Question

An object 3 cm high is held at a distance of 50cm from a diverging mirror of focal length 25 cm. find the nature,position, and size of image formed?

Answer 1

Hey... I think this can be ur answer dear!!

h=3cm
u=-50cm
f=25cm

1/v=1/f-1/u
1/v=1/25-1/50
1/v=2+1/50
v=50/3=16.67cm

h'/h=-v/u
h'=-50/3*3/-50
h'=1cm

Nature - virtual, erect and diminished .....

Answer 2

Image distance = $16.67 cm$ $&$& size of the image $= 1 cm .$

Given,

Size of the object $h_{1} = 3cm$

Size of the object $h_{2}$ $=$ $?$

object distance $u = -50cm$

focal length $f= 25cm$

Image distance $v = ?$

Now,

According to mirror formula  $\frac{1}{v} +\frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} =\frac{1}{f} -\frac{1}{u} =\frac{1}{25} -\frac{1}{-50} =\frac{3}{30}$

or  $v=\frac{50}{3} =16.67 cm$

As $v$ is positive,  image is behind the mirror . It is virtual and erect .

Now, using magnification formula  $m = \frac{h_{2} }{h_{1} }$$= -\frac{v}{u}$

$\frac{h_{2} }{3}$  $= \frac{-50/3}{-50}$$=\frac{1}{3}$

or size of the image is $h_{2}$ $= 1 cm .$

#SPJ2