Question

The image of an object formed by a mirror is real, inverted and is of magnification -1 . If the image is at the distance of 30cm from the mirror, where is the object placed ? Find the position of the image if the object is now moved 20cm towards the mirror . What is the nature of the image obtained ? Justify your answer with the help of ray diagram.

Answer 1

m=-v/u

-1=-(-30)/u

-1=30/u

u=-30

so object is placed at a distance of 30 cm from the mirror

now, 1/f=1/v+ 1/u

so, 1/f=-1/30+(-1/30)

1/f= -1/30-1/30

1/f=-1-1/30

1/f=-2/30

f=-30/2

f=-15cm

now, if object is placed 20 cm closer to mirror

then,u=-10cm

1/f=1/v+(-1/10)

-1/15= 1/v-1/10

-1/15+1/10= 1/v

-2+3/30=1/v

-1/30=1/v

v=-30

so image is real and inverted.

Answer 2

Answer: The  image is formed at 30 cm behind the mirror and image is virtual.

Explanation:

Given that,

Magnification m = -1

Distance of the image v = -30

We know that,

The magnification is the ratio of the distance of the image and distance of the object.

$m =-\dfrac{v}{u}$

$-1 = -\dfrac{-30}{u}$

$u =-30 cm$

The focal length is

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

$\dfrac{1}{f}=\dfrac{-2}{30}$

$f = -15 cm$

If the object is moved 20 cm towards the mirror.

$u = -10 cm$

Using again mirror's formula

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

$\dfrac{1}{-15}= \dfrac{1}{v}+\dfrac{1}{-10}$

$\dfrac{1}{v}=\dfrac{1}{30}$

$v =30 cm$

Hence, the  image is formed at 30 cm behind the mirror and image is virtual.