Question

The image of an object formed by a mirror is real, inverted and is of magnification -1. If the image is at a distance of 30cm from the mirror, where is the object placed, where could the image be if the object is moved 20cm towards the mirror, state reason and draw a ray diagram.

Answer 1

$v = - 30cm \\ u = to \: find \\ m = - 1 \\ m = \frac{ - v}{u} \\ - 1 = \frac{ - ( - 30)}{u} \\ u = - 30cm$
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\ \frac{1}{f} = \frac{1}{ - 30} + \frac{1}{ - 30} \\ \frac{1}{f} = \frac{1 + 1}{ - 30} \\ \frac{1}{f} = \frac{2}{ - 30} \\ f = - 15cm$


if the object is moved 20 cm away.Then

u=-30+20
u=-10cm
v=to find
f=-15cm
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\ \frac{1}{ - 15} = \frac{1}{v} + \frac{1}{ - 10} \\ \frac{1}{v} = \frac{1}{ - 15} + \frac{1}{10} \\ \frac{1}{v} = \frac{2 - 3}{ - 30} \\ v = - 30cm$

thnxx
Hope it helps you
Please mark the answer as Brainliest.

Answer 2

Answer:

+ 30 cm

Explanation:

Given :

Magnification = - 1

Image distance = 30 cm

We know

m = - v / u

- 1 = 30 / u

u = - 30 cm

Now we have ,

1 / f = 1 / v + 1 / u

1 /  f = 1 / - 30 - 1 / 30

f = - 15 cm

When object moved 20 cm

u' = - 30 + 20 = - 10 cm

We have f = - 15 cm

We have to find new image distance v'

Again ,

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

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

1 / v' = 1 / 30

v' = 30.

Nature of image as :

Virtual

Erect

Magnified.