Question

b) Calculate the magnification if an object of height 2 cm kept in front of a concave mirror, give a new image of size 6 cm.

​

Answer 1

Answer:

Given that,

Focal length f=

2

R

Radius of curvature R=30

Object distance u=10cm

Height of object h

o

=2cm

From the mirror formula,

f

1

=

v

1

+

u

1

−

v

1

=

u

1

−

f

1

−

v

1

=

10

1

−

30

2

−

v

1

=

30

3−2

v=−30cm

Now, the magnification is

m=

u

−v

m=

10

30

m=3

Now, the height of image

m=

h

o

h

i

3=

2

h

i

h

i

=6cm

Hence, the image distance is −30 cm, height of image is 6 cm and magnification is 3

Answer 2

Explanation:

Given

Focal length of concave mirror (f) = -10 cm

Height of the object (ho) = 2 cm

Height of the image (hi) = 6 cm

We need to find the object distance

So we know that

Magnification m = hi/ho

hi/ho = 6/2

hi/ho = 3 (erect image)

m = -v/u

3 = -v/u

⇒ v = -3u

Let us now consider Mirror formula

1/v + 1/u = 1/f

1/(-3u) + 1/u = 1/(-10)

⇒ u = -20/3

u = -6.7 cm

So At -6.7cm distance from a concave mirror of focal length 10 cm should an object 2 cm long be placed in order to get an erect image 6 cm tall.