Question

A diverging mirror of radius of curvature 40 cm forms an image which is half the height of the object . Find the object and image positions.

Answer 1

Answer:

R = 40cm

f = R/2 = 40/2 = 20cm

Image is half the height of the object.

i.e. m = - (v/u) = h2/h1 = 1/2

⇒ u = -2v

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

⇒ 1/v + 1/(-2v) = 1/20

⇒ 1/v – 1/2v = 1/20

⇒ 1/2v = 1/20

∴ v = 10 cm

u = -2v = -2 × 10 = -20 cm

So, the object is placed 20 cm in front of the mirror and the image is formed 10 cm behind the mirror.

Answer 2

R = 40cm

f = R/2 = 40/2 = 20cm

Image is half the height of the object.

i.e. m = - (v/u) = h2/h1 = 1/2

⇒ u = -2v

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

⇒ 1/v + 1/(-2v) = 1/20

⇒ 1/v – 1/2v = 1/20

⇒ 1/2v = 1/20

∴ v = 10 cm

u = -2v = -2 × 10 = -20 cm

So, the object is placed 20 cm in front of the mirror and the image is formed 10 cm behind the mirror.