Question

the diverginng mirror of a radius of curvature 40 cm forms image at 10 cmfrom it pole the object distance is​

Answer 1

Explanation:

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

Answer:

Explanation:

Data:

Radius of curvature = R = 40 cm

We have to find the position of object and image.

Let U be the position of an object and V be the position of an image.

Then,

U = ?

V = ?

Also suppose that, h₁ and h₂ be the height of object and height of image respectively.

Then, according to the given condition, the height of image is half of the height of object.

Therefore, h₂ = h₁ / 2

Also, we know that,  

- V / U = h₂ / h₁

- V / U = (h₁ / 2) / h₁

- V / U = 1 / 2

2 V = - U  

V = - U / 2   ......... (1)

Now we use the mirror formula,

1 / V + 1 / U = 1 / f

1 / V + 1 / U = 2 / R

1 / (-U/2) + 1 / U = 2 / R  ,   Using equation (1)

- 2 / U + 1 / U = 2 / 40

- 1 / U = 1 / 20

U = - 20 cm

Now,

V = - U / 2 = - (-20) / 2

V = 10 cm

which is the required answer.

Hope it helps. Thanks.