Question

A converging mirror forms a three times magnified virtual image when an object is placed at a distance of 8 cm from it.Calculate : ( I ) the position of the image ( ii) the focal length of the mirror​

Answer 1

Answer :-

(i) Position of the image is 24 cm .

(ii) Focal length of the mirror is 12 cm .

Explanation :-

We have :-

→ A converging [concave] mirror.

→ Image is three times, magnified .

→ Distance of the object = 8 cm

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From sign convention :-

• u = -8 cm

From the formula of magnification, we have :-

m = -(v/u) = hᵢ/hₒ

⇒ -(v/-8) = 3hₒ/hₒ

⇒ -(v/-8) = 3

⇒ v/8 = 3

⇒ v = 3(8)

⇒ v = 24 cm

[Positive sign of 'v' indicates that the image is formed behind the mirror.]

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Now, according to mirror formula :-

1/v + 1/u = 1/f

⇒ 1/24 + 1/(-8) = 1/f

⇒ 1/24 + (-1)/8 = 1/f

⇒ (1 - 3)/24 = 1/f

⇒ -2/24 = 1/f

⇒ -2f = 24

⇒ f = 24/-2

⇒ f = -12 cm

[We got value of focal length as negative as focal length is always negative for a concave mirror.]

Answer 2

Given :-

A converging mirror forms a three times magnified virtual image when an object is placed at a distance of 8 cm from it.

To Find :-

(I ) the position of the image ( ii) the focal length of the mirror​

Solution :-

We know that

m = -v/u = hi/ho

-v/-8 = hi/ho

v/8 = 3 × ho/ho

v/8 = 3

v = 8 × 3

v = 24 cm

Now

1/u + 1/v = 1/f

1/(-8) + 1/24 = 1/f

-1/8 + 1/24 = 1/f

1/8 - 1/24 = 1/f

3 - 1/24 = 1/f

2/24 = 1/f

1/12 = 1/f

12 × 1 = f

12 = f

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