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If a concave mirror has a focal length of $10 cm$ , find the positions where an object can be placed to give, in each case, an image twice the height of the object.

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- 5 cm

- 10 cm

- 15 cm

Both 1 and 3

answer is D.

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Detailed Solution

The object should be placed at a distance of

5 cm

and

15 cm

in front of the concave mirror.
Given,
Focal length,

f = - 10 cm

(concave mirror)
Case-1
The image is real, and its magnification,

m = - 2

From the relation of magnification, distance of object and image

m = - \frac{v}{u}

Here

\Rightarrow m = - 2

\Rightarrow - 2 = - \frac{v}{u}

\Rightarrow v = 2 u

The relation between image distance, object distance and focal length is given below

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

\Rightarrow \frac{1}{2 u} + \frac{1}{u} = \frac{1}{- 10}

\Rightarrow \frac{1 + 2}{2 u} = \frac{1}{- 10}

\Rightarrow \frac{3}{2 u} = \frac{1}{- 10}

\Rightarrow 2 u = - 10 \times 3

\Rightarrow u = \frac{- 10 \times 3}{2}

\Rightarrow u = - 15 cm

Thus, the object should be placed at

15 cm

in front of the concave mirror.
Case-2
The image is virtual, and its magnification,

m = + 2

From the relation of magnification, distance of object and image

m = - \frac{v}{u}

Here

\Rightarrow m = + 2

\Rightarrow + 2 = - \frac{v}{u}

\Rightarrow 2 u = - v

\Rightarrow v = - 2 u

The relation between image distance, object distance and focal length is given below

\Rightarrow \frac{1}{v} + \frac{1}{u} = \frac{1}{f}

\Rightarrow \frac{1}{- 2 u} + \frac{1}{u} = \frac{1}{- 10}

\Rightarrow \frac{- 1 + 2}{2 u} = \frac{1}{- 10}

\Rightarrow \frac{1}{2 u} = \frac{1}{- 10}

\Rightarrow 2 u = - 10

\Rightarrow u = - 5 cm

Thus, the object should be placed at

5 cm

in front of the concave mirror.

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