A concave mirror of focal length $15$ cm forms an image having twice the linear dimensions of the object. The position of the object, when the image is virtual, will be :

22.5

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7.5

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30

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45

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Solution

The correct option is B $7.5$ cm
$\begin{matrix} f = - 15 c m m = + 2 (p o s i t i o n b e c a u s e i m a g e v i r t u a l) m = - \frac{v}{u} \Rightarrow v = - 2 u b y u s i n g m i r r o r f o r m u l a \frac{1}{- 15} = \frac{1}{(- 2 u)} + \frac{1}{u} \Rightarrow u = - 7.5 c m p o s i t i o n o f o b j e c t i s 7.5 c m \end{matrix}$
Hence,
option $(B)$ is correct answer.

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