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A concave mirror of focal length $20$ cm produces an image twice the height of the object. If the image is real, then the distance of the object from the mirror is
(A) $20$cm
(B) $60$cm
(C) \[10\]cm
(D) \[30\]cm

Last updated date: 03rd Mar 2024
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Hint: The ratio of height of image to that of the object is equal to the magnification. From the given information, we will get the magnification equal to two. Then using the magnification formula for a mirror, we will get the image distance in terms of the object distance, which on substituting into the mirror formula, along with the given focal length, will yield the required object distance.
Formula used: The formulae used to solve this question are
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
$m = \dfrac{{h'}}{h}$
$m = - \dfrac{v}{u}$

Complete step-by-step solution:
We know that the magnification produced by an optical instrument is equal to the ratio of the height of the image to that of the object. So it is given by
$m = \dfrac{{h'}}{h}$
According to the question, $h' = 2h$. Putting this above we get
$m = \dfrac{{2h}}{h}$
$ \Rightarrow m = 2$
For a mirror, magnification is also given by
$m = - \dfrac{v}{u}$
$ \Rightarrow 2 = - \dfrac{v}{u}$
Multiplying $u$ both the sides, we get
$2u = - v$
$ \Rightarrow v = - 2u$ (1)
Now, we have the mirror formula
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
According to the question, the mirror is concave having a focal length of $20$ cm. Since the focal length of a concave mirror is taken as positive. Therefore, we substitute $f = 20cm$ and the equation (1) above to get
$\dfrac{1}{{20}} = \dfrac{1}{{ - 2u}} + \dfrac{1}{u}$
$ \Rightarrow \dfrac{1}{{2u}} = \dfrac{1}{{20}}$
Taking the reciprocal of both the sides, we get
$2u = 20$
Finally, dividing both sides by $2$ we get
$u = 10cm$
Thus, the distance of the object from the mirror is equal to $10cm$.

Hence, the correct answer is option C.

Note: Take proper care of the sign convention for the focal length, as defined by the Cartesian sign convention. Do not forget the negative sign which occurs in the magnification formula for a mirror, given by $m = - \dfrac{v}{u}$.
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