Question

To form an image twice the size of the object, using a concave mirror of focal length 20 cm, the object distance must be
(A) $> 40cm$
(B) $< 20cm$
(C) Between 20 cm and 60 cm
(D) $< 40cm$

Answer 1

Hint
By convention, the focal length is considered positive for concave mirrors, and image distance is considered to be positive if real, and negative if virtual. The Magnification for a real image of a concave mirror is always negative but for a virtual image, the magnification is positive. So on substituting the values in the formula $\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$ we will get the answer.

Formula used: In this solution we will be using the following formula,
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$ where $f$ is the focal length, $u$ is the object distance and $v$ is image distance. $M = \dfrac{{{h_i}}}{{{h_o}}} = - \dfrac{v}{u}$ where $M$ is the magnification, ${h_i}$ is the image height (or size) and ${h_o}$ is the object height.

Complete step by step answer
For a concave mirror, there are two types of images that can be formed, virtual images and real images. For an object placed at some distance, it produces real images, at other distances it produces virtual images. Hence, we need to investigate both possibilities.
Recalling the mirror equation given by
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$
Also, the formula for Magnification is given by
$\Rightarrow M = \dfrac{{{h_i}}}{{{h_o}}} = - \dfrac{v}{u}$
According to the question, the height (size) of the image is twice that of the object. Since, for a real image we have the magnification as negative, so we get on substituting,
$\Rightarrow - 2 = \dfrac{{{h_i}}}{{{h_o}}} = - \dfrac{v}{u}$
$\Rightarrow v = 2u$
Inserting this value into the mirror equation, we have
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{{2u}}$.
By adding the right hand side of the equation we have
$\Rightarrow \dfrac{1}{f} = \dfrac{{2 + 1}}{{2u}} = \dfrac{3}{{2u}}$
Hence, making $u$ subject of the formula and on substituting the value of the focal length from the question, we get
$\Rightarrow u = \dfrac{3}{2}f = \dfrac{3}{2} \times 20 = 30cm$
Investigating for virtual images, the magnification becomes
$\Rightarrow 2 = \dfrac{{{h_i}}}{{{h_o}}} = - \dfrac{v}{u}$ (magnification for virtual images are positive)
Hence we get,
$\Rightarrow v = - 2u$
Inserting the values into the mirror equation, we have that
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{{( - 2u)}} = \dfrac{1}{f} = \dfrac{1}{u} - \dfrac{1}{{2u}}$
By solving RHS we have
$\Rightarrow \dfrac{1}{f} = \dfrac{{2 - 1}}{{2u}} = \dfrac{1}{{2u}}$
Hence again on substituting the value of the focal length we get,
$\Rightarrow u = \dfrac{1}{2}f = \dfrac{1}{2} \times 20 = 10 cm$
Hence for both types of images produced by concave mirrors, we have one less than 20 cm and another between 20 and 60cm, however both are less than 40 cm. Thus, the object must be somewhere less than 40 cm.
Hence the correct option is (D).

Note
We can make a mistake by not investigating both situations and opting for an answer. For such a case, one may opt for option B or C. For most distances, except for when the object is between the focal length and pole, concave mirrors produce real images. So we should keep this property in mind and not incorrectly generalize without noting this exception.