Question

What is the power of a concave mirror if it produces a real image twice the height of the object placed at 30 cm from the mirror?
A) $5D$
B) $- 5D$
C) $0.05D$
D) $- 0.05D$

Answer 1

In this solution, we will use the concepts of magnification and the mirror formula to determine the focal length of the mirror. The power of a mirror is the inverse of its focal length. We will use these relations to determine the power of the concave mirror.
Formula used: In this solution we will be using the following formula,
Mirror formula: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ where $v$ is the image distance, $u$ is the object distance and $f$ is the focal length of the mirror
Magnification of a mirror: $m = \dfrac{v}{u}$

Complete step by step answer:
We’ve been given that the concave mirror produces an image twice the height of the object. So, we can say that
$m = \dfrac{v}{u} = 2$
Which gives us $v = 2u$
Now since $u = - 30\,cm$ (assuming it is placed on the left of the mirror), we will get
$v = - 60\,cm$
Using the mirror formula, we can write that
$\dfrac{1}{{ - 60}} + \dfrac{1}{{ - 30}} = \dfrac{1}{f}$
Solving for the focal length, we get
$\dfrac{1}{f} = \dfrac{3}{{ - 60}}\,cm$
On taking inverse
$f = - 20\,cm$
Now the power of a mirror is the inverse of its focal length. The focal length of the mirror in question is $f = - 20\,cm$.
This means the power will be
$P = - 1/20$
$\Rightarrow P = - 0.05\,D$ which corresponds to option (D).

Note: While using the mirror formula, we should be careful in taking the sign convention of the object image and the focal length of the system. By convention, the object is usually assumed to be placed left of the mirror as we have done in the question. So, the reflective surface of the mirror should also be facing the object which means for a concave lens, the focal length will also be negative which can help us in narrowing the choices down to options (B) and (D).