Question

Object when placed at_______ in front of concave mirror magnification of -1 is obtained.
A. Infinity
B. Center of curvature
C. Between focus and center of curvature
D. Principal focus

Answer 1

Hint: When magnification produced by the mirror is negative then it implies that the image formed is below the principle axis and on the same side of the mirror as the object is. Therefore the image formed when magnification is negative shows the image is real and inverted.

Complete step by step solution:
Let the focal length of the concave mirror is $f$
The object is kept at distance $u$ from the concave mirror.
Let $v$ is the image distance from the mirror.
As we know that the magnification is the ratio of the image height to the height of the object.
Mathematically,
$m=\dfrac{{{H}_{i}}}{{{H}_{o}}}=-\left( \dfrac{v}{u} \right)$
Where,
${{H}_{i}}=$Height of the image
${{H}_{o}}=$Height of the object
As it is given that the magnification is -1,
$\begin{align}
  & -\left( \dfrac{v}{u} \right)=-1 \\
 & v=u \\
\end{align}$
Using mirror formula,
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
Putting the value of $v$we get,
$\begin{align}
  & \dfrac{1}{u}+\dfrac{1}{u}=\dfrac{1}{f} \\
 & \dfrac{2}{u}=\dfrac{1}{f} \\
 & u=2f \\
\end{align}$
As we know that the focal length of a spherical mirror is the half of radius of curvature,
$\begin{align}
  & f=\dfrac{R}{2} \\
 & 2f=R \\
\end{align}$

Therefore the object is kept at a distance equal to the radius of the curvature. Hence, the object is at center of curvature.

Note: - For concave mirror, the focal length is negative.
- For convex mirrors, the focal length is positive.