The image formed by a convex mirror of focal length $f$ is $(1/n)$ times the size of the object. The distance of the object from the mirror is:

$A$ thin cylindrical rod of length $10\,cm$ is placed horizontally on the principal axis of a concave mirror of focal length $20\,cm$. The rod is placed in such a way that the midpoint of the rod is at $40\,cm$ from the pole of the mirror. The length of the image formed by the mirror will be $\frac{x}{3}\,cm$. The value of $x$ is $............$.

An object of height $3.6 \ cm$ is placed normally on the principal axis of a concave mirror of radius of curvature $30 \ cm$. If the object is at a distance of $10 \ cm$ from the principal focus of the mirror, then the height of the real image formed due to the mirror is (in $cm$)

Use the mirror equation to deduce that:
$(a)$ an object placed between $f$ and $2f$ of a concave mirror produces a real image beyond $2f$.
$(b)$ a convex mirror always produces a virtual image independent of the location of the object.
$(c)$ the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
$(d)$ an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

$A$ thin rod of $5 \, cm$ length is kept along the axis of a concave mirror of $10 \, cm$ focal length such that its image is real and magnified. If one end of the rod is at a distance of $2f$ from the mirror,what will be its magnification?

$A$ concave mirror is used to focus the image of a flower on a nearby wall $120 \ cm$ from the flower. If a lateral magnification of $16$ is desired,the distance of the flower from the mirror should be $... \ cm$.