Question

Define the focal length of a spherical mirror. A concave mirror produces $3$ times a real image of an object placed at a distance of $10\,cm$ in front of it. Find the radius of curvature of the mirror.

Answer 1

Hint
Focal length of the spherical mirror is the distance from the pole of the mirror to the focus point of the object. The radius of the curvature of the mirror is determined by the two formulas, one is magnification formula and other is focal length formula.
The magnification of the mirror is given by,
$\Rightarrow m = \dfrac{{ - v}}{u}$
Where, $m$ is the magnification, $v$ distance of the image and $u$ is the distance of the object.
The focal length of the mirror is given by,
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
Where, $f$ is the focal length, $v$ distance of the image and $u$ is the distance of the object.

Complete step by step answer
Given that,
The magnification of the mirror is, $m = 3$,
The distance of the object of a real image is, $u = 10\,cm$.
Now,
The magnification of the mirror is given by,
$\Rightarrow m = \dfrac{{ - v}}{u}\,.....................\left( 1 \right)$
By substituting the magnification and the distance of the object value in the above equation (1), then the above equation is written as,
$\Rightarrow 3 = \dfrac{{ - v}}{{10}}$
By keeping the term $v$ in one side and the other terms in other side, then
$\Rightarrow - v = 3 \times 10$
On multiplying the terms in the above equation, then
$\Rightarrow v = - 30\,cm$
The above equation is written as,
$\Rightarrow v = 30\,cm$
Now,
The focal length of the mirror is given by,
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\,...............\left( 2 \right)$
By substituting the distance of the object and the distance of the image in the above equation, then
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{{30}} + \dfrac{1}{{10}}$
By cross multiplying the terms in RHS, then
$\Rightarrow \dfrac{1}{f} = \dfrac{{10 + 30}}{{30 \times 10}}$
On further simplification, then
$\Rightarrow \dfrac{1}{f} = \dfrac{{40}}{{300}}$
By taking reciprocal on both sides, then
$\Rightarrow f = \dfrac{{300}}{{40}}$
On dividing the above equation, then
$\Rightarrow f = 7.5\,cm$
The radius of curvature is equal to the two times of the focal length.
$\Rightarrow R = 2f$
By substituting the focal length value in the above equation, then
$\Rightarrow R = 2 \times 7.5$
On multiplying the above equation, then
$\Rightarrow R = 15\,cm$
The radius of the curvature is $15\,cm$.

Note
The negative values show that the concave mirror reduces the distance of the image focused. The main function of the concave mirror is to decrease the length of the image focused. And the radius of the curvature is equal to the two times of the focal length.