Question

What will be the radius of curvature of a concave mirror whose focal length is $25cm$?

Answer 1

Hint: For a uniform spherical lens or spherical mirror, the radius of curvature is twice in length of the focal length of the mirror or lens. Hence, this problem can be solved by using this fact. This relation comes as a result that the object distance and image distance are the same (equal to the radius of curvature), when the object is kept at the center of curvature. Putting this information in the mirror or lens formula, we can get this relation.

Formula used:
For a mirror, the mirror formula is given by,
$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$ (Mirror formula)
where $u$,$v$ are the object and image distances respectively and $f$ is the focal length of the mirror.
Proper sign conventions must be applied.
The focal length $f$ of a uniform spherical lens or mirror is given by
$f=\dfrac{R}{2}$
where $R$ is the radius of curvature, that is, the distance between the pole of the mirror or lens to the center of curvature.

Complete step by step answer:
A concave mirror produces a real image when the object distance is more than the focal length. When the object is at the center of curvature, an inverted real image of the same size is formed at the same location.
The distance between the pole of the mirror and the center of curvature is the radius of curvature of the spherical mirror.
For a mirror, the mirror formula is given by,
$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$ --(1) (Mirror formula)
where $u$,$v$ are the object and image distances respectively and $f$ is the focal length of the mirror.
Proper sign conventions must be applied.
Now, let us analyze the question.
Let the radius of curvature be $R$.
Now if the object is at the center of curvature, by applying proper sign conventions, we get,
$u=-R$
$v=-R$
Using (1), we get,
$\dfrac{1}{-R}+\dfrac{1}{-R}=\dfrac{1}{f}$
$\therefore -\dfrac{2}{R}=\dfrac{1}{f}$
$\therefore R=-2f$
The negative sign is due to the sign convention. The focus and center of curvature are on the side of the object.
Considering only the magnitude, we get,
$R=2f$
Now, it is given that the magnitude of the focal length is $25cm$, the radius of curvature will be
$R=2\times 25=50cm$
Hence, the required radius of curvature is $50cm$.

Note: This problem could also have been solved without the long derivation using the known relation that the radius of curvature is twice the focal length directly. Students should also be aware of always putting the proper sign conventions. The sign conventions are very important since they give information about the nature of the image and also whether they are inverted, real, virtual, magnified, diminished, etc.