Question

An object of size $3cm$ is placed at $30cm$ in front of a concave mirror of radius of curvature $40cm$. Find the distance and size of the image from the mirror and describe the nature of the image.

Answer 1

This problem can be solved by using the mirror formula after finding out the focal length from the radius of curvature. The image size can be found out from the magnification obtained with the formula involving the object and image distance after getting the image distance.
Formula used:
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
$m=-\dfrac{v}{u}$
$m=\dfrac{{{h}_{i}}}{{{h}_{o}}}$
$f=\dfrac{R}{2}$

Complete answer:
We will first find out the focal length of the mirror from its radius of curvature and use it in the mirror formula to get the image distance which will be used to get the magnification. The image height can then be found out from the magnification.
The focal length $f$ of a spherical mirror of radius of curvature $R$ is given by
$f=\dfrac{R}{2}$ --(1)
The mirror formula relates the image distance $v$, object distance $u$ and the focal length $f$ for a mirror as
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ --(2)
The magnification $m$ of the image produced by a mirror is given by
$m=-\dfrac{v}{u}$ --(3)
where $v,u$ are the image distance and object distances respectively.
Also, the magnification $m$ can be written as
$m=\dfrac{{{h}_{i}}}{{{h}_{o}}}$ --(4)
Where ${{h}_{i}},{{h}_{o}}$ are the image and object sizes respectively.
Now, let us analyze the question.
The radius of curvature of the mirror is $R=-40cm$ (For a concave mirror, the radius of curvature is negative according to sign convention).
Let the focal length of the mirror be $f$.
The object distance is $u=-30cm$ (The object distance is negative by sign convention)
Let the image distance be $v$.
The size of the object is ${{h}_{o}}=3cm$.
Let the size of the image be ${{h}_{i}}$.
Let the magnification of the image produced by the mirror be $m$.
Now, using (1), we get the focal length as
$f=\dfrac{-40}{2}=-20cm$ --(5)
Using the mirror formula (2) for the situation, we get
$\dfrac{1}{v}+\dfrac{1}{-30}=\dfrac{1}{-20}$ [Using (5)]
$\therefore \dfrac{1}{v}=-\dfrac{1}{20}+\dfrac{1}{30}=\dfrac{-3+2}{60}=\dfrac{-1}{60}$
$\therefore v=-60cm$ --(6)

Now, we get the magnification of the image produced using (3) as
$m=-\dfrac{60}{-30}=-2$ [Using (6)] --(7)
Therefore, using (4) and from this magnification, we get the image height as
$-2=\dfrac{{{h}_{i}}}{3}$
$\therefore {{h}_{i}}=-2\times 3=-6cm$
Therefore, the image distance, that is, the distance of the image from the mirror is $v=-60cm$. The negative sign implies that the image is formed on the same side of the mirror as that of the object. It also implies that the object is a real image.
The size of the image is $-6cm$. The negative sign implies that the image is inverted with respect to the object.
Also the magnification is $-2$ which implies that the image is inverted and twice as large as the object.
Hence, the image obtained is real, inverted and enlarged.

Note:
Students must always take care of the sign conventions especially while solving such numerical problems in optics as the signs hold a lot of meaning with respect to the orientations of the mirrors, the objects and their images. For example, if we had not taken the focal length of the concave mirror as negative (by considering the radius of curvature as negative), we would have obtained a completely different and wrong answer for the image distance using the mirror formula and subsequently all our following calculations and answers would have been wrong.