Question

An object \[2\,{\text{cm}}\] high is placed at a distance of \[16\,{\text{cm}}\] from a concave mirror, which produces a real image \[3\,{\text{cm}}\] high. What is the focal length of the mirror?
A. \[ - 9.6\,{\text{cm}}\]
B. \[ - 3.6\,{\text{cm}}\]
C. \[ - 6.3\,{\text{cm}}\]
D. \[ - 8.3\,{\text{cm}}\]

Answer 1

Hint: First of all, we will find the image distance by using the magnification formula. After that we will use the mirror formula with appropriate sign convention and substitute the required values and manipulate accordingly.

Complete step by step answer:

In the given problem we are supplied the following data:
The height of the object is \[2\,{\text{cm}}\] .
The height of the image is \[3\,{\text{cm}}\] .
The distance of the object from the mirror is \[16\,{\text{cm}}\] .
We are asked to find the focal length of the mirror.
We know, in the case of a concave mirror, when the image is real then it has to be in an inverted position i.e. a concave mirror always gives a real and inverted image except in one condition when the object is inside the focus.

Now we will use the magnification formula in order to find the image distance, which is given as:
\[m = \dfrac{{ - v}}{u}\] and
\[m = \dfrac{{h'}}{h}\]
From these two formulae we can write:
\[\dfrac{{ - v}}{u} = \dfrac{{h'}}{h}\] …… (1)
Where,
\[v\] indicates image distance.
\[u\] indicates object distance.
\[h'\] indicates the height of the image.
\[h\] indicates the height of the object.

Substituting the require values in the equation (1), we get:
$ \dfrac{{ - v}}{u} = \dfrac{{h'}}{h} \\ $
$ \implies \dfrac{{ - v}}{{ - 16}} = \dfrac{{ - 3}}{2} \\ $
$\implies v = \dfrac{{ - 16 \times 3}}{2} \\ $
$ v = - 24\,{\text{cm}} \\ $
Therefore, the image distance comes out to be \[24\,{\text{cm}}\] (negative).

Again, we apply the mirror formula, to find the focal length, which is given by:
\[\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\] …… (2)
Where,
\[v\] indicates image distance.
\[u\] indicates object distance.
\[f\] indicates focal length.

Substituting the required values in the equation (2), we get:
 $ \dfrac{1}{{ - 24}} + \dfrac{1}{{ - 16}} = \dfrac{1}{f} \\ - \left( {\dfrac{1}{{24}} + \dfrac{1}{{16}}} \right) = \dfrac{1}{f} \\ $
$ \implies \dfrac{1}{f} = - \left( {\dfrac{{2 + 3}}{{48}}} \right) \\ $
$ \implies \dfrac{1}{f} = - \dfrac{5}{{48}} \\ $
Again, further simplifying the above expression, we get:
$ f = \dfrac{{ - 48}}{5} \\ $
$ f = - 9.6\,{\text{cm}} \\ $
Hence, the focal length comes out to be \[ - 9.6\,{\text{cm}}\] .

So, the correct answer is “Option A”.

Note:
While solving this problem, you should have a good knowledge of real image and virtual image along with their sign convention. Whenever a mirror forms a real image, then it is formed on the same side of the object and it will be inverted in nature. In contrast, a convex mirror forms a virtual and erect image as it is a diverging mirror.