Question

A concave mirror has a radius of curvature of \[24\,cm\], how far is an object from the mirror if the image formed is $3$ times the size of the object.

Answer 1

Hint: Learn the mirror’s formula to find the distance of the object from the mirror. The mirror’s formula is the relation between the image distance object distance and the focal length of the mirror.

Formula used:
Mirror’s formula is given by,
\[\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\]
where, \[v\]is the image distance, \[u\] is the object distance and \[f\] is the focal length of the mirror.
Magnification in a mirror is given by,
\[m = \dfrac{{{h_i}}}{{{h_o}}} = - \dfrac{v}{u}\]
where, \[m\] is the magnification of the object, \[{h_i}\] is the image height and \[{h_o}\] is the object height.

Complete step by step answer:
We have given here a concave mirror of radius curvature \[24\,cm\]. The size of the image is 3 times the size of the object. We have to find the distance of the object from the mirror.Now, let’s at first say that the image is real and the distance of the object is \[ - x\,cm\] away from the object.Now, the magnification can be written as,
\[\left| {\dfrac{v}{{ - x}}} \right| = 3\]
\[\Rightarrow \left| v \right| = 3x\].
We know that the mirror’s formula is given by, \[\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\]

Also, we know that the radius of the curvature is related to the focal length of the mirror as, \[f = \dfrac{r}{2}\].
Hence, putting the values in the mirror’s formula we will have,
\[\dfrac{1}{{ - 3x}} + \dfrac{1}{{ - x}} = \dfrac{2}{{ - 24}}\].....(since the focal length is negative for concave mirror and the image is real )
\[\dfrac{4}{{3x}} = \dfrac{1}{{12}}\]
\[\therefore x = 16\]
Hence, the image will be at a distance of \[16\,cm\] away from the mirror for a real image.

Now, if the image is virtual then the image distance will be positive hence, object distance will be \[ - x\,cm\] and the image distance will be \[3x\,cm\]
Hence, putting the values in the mirror’s formula we will have,
\[\dfrac{1}{{3x}} - \dfrac{1}{x} = \dfrac{2}{{ - 24}}\]
\[\Rightarrow \dfrac{2}{{3x}} = \dfrac{1}{{12}}\]
\[\therefore x = 8\]

So, for a virtual image the object will be at a distance of \[8\,cm\] away from the mirror.

Hence, the object can be either at \[16\,cm\] or at \[8\,cm\] depending on the type of the image (real or virtual).

Note:The negative sign implies that the distance is in the left of the mirror. The distance measured to the left of the incident ray is taken to be negative and the distance measured to the right of the incident ray is taken to be always positive. . If the image is formed behind the mirror the image will said to be a virtual image.Here, since nothing is specified about the image if it is real or virtual we need to find out the image for both the cases.Since the magnification is negative for the real image the image is inverted and for virtual since it is positive the image is erect.