Question

At what distance from a concave mirror of focal length $ 10\;cm $ should an object $ 2\;cm $ long be placed to get an erect image $ 6cm $ tall?

Answer 1

Hint: Concave mirrors can produce images that are real as well as virtual. They can be erect (if virtual) or inverted (if real). Behind the mirror (if virtual) or in front of the mirror (if real). They can also be bigger, smaller, or the same size as the object.

Formula Used
Magnification formula:
 $ - \dfrac{v}{u} = \dfrac{{{h_{\text{i}}}}}{{{h_0}}} $
Mirror formula:
 $ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{{\text{u}}} $
Where
 $ v $ is the image distance,
 $ u $ is the object distance,
 $ {h_i} $ is the height of the image,
 $ {h_o} $ is the height of the object,
 $ f $ is the focal length.

Complete Answer:
We know that magnification $ = \dfrac{{{\text{ height of the object }}}}{{{\text{ height of the image }}}} $
Magnification is also defined as $ \dfrac{{{\text{ - image distance }}}}{{{\text{ object distance }}}} $
So, we can equate the above two definitions such that
 $ - \dfrac{v}{u} = \dfrac{{{{\text{h}}_{\text{i}}}}}{{{{\text{h}}_0}}} = \dfrac{6}{2} = 3 $
 $ - v = 3u $
We will now replace $ v $ with $ 3u $ in the mirror formula
 $ \dfrac{{ - 1}}{{10}} = \dfrac{1}{{ - 3u}} + \dfrac{1}{u} $
 $ \Rightarrow \dfrac{{ - 1}}{{10}} = \dfrac{2}{{3u}} $
Hence we get,
 $ \Rightarrow - 10 = \dfrac{{3u}}{2} $
 $ \Rightarrow u = \dfrac{{ - 20}}{3} = 6.7\;m $ .

Additional Information:
Concave mirrors are capable of having real images. The image is inverted and real if the object is further away from the mirror than the focal point, which implies that the image appears on the same side of the mirror as the object.

Note:
A concave mirror has a reflective surface that is curved inward and away from the light source. Unlike convex mirrors, depending on the distance between the object and the mirror, the image formed by a concave mirror displays different image types.