Question

A man standing $1.52m$ in front of a shaving mirror produces an inverted image $18.0cm$ in front of it. How close the mirror should he stand if he wants to form an upright image of his chin that is twice the chin's actual size.

Answer 1

Hint: If we have to solve this question, we need to find the focus of the mirror i.e., a concave mirror here, with the help of mirror formula with proper sign convention. Then, we need to put the given conditions in a magnification formula then proceed further.
Formula used: Mirror formula: $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$where$f = $focal length$v = $ image distance,$u = $ object distance
Magnification formula, $M = - \left( {\dfrac{v}{u}} \right)$

Complete step by step answer:
We have to use proper sign convention for concave mirrors.
We know that we have to put a negative sign for $u$ and $v$ as both lie on the left side of the pole.
So,
$u = - 1.52m$, $v = - 18cm$ or $0.18m$
Now, we have all the data and after putting it in mirror formula we get:
$ \Rightarrow \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} \\
   \Rightarrow \dfrac{1}{f} = - \dfrac{1}{{1.52}} + \left( {\dfrac{{ - 1}}{{0.18}}} \right) \\
   \Rightarrow f = 0.1609m $
Or $f = 16.1cm$
Now,
Magnification,$M = - \left( {\dfrac{v}{u}} \right)$
$\therefore v = - 2u$
Now, putting the above relation and obtained focus in mirror formula we get:
$ \Rightarrow \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} \\
   \Rightarrow \dfrac{1}{{16.1}} = \dfrac{{ - 1}}{{2u}} + \dfrac{1}{u} \\
  \Rightarrow u = 8.05cm $
Final answer: He should stand $8.05cm$ apart from the mirror to get an upright double sized image of himself.

Note: Some points to be kept in the mind while solving these kinds of questions:
> Sign convention must be taken care of to put the data in the formula.
> All the basic formulas must be remembered.
> Given conditions must be followed carefully .