Question

A plane mirror produces a magnification of
A. $-1$
B. $+1$
C. Zero
D. Infinity

Answer 1

Hint: Magnification is defined as the ratio of the height of the image to the height of the object. Use the fact that the image formed due to a plane mirror is always erect and the size of the image is the same as the size of the object.

Formula used:
1) $m=\dfrac{{{h}_{i}}}{{{h}_{o}}}$
m is magnification, ${{h}_{i}}$ and ${{h}_{o}}$ are the heights of the image and the object respectively.
2) $m=-\dfrac{v}{u}$
v and u are the positions of the image and the object with respect to the mirror.

Complete step by step answer:
Let us first understand what is the meant by the magnification.
Magnification is a term used in the cases of mirror and lenses, to understand the nature and size of the image formed.
Magnification is defined as the ratio of the height of the image to the height of the object.
i.e. $m=\dfrac{{{h}_{i}}}{{{h}_{o}}}$ …. (i),
where m is magnification, ${{h}_{i}}$ and ${{h}_{o}}$ are the heights of the image and the object respectively.

The heights that are above the principal axis are considered to be positive and the heights that are below the principal axis are considered to be negative.Therefore, the value of magnification can be positive as well as negative. In the case of a plane mirror, we know that the image formed is erect. The size of the image is the same as the size of the object. Therefore, the height of the image is equal to the height of the object.
Also, both the heights are positive.
${{h}_{i}}={{h}_{o}}$
Substitute this value in equation (i).
$\therefore m=\dfrac{{{h}_{0}}}{{{h}_{o}}}=1$.

This means that the magnification in case of a plane mirror is +1. Hence, the correct option is B.

Note:The magnification for mirrors is also given as $m=-\dfrac{v}{u}$ …. (ii),
where v and u are the positions of the image and the object with respect to the mirror.
Suppose the object is placed in front of a plane mirror at distance d. The image is formed inside the mirror at a distance d from the mirror. Therefore, in this case u is negative and since the image is formed inside the mirror, v is positive.
$ u=-d$ and $ v=d$.
Substitute the values of u and v in (ii).
$\therefore m=-\dfrac{d}{(-d)}=1$