Question

Match the corresponding entries of column $1$ with column $2$ where $m$ is the magnification produced by the mirror for the real objects.
$\begin{array}{cc}\text{Column}1& \text{Column}2\\ \left(A\right)m=-2& \text{(p) Convex mirror}\\ \left(B\right)m=-\frac{1}{2}& \text{(q) Concave mirror}\\ \left(C\right)m=+2& \text{(r) Real image}\\ \left(D\right)m=+\frac{1}{2}& \text{(s) Virtual image}\end{array}$

• A. $A\to r$, $s$; $B\to q$, $s$; $C\to q$, $r$; $D\to p$, $s$
• B. $A\to q$, $r$; $B\to q$, $r$; $C\to q$, $s$; $D\to p$, $s$
• C. $A\to p$, $r$; $B\to p$, $s$; $C\to p$, $q$; $D\to r$, $s$
• D. $A\to p$, $s$; $B\to q$, $r$; $C\to q$, $s$; $D\to q$, $r$

Answer 1

The correct option is B $A\to q$, $r$; $B\to q$, $r$; $C\to q$, $s$; $D\to p$, $s$

Convex mirror always produces virtual and diminished image of a real object.
Thus, $m=+ve$ and $m<1$.
Real images are always inverted, hence $m=-ve$
Virtual images are always erect, hence $m=+ve$
If images are enlarged then $m>1$ and if diminished then $m<1$.
For a concave mirror, a virtual and enlarged image will be formed for position of the object between Pole $\left(P\right)$ and focus $\left(F\right)$ and in all other cases a real and inverted image is formed.


Now,
$\left(A\right)m=-2$, indicates an enlarged real image formed by a concave mirror.
$\left(B\right)m=-\frac{1}{2}$, indicates a diminished real image produced by concave mirror.
$\left(C\right)m=+2$, indicates a virtual and enlarge image produced by concave mirror.
$\left(D\right)m=+\frac{1}{2}$, indicates a diminished virtual image produced by a convex mirror.
$\therefore A\to q$, $r$; $B\to q$, $r$; $C\to q$, $s$; $D\to p$, $s$