Question

Match the Column-I and Column-II

Column I		Column II
A	An object is placed at focus of a convex mirror	1	Magnification is $-\infty$
B	An object is placed at centre of curvature before a concave mirror	2	Magnification is 0.5
C	An object is placed at focus before a concave mirror	3	Magnification is +1
D	An object is placed at centre of curvature before a convex mirror	4	Magnification is -1
		5	Magnification is 0.33

• A. (A) $\rightarrow$(2) ; (B) $\rightarrow$ (3) ; (C) $\rightarrow$ (4); (D) $\rightarrow$ (5)
• B. (A) $\rightarrow$ (5); (B) $\rightarrow$ (4); (C) $\rightarrow$ (3); (D) $\rightarrow$ (1)
• C. (A) $\rightarrow$ (2); (B) $\rightarrow$ (4); (C) $\rightarrow$ (1); (D) $\rightarrow$ (5)
• D. (A) $\rightarrow(3) ;$ (B) $\rightarrow(5) ;$ (C) $\rightarrow(2) ;$ (D) $\rightarrow$ (4)

Answer 1

Answer: (C)

(A) $\rightarrow$ (2); (B) $\rightarrow$ (4); (C) $\rightarrow$ (1); (D) $\rightarrow$ (5)
(A) For $u=-f$,
$ \frac{1}{v}+\frac{1}{-f}=\frac{1}{f} $
$\therefore v =\frac{f}{2}$
and $M=-\frac{v}{u}=-\frac{f / 2}{(-f)}=0.5$.
(B) $u=-2 f$, so $v=-2 f$
$M=-\frac{v}{u}=-\left(\frac{-2 f}{-2 f}\right)=-1$
(C) In concave mirror, $u=-2 f, v=-\infty$
$\therefore M=-\frac{v}{u}=-\infty \text {. }$
(D) In convex mirror $ u=-2 f$
so $\frac{1}{v}+\frac{1}{-2 f}=\frac{1}{f} \Rightarrow v=\frac{2 f}{3}$.
Now $M=-\frac{v}{u}=\frac{1}{3}$.