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

Answer 1

Answer: (C)

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