Question

Which of the following option has the correct combination considering column-I and column-II ?

• A. A → S
• B. B → R
• C. C → P
• D. D → Q

Answer 1

The correct option is D D → Q

(A) Concave mirror:

$v=\frac{uf}{u-f},f\text{is negative, let}f=-F_0$

$u$	$-\infty$	$-2F_0$	$-F_0$	$0$	$F_0$	$\infty$
$v$	$-F_0$	$-2F_0$	$\pm \infty$	$0$	$-\frac{F_0}{2}$	$F_0$

$v=-\frac{u{F}_0}{u+F_0}$


(B) Convex lens:
$v=\frac{uf}{u+f},f\text{is positive, let}f=F_0$
$v=\frac{u{F}_0}{u+F_0}$

$u$	$-\infty$	$-2F_0$	$-F_0$	$0$	$F_0$	$\infty$
$v$	$-F_0$	$-2F_0$	$\pm \infty$	$0$	$-\frac{F_0}{2}$	$F_0$

(C) Convex Mirror:
$v=\frac{uf}{u+f},f\text{is positive, let}f=F_0$
$v=\frac{u{F}_0}{u-F_0}$

$u$	$-\infty$	$-F_0$	$0$	$F_0$	$2F_0$	$\infty$
$v$	$F_0$	$\frac{F_0}{2}$	$0$	$\pm \infty$	$2F$	$F_0$

(D) Concave lens:


$v=\frac{uf}{u+f},f\text{is negative left}f=-F_0$
$v=\frac{u{F}_0}{u-F_0}$

$u$	$-\infty$	$-F_0$	$0$	$F_0$	$2F_0$
$v$	$-F_0$	$\frac{-F_0}{2}$	$0$	$\pm \infty$	$-2F_0$