For a given convex lens, if the object distance is:

(A) $u = -∞$
(B) $u = -2f$
(C) $u = -f$
(D) $u = -f/3$

The distance of the image is arranged in decreasing order:

Choose the correct answer from the options given below:

(C), (B), (A), (D)

The correct answer is Option (4) → (C), (B), (A), (D)

Using lens formula:

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

⟹ $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$

(A) For $u = -\infty$ :

$\frac{1}{v} = \frac{1}{f} + 0$

$v = f$

(B) For $u = -2f$ :

$\frac{1}{v} = \frac{1}{f} - \frac{1}{2f} = \frac{1}{2f}$

$v = 2f$

(C) For $u = -f$ :

$\frac{1}{v} = \frac{1}{f} - \frac{1}{f} = 0$

$v = \infty$

(D) For $u = -\frac{f}{3}$ :

$\frac{1}{v} = \frac{1}{f} - \frac{3}{f} = -\frac{2}{f}$

$v = -\frac{f}{2}$ (virtual, on same side as object)

Hence, in decreasing order of image distance (considering magnitude):

$\infty > 2f > f > \frac{f}{2}$

Final Order: (C), (B), (A), (D)