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:
Answer & explanation
Correct answer: option 4
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)