For a given convex lens, if the object distance is:
(A) \( u = -\infty \)
(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)
Use the lens formula \( 1/v - 1/u = 1/f \) with the Cartesian sign convention for each object distance, and compare the magnitude of the image distance \( v \).
\( \displaystyle \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \)
\( (A)\ u = -\infty: \; v = f \)
\( \displaystyle (B)\ u = -2f: \; \frac{1}{v} = \frac{1}{f} - \frac{1}{2f} = \frac{1}{2f} \;\Rightarrow\; v = 2f \)
\( \displaystyle (C)\ u = -f: \; \frac{1}{v} = \frac{1}{f} - \frac{1}{f} = 0 \;\Rightarrow\; v = \infty \)
\( \displaystyle (D)\ u = -\frac{f}{3}: \; \frac{1}{v} = \frac{1}{f} - \frac{3}{f} = -\frac{2}{f} \;\Rightarrow\; v = -\frac{f}{2} \)
In (D) the image is virtual, on the same side as the object, at a distance \( f/2 \). In decreasing order of distance, \( \infty \gt 2f \gt f \gt f/2 \), that is (C), (B), (A), (D).