The minimum distance between real object and real image in the context of convex lens of focal length 'f' is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 4f \)
Use the lens formula \( \displaystyle\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \) with the Cartesian sign convention. For a real object \( u = -x \) and for a real image \( v \) is positive, so with the object distance \( x \) and the image distance \( v \) both as positive numbers
\( \displaystyle \frac{1}{v} + \frac{1}{x} = \frac{1}{f} \)
The object and the image are on opposite sides of the lens, so the distance between them is \( D = x + v \). Putting \( v = D - x \):
\( \displaystyle \frac{1}{D - x} + \frac{1}{x} = \frac{1}{f} \;\Rightarrow\; \frac{D}{x(D - x)} = \frac{1}{f} \)
\( \displaystyle x^{2} - D\,x + D f = 0 \)
A real object position \( x \) exists only if this quadratic has real roots, so its discriminant cannot be negative:
\( \displaystyle D^{2} - 4Df \ge 0 \;\Rightarrow\; D \ge 4f \)
The minimum distance is \( 4f \), reached when \( x = D/2 = 2f \), that is, object and image both at \( 2f \) from the lens.
Shortcut: The object-image distance is least for the symmetric position \( u = v = 2f \), giving \( 2f + 2f = 4f \).