The thin lens formula \( \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \) is derived for a real image formed by a convex lens. A student wants to use it for a concave lens that forms a virtual image of a real object. Which of the following is correct?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → The formula can be used, with \(f\) negative and \(v\) negative.
The thin lens formula works for any thin lens and any image when the Cartesian sign convention is used. The signs of \(u\), \(v\) and \(f\) then show the type of lens and the type of image.
Option (1) The formula cannot be used, because it holds for a convex lens. — Incorrect. A real image of a convex lens is only the case used to derive the formula. With the sign convention, the same formula also works for a concave lens.
Option (2) The formula can be used, with \(f\) negative and \(v\) positive. — Incorrect. A positive \(v\) means the image lies on the far side of the lens. The virtual image of a concave lens forms on the same side as the object, so \(v\) is negative.
Option (3) The formula can be used, with \(f\) negative and \(v\) negative. — Correct. A concave lens is a diverging lens, so \(f\) is negative. Its virtual image lies on the side of the object, against the incident light, so \(v\) is also negative.
Option (4) The formula can be used, with \(f\) positive and \(v\) negative. — Incorrect. A positive \(f\) belongs to a converging lens. Using it for a concave lens would treat the lens as a convex one.
NCERT (2026-27):
Refraction by a lens – “Equation (9.23) is the familiar thin lens formula. Though we derived it for a real image formed by a convex lens, the formula is valid for both convex as well as concave lenses and for both real and virtual images.”
\( \displaystyle \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \qquad (9.23) \)
Summary – “… f is positive for a converging lens; f is negative for a diverging lens. …”
Summary – “3. Cartesian sign convention: Distances measured in the same direction as the incident light are positive; those measured in the opposite direction are negative. …”