Each lens used below has a focal length of magnitude 20 cm. A real object is placed in front of it. Match List-I (lens and object distance) with List-II (magnification).
| List-I | List-II |
|---|---|
| (A) Convex lens, object 30 cm from the lens | (I) \(-\frac{1}{2}\) |
| (B) Convex lens, object 60 cm from the lens | (II) \(+2\) |
| (C) Convex lens, object 10 cm from the lens | (III) \(-2\) |
| (D) Concave lens, object 20 cm from the lens | (IV) \(+\frac{1}{2}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
(A) Convex lens, object 30 cm from the lens — This matches (III). With \(u=-30\ \mathrm{cm}\) and \(f=+20\ \mathrm{cm}\), the lens formula \( \frac{1}{v}=\frac{1}{f}+\frac{1}{u} \) gives \(v=+60\ \mathrm{cm}\). Since \( m=\frac{v}{u} \), we get \(m=-2\), a real image twice as large.
(B) Convex lens, object 60 cm from the lens — This matches (I). With \(u=-60\ \mathrm{cm}\), the formula gives \(v=+30\ \mathrm{cm}\). So \(m=-\frac{1}{2}\).
(C) Convex lens, object 10 cm from the lens — This matches (II). The object is inside the focal length. With \(u=-10\ \mathrm{cm}\), the formula gives \(v=-20\ \mathrm{cm}\), a virtual image. So \(m=+2\).
(D) Concave lens, object 20 cm from the lens — This matches (IV). Here \(f=-20\ \mathrm{cm}\) and \(u=-20\ \mathrm{cm}\), so the formula gives \(v=-10\ \mathrm{cm}\). So \(m=+\frac{1}{2}\), an erect image half the size.
NCERT (2026-27):
Refraction by a lens – “… Applying the sign convention, BO = – u, DI = +v, we get”
\( \displaystyle \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \)\( \qquad (9.23) \)
Refraction by a lens – “… 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.”
Refraction by a lens – “… it is easily seen that for a lens”
\( \displaystyle \frac{h^{\prime}}{h}=\frac{v}{u} \)\( \qquad (9.24) \)