A thin equiconvex glass lens (refractive index 1.5) has faces of radius of curvature 20 cm. A distant object lies on its principal axis. If refraction at the first face alone is considered, the image I1 formed by this face is at a distance from the lens of
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 60 cm
In the derivation of the lens maker’s formula, the image is found in two steps. The first face forms an image \(I_1\), and \(I_1\) then acts as the object for the second face. At the first face, light goes from air into glass. The object is very far, so \(u\to-\infty\) and the term \(n_1/u\) becomes zero.
\( n_1=1 \)
\( n_2=1.5 \)
\( R=+20\ \mathrm{cm} \)
\( \begin{aligned} \frac{n_2}{v_1}-\frac{n_1}{u} &= \frac{n_2-n_1}{R} \\ \frac{1.5}{v_1}-0 &= \frac{0.5}{20} \\ v_1 &= \frac{1.5\times20}{0.5} \\ &= 60\ \mathrm{cm} \end{aligned} \)
The second face then bends these rays further, and they meet at the focus of the lens, 20 cm away. Hence, the first face alone forms \(I_1\) at 60 cm from the lens.
NCERT (2026-27):
Refraction by a lens – “… The image formation can be seen in terms of two steps:”
“(i) The first refracting surface forms the image I1 of the object O … The image I1 acts as a virtual object for the second surface that forms the image at I …”
Refraction at a spherical surface – “… Substituting these in … we get”
\( \displaystyle \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \)\( \qquad (9.16) \)