A point object in air is placed 30 cm in front of a convex spherical surface of glass (refractive index 1.5). A real image is formed inside the glass, 90 cm from the surface. The radius of curvature of the surface is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 10 cm
The object and the image are both known here, so Eq. (9.16) gives the radius of curvature. Light goes from air (\(n_1=1\)) into glass (\(n_2=1.5\)). The real image lies inside the glass, so \(v\) is positive.
\( u=-30\ \mathrm{cm} \)
\( v=+90\ \mathrm{cm} \)
\( \begin{aligned} \frac{n_2-n_1}{R} &= \frac{n_2}{v}-\frac{n_1}{u} \\ \frac{0.5}{R} &= \frac{1.5}{90}+\frac{1}{30} \\ &= \frac{1}{60}+\frac{2}{60}=\frac{1}{20} \end{aligned} \)
\( \begin{aligned} R &= 0.5\times20 \\ &= 10\ \mathrm{cm} \end{aligned} \)
Therefore, the radius of curvature of the surface is 10 cm.
NCERT (2026-27):
Refraction at a spherical surface – “… Substituting these in … we get”
\( \displaystyle \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \)\( \qquad (9.16) \)
“Equation (9.16) gives us a relation between object and image distance in terms of refractive index of the medium and the radius of curvature of the curved spherical surface. It holds for any curved spherical surface.”