A point object in air is 60 cm from the convex spherical surface of a transparent plastic block. The radius of curvature of the surface is 20 cm. A real image is formed inside the plastic, 120 cm from the surface. The refractive index of the plastic is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.6
Eq. (9.16) links \(u\), \(v\), \(R\) and the two refractive indices. Here only the refractive index \(n\) of the plastic is unknown. Light goes from air (\(n_1=1\)) into the plastic (\(n_2=n\)). The real image is inside the plastic, so \(v\) is positive.
\( u=-60\ \mathrm{cm} \)
\( v=+120\ \mathrm{cm} \)
\( R=+20\ \mathrm{cm} \)
\( \begin{aligned} \frac{n_2}{v}-\frac{n_1}{u} &= \frac{n_2-n_1}{R} \\ \frac{n}{120}+\frac{1}{60} &= \frac{n-1}{20} \end{aligned} \)
\( n+2=6n-6 \)
\( 5n=8 \)
\( n=1.6 \)
Therefore, the refractive index of the plastic is 1.6.
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.”