One end of a long glass rod (refractive index 1.5) is ground into a convex hemispherical surface of radius 10 cm. The rod is kept in water (refractive index 4/3). A point object in the water lies on the axis of the rod, 40 cm from the curved end. The image is formed
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 90 cm from the curved end, in the water
Eq. (9.16) holds for any two media. Here light goes from water into glass, so \(n_1=4/3\) and \(n_2=1.5\). The curved end is convex towards the object. Its centre of curvature lies inside the rod, so \(R\) is positive.
\( u=-40\ \mathrm{cm} \)
\( R=+10\ \mathrm{cm} \)
\( \begin{aligned} \frac{n_2}{v}-\frac{n_1}{u} &= \frac{n_2-n_1}{R} \\ \frac{1.5}{v}+\frac{4/3}{40} &= \frac{1.5-4/3}{10} \\ \frac{1.5}{v}+\frac{1}{30} &= \frac{1}{60} \\ \frac{1.5}{v} &= -\frac{1}{60} \end{aligned} \)
\( v=-90\ \mathrm{cm} \)
The image distance is negative, so the image lies in the water, on the side of the object. It is a virtual image. Hence, the image is formed 90 cm from the curved end, in the water.
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.”