A small light source is fixed inside a long glass rod (refractive index 1.5), on its axis. One end of the rod is ground into a convex spherical surface of radius of curvature 10 cm. The source is 45 cm from this end. The image of the source formed by refraction at this end is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 60 cm from the end, in air
Here the light starts inside the glass and comes out into air through the curved end. So \(n_1=1.5\) and \(n_2=1\) in Eq. (9.16). The centre of curvature of the end lies inside the rod, on the side from which the light comes. So \(R\) is negative.
\( u=-45\ \mathrm{cm} \)
\( R=-10\ \mathrm{cm} \)
\( \begin{aligned} \frac{n_2}{v}-\frac{n_1}{u} &= \frac{n_2-n_1}{R} \\ \frac{1}{v}+\frac{1.5}{45} &= \frac{1-1.5}{-10} \\ \frac{1}{v}+\frac{1}{30} &= \frac{1}{20} \\ \frac{1}{v} &= \frac{1}{60} \end{aligned} \)
\( v=+60\ \mathrm{cm} \)
The value of \(v\) is positive, so the image is real and lies in the air outside the rod. Hence, the image is formed 60 cm from the end, in air.
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.”