Read the following passage and answer the questions.
In a school workshop, Riya takes a long glass rod of refractive index 1.5. She grinds one end of the rod into a convex spherical surface of radius of curvature 10 cm. The centre of curvature lies inside the glass. She places a small lamp in air on the axis of the rod, facing the curved end. Then she moves the lamp to different distances from this end. For some positions, the light comes to a focus inside the glass. For other positions, the light spreads out inside the rod. Next, she cuts a thin disc from the curved end and polishes its back face flat. This gives her a thin plano-convex lens. From the same glass, she also grinds a thin double convex lens with both faces of radius 10 cm, and a thin double concave lens. Her teacher asks her to explain what she sees with the relation for refraction at a single spherical surface and the lens maker’s formula.
At what distance from the curved end should the lamp be placed so that the light inside the rod travels parallel to the axis?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 20 cm
Light that travels parallel to the axis inside the glass forms its image at infinity. So the image distance in the relation for a single spherical surface is \( v=\infty \), and the term \( n_2/v \) becomes zero.
Given:
\( n_1=1 \)
\( n_2=1.5 \)
\( R=+10\ \mathrm{cm} \)
\( v=\infty \)
\( \begin{aligned} \frac{n_2}{v}-\frac{n_1}{u} &= \frac{n_2-n_1}{R} \\ 0-\frac{1}{u} &= \frac{1.5-1}{10} \\ \frac{1}{u} &= -\frac{1}{20} \end{aligned} \)
\( u=-20\ \mathrm{cm} \)
The negative sign shows that the lamp is in air, in front of the curved end. Hence, the lamp must be placed 20 cm from the curved end.
NCERT (2026-27):
Refraction at a spherical surface – “… Applying the Cartesian sign convention, OM = –u, MI = +v, MC = +R Substituting these …, we get”
\( \displaystyle \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \)\( \qquad (9.16) \)
“… It holds for any curved spherical surface.”