Light from point source in air falls on a spherical glass surface (\( n = 1.5 \) and radius of curvature = 10 cm). The distance of the light source from the glass surface is 50 cm. The position at which the image is formed is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 50 cm in the glass surface
Use the formula for refraction at a spherical surface, with the Cartesian sign convention (distances measured from the pole, positive in the direction of the incident light).
\( \displaystyle \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \)
\( n_1 = 1 \)
\( n_2 = 1.5 \)
\( u = -50\,\mathrm{cm} \)
\( R = +10\,\mathrm{cm} \)
The surface is convex towards the source, so its centre of curvature lies in the glass and \( R \) is positive.
\( \displaystyle \frac{1.5}{v} + \frac{1}{50} = \frac{1.5 - 1}{10} = \frac{1}{20} \)
\( \displaystyle \frac{1.5}{v} = \frac{1}{20} - \frac{1}{50} = \frac{3}{100} \)
\( \displaystyle v = \frac{1.5 \times 100}{3} = +50\,\mathrm{cm} \)
The positive sign means the image is real and forms in the glass, 50 cm from the surface.
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Refraction at a Spherical Surface – …
