Light from a point source in air falls on a spherical glass surface. If \( \mu = 1.5 \), and radius of curvature = 20 cm. The distance of light source from the glass surface is 100 cm, at what position will the image be formed?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 100 cm
For refraction at a single spherical surface, with light going from a medium of index \( n_1 \) to one of index \( n_2 \):
\( \displaystyle \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \)
Take the surface to be convex towards the source. Distances are measured from the surface, positive along the direction of the incident light, so \( u = -100\,\mathrm{cm} \), \( R = +20\,\mathrm{cm} \), \( n_1 = 1 \) (air) and \( n_2 = 1.5 \) (glass).
\( \displaystyle \frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20} \)
\( \displaystyle \frac{1.5}{v} = \frac{1}{40} - \frac{1}{100} = \frac{5 - 2}{200} = \frac{3}{200} \)
\( v = +100\,\mathrm{cm} \)
The positive sign shows that a real image is formed inside the glass, 100 cm from the surface.
Note: A similar question was also asked in CUET (UG) 2025 on 29-05-2025 (Slot 2).
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Refraction at a Spherical Surface – …
