A light source is kept in air \( (n \approx 1) \) at a distance of 200 cm from a convex spherical glass surface \( \left(n = \displaystyle\frac{3}{2}\right) \) such that light falls on it. If the radius of curvature of the surface is 40 cm, the position of the image is at
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( +200\,\mathrm{cm} \)
For refraction at a single spherical surface from a medium \( n_1 \) into a medium \( n_2 \):
\( \displaystyle \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \)
Sign convention: distances are measured from the pole, positive in the direction of the incident light. The source is in front of the surface, so \( u \) is negative; the surface is convex towards the source, so its centre of curvature lies in the glass and \( R \) is positive.
\( n_1 = 1 \)
\( n_2 = 1.5 \)
\( u = -200\,\mathrm{cm} \)
\( R = +40\,\mathrm{cm} \)
\( \displaystyle \frac{1.5}{v} - \frac{1}{-200} = \frac{1.5 - 1}{40} \)
\( \displaystyle \frac{1.5}{v} = \frac{1}{80} - \frac{1}{200} = \frac{5 - 2}{400} = \frac{3}{400} \)
\( \displaystyle v = \frac{1.5 \times 400}{3} = +200\,\mathrm{cm} \)
The positive sign means a real image 200 cm inside the glass.
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 – …
