An equiconvex lens of focal length 10 cm is made up of material with refractive index 1.5. The radius of curvature of the each surface is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 10 cm
The lens maker's formula relates the focal length to the radii of curvature of the two surfaces.
\( \displaystyle \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
For an equiconvex lens \( R_1 = +R \) and \( R_2 = -R \):
\( \displaystyle \frac{1}{f} = (\mu - 1)\,\frac{2}{R} = (1.5 - 1)\,\frac{2}{R} = \frac{1}{R} \)
\( \displaystyle R = f = 10\,\mathrm{cm} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
