A convex lens of refractive index 1.55, with both the surfaces of the same radius of curvature has a focal length of 20 cm. The radius of curvature of the surface will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 22\,\mathrm{cm} \)
The focal length of a thin lens is given by the lens maker's formula. For an equiconvex lens, with the Cartesian sign convention (light travelling left to right), \( R_1 = +R \) and \( R_2 = -R \).
\( \displaystyle \frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (n - 1)\left(\frac{1}{R} + \frac{1}{R}\right) = \frac{2(n - 1)}{R} \)
\( \displaystyle R = 2(n - 1)\,f \)
\( n = 1.55 \)
\( f = 20\,\mathrm{cm} \)
\( \displaystyle R = 2 \times 0.55 \times 20\,\mathrm{cm} = 22\,\mathrm{cm} \)
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):
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