The radii of curvature of a double convex lens are 15 cm and 10 cm and its refractive index is 1.3. Its focal length will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( +20\,\mathrm{cm} \)
The focal length of a thin lens in air follows from the lens maker's formula. With the Cartesian sign convention (light from left to right), a double convex lens has \( R_1 \) positive and \( R_2 \) negative.
\( \displaystyle \frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
\( R_1 = +15\,\mathrm{cm} \)
\( R_2 = -10\,\mathrm{cm} \)
\( n = 1.3 \)
\( \displaystyle \frac{1}{f} = (1.3 - 1)\left(\frac{1}{15} + \frac{1}{10}\right) = 0.3 \times \frac{2 + 3}{30} = 0.3 \times \frac{1}{6} = \frac{1}{20}\,\mathrm{cm^{-1}} \)
\( f = +20\,\mathrm{cm} \)
The positive sign shows a converging lens, as expected for a convex lens of \( n \gt 1 \) in air.