The radii of curvature of the faces of a double convex lens are 10 cm and 12 cm, respectively. If the refractive index of the material of the lens is 1.5, the focal length of the lens is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → ~ 10.9 cm
Use the lens maker’s formula with the sign convention: for a double convex lens the first surface has R1 = +10 cm and the second R2 = −12 cm.
\( \displaystyle \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
\( \displaystyle \frac{1}{f} = (1.5 - 1)\left(\frac{1}{10} + \frac{1}{12}\right) \)
\( \displaystyle \frac{1}{f} = 0.5 \times \frac{11}{60} = \frac{11}{120} \)
\( \displaystyle f = \frac{120}{11} \approx 10.9\,\mathrm{cm} \)