An equiconvex lens has a focal length of 20 cm and its radius of curvature is 30 cm. Its refractive index is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 7/4
Use the lens maker's formula. With the Cartesian sign convention (light from the left), the first surface of an equiconvex lens has \( R_1 = +R \) and the second \( R_2 = -R \).
\( \displaystyle \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (\mu - 1)\,\frac{2}{R} \)
\( f = 20\,\mathrm{cm} \)
\( R = 30\,\mathrm{cm} \)
\( \displaystyle \frac{1}{20} = (\mu - 1)\,\frac{2}{30} \;\Rightarrow\; \mu - 1 = \frac{30}{40} = \frac{3}{4} \)
\( \mu = 1 + 3/4 = 7/4 \)