An equiconvex lens has a focal length of 20 cm and its radius of curvature is 30 cm. Its refractive index is

7/4

The correct answer is Option (3) → 7/4

$\text{Lens maker formula: } \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

$R_1 = +30,\; R_2 = -30,\; f = 20$

$\frac{1}{20} = (\mu - 1)\left(\frac{1}{30} - \frac{-1}{30}\right)$

$\frac{1}{20} = (\mu - 1)\left(\frac{2}{30}\right)$

$\frac{1}{20} = (\mu - 1)\cdot \frac{1}{15}$

$\mu - 1 = \frac{15}{20} = \frac{3}{4}$

$\mu = 1 + \frac{3}{4} = \frac{7}{4}$

Final Answer: $\frac{7}{4}$