An equiconvex lens has a focal length $\frac{2}{3}$ times the radius of curvature of either surface. The refractive index of the material of the lens is:

1.75

The correct answer is Option (4) → 1.75

For a thin lens, lens maker's formula is:

$\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

For an equiconvex lens: $R_1 = R$, $R_2 = -R$ (convex surfaces opposite direction)

So, $\frac{1}{f} = (n - 1) \left( \frac{1}{R} - \frac{-1}{R} \right) = (n - 1) \left( \frac{2}{R} \right)$

Given $f = \frac{2}{3} R$, then $\frac{1}{f} = \frac{3}{2R}$

Equating:

$\frac{3}{2R} = (n - 1) \frac{2}{R} \text{ implies } n - 1 = \frac{3}{4} \text{ implies } n = \frac{7}{4} = 1.75$

Final Answer: $n = 1.75$