The focal length of an equiconcave lens is $\dfrac{3}{4}$ times of radius of curvature of its surfaces. Find the refractive index of the material of the lens.

Lens makes formula (equiconcave lens)

$\dfrac{1}{f} = (n - 1) \dfrac{2}{R} , \,\, n = \mu$ of the lens

Given $f = \dfrac{3}{4} R$

$\dfrac{4}{2R} = \dfrac{2}{R} (n - 1)$

$n = \dfrac{5}{3}$