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:
Answer & explanation
Correct answer: option 4
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: $n = 1.75$