Q 12-09-229JEE MainJEE Main 2025 (3 Apr, Shift 1)Easy
The radii of curvature for a thin convex lens are $10\ \text{cm}$ and $15\ \text{cm}$ respectively. The focal length of the lens is $12\ \text{cm}$. The refractive index of the lens material is:
Answer: (C) $1.5$
For a biconvex lens, $R_1 = +10\ \text{cm}$ and $R_2 = -15\ \text{cm}$:
$$\frac1f = (\mu - 1)\left(\frac1{10} + \frac1{15}\right) = (\mu - 1)\frac16$$
$$\frac1{12} = \frac{\mu - 1}{6} \Rightarrow \mu - 1 = 0.5 \Rightarrow \mu = 1.5$$
Solution by Sreeraj P, M.Sc Physics