Find Refractive Index of Convex Lens Burning Paper

To find the value of $\alpha$, we analyze the given information step-by-step:

1. Focal Length of the Lens ($f$): Sunlight consists of nearly parallel light rays coming from infinity ($u = \infty$). A convex lens focuses these parallel rays at its focal plane. To burn the paper in the shortest time, the light intensity must be maximized at the position of the paper, meaning the paper is placed precisely at the focal point of the lens. Thus, the focal length of the convex lens is: $f = 30\text{ cm}$

2. Lens Maker's Formula: The Lens Maker's formula relates the focal length ($f$), the refractive index ($\mu$), and the radii of curvature ($R_1$ and $R_2$) of a lens: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

   For a symmetric equiconvex lens with a radius of curvature $R = 60\text{ cm}$: $R_1 = +60\text{ cm} \quad \text{and} \quad R_2 = -60\text{ cm}$

   Substituting these values into the Lens Maker's formula gives: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \left(-\frac{1}{R}\right) \right) = (\mu - 1) \frac{2}{R}$

3. Calculating the Refractive Index ($\mu$): Substitute $f = 30\text{ cm}$ and $R = 60\text{ cm}$ into the formula: $\frac{1}{30} = (\mu - 1) \frac{2}{60}$ $\frac{1}{30} = \frac{\mu - 1}{30}$ $\mu - 1 = 1 \implies \mu = 2$

4. Finding $\alpha$: Given that the refractive index is $\mu = \frac{\alpha}{10}$: $\frac{\alpha}{10} = 2 \implies \alpha = 20$