If sunlight is focused on a paper using convex lens, it starts burning the paper in shortest time when the lens is kept at 30 cm above the paper. If the radius of curvature of the lens is 60 cm then the refractive index of the lens material is $$\frac{\alpha}{10}$$. The value of $$\alpha$$ is _________.

The paper is burnt fastest when the Sun (an object at infinity) is imaged as the smallest, brightest spot on the paper. For a thin lens, an object at infinity is brought to focus at the focal plane. Hence the distance between the lens and the paper equals the focal length of the lens.

Given distance between lens and paper: $$f = 30 \text{ cm}$$.

The lens is symmetric and convex with radius of curvature $$R = 60 \text{ cm}$$ on both faces. Using the lens-maker’s formula for a thin lens in air:

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

For a double-convex lens:
  • First surface: centre of curvature on the right ⇒ $$R_1 = +60 \text{ cm}$$.
  • Second surface: centre of curvature on the left ⇒ $$R_2 = -60 \text{ cm}$$.

Substituting these values:

$$\frac{1}{30} = (n-1)\left(\frac{1}{60} - \left(-\frac{1}{60}\right)\right)$$

$$\frac{1}{30} = (n-1)\left(\frac{2}{60}\right) = (n-1)\left(\frac{1}{30}\right)$$

Multiplying both sides by $$30$$:

$$1 = n - 1 \quad \Longrightarrow \quad n = 2$$

The refractive index is written as $$n = \dfrac{\alpha}{10}$$, so

$$\alpha = 10n = 10 \times 2 = 20$$

Hence, the required value of $$\alpha$$ is 20.