Question

If sunlight is focused on a paper using a convex lens, it starts burning the paper in the 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 _______.

Answer 1

Correct Answer: 20

Step 1: Understanding the Concept:
Sunlight consists of parallel rays from infinity. When a convex lens focuses parallel rays, they converge exactly at the principal focus.
Since the burning is fastest when focused perfectly, the distance between the lens and the paper is the focal length \(f\).
We can use the Lens Maker's Formula to relate the focal length, radii of curvature, and the refractive index.
Step 2: Key Formula or Approach:
Lens Maker's Formula: \(\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\).
For an equiconvex lens (implied when only one radius is given), \(R_1 = R\) and \(R_2 = -R\).
Step 3: Detailed Explanation:
From the problem description, the focal length is \(f = 30 \text{ cm}\).
The radius of curvature is given as \(R = 60 \text{ cm}\). So \(R_1 = 60 \text{ cm}\) and \(R_2 = -60 \text{ cm}\).
Apply the Lens Maker's Formula:
\[ \frac{1}{30} = (\mu - 1) \left(\frac{1}{60} - \frac{1}{-60}\right) \] \[ \frac{1}{30} = (\mu - 1) \left(\frac{1}{60} + \frac{1}{60}\right) \] \[ \frac{1}{30} = (\mu - 1) \left(\frac{2}{60}\right) \] \[ \frac{1}{30} = \frac{\mu - 1}{30} \] Multiply both sides by 30:
\[ 1 = \mu - 1 \implies \mu = 2 \] The problem states the refractive index is \(\frac{\alpha}{10}\).
Equating the values:
\[ \frac{\alpha}{10} = 2 \implies \alpha = 20 \] Step 4: Final Answer:
The value of \(\alpha\) is \(20\).