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 _______.
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 _______.
Updated On: Apr 10, 2026
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Correct Answer: 20
Solution and Explanation
Step 1: Understanding the given information.
The lens is a convex lens, and sunlight is focused on a paper, causing it to burn when the lens is at a specific distance (30 cm) from the paper. The radius of curvature \( R \) of the lens is given as 60 cm. The distance at which the paper starts burning corresponds to the focal length \( f \) of the lens, and the focal length of a convex lens is related to the radius of curvature \( R \) by the formula: \[ f = \frac{R}{2}. \] Substituting the value of \( R \): \[ f = \frac{60}{2} = 30 \, \text{cm}. \] Thus, the focal length of the lens is 30 cm, which matches the distance from the lens to the paper.
Step 2: Using the lens formula.
The lens formula relates the focal length \( f \), the object distance \( u \), and the image distance \( v \) as: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v}. \] In this case, the object is at infinity (as sunlight is parallel), and the image is formed at the focal point of the lens. Therefore, the image distance \( v \) is equal to the focal length \( f \). So, we have: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{f}. \] Since the object distance is at infinity, the equation simplifies to: \[ \frac{1}{f} = \frac{1}{f} \quad \text{(valid)}. \]
Step 3: Refractive index calculation.
The refractive index \( n \) of the lens material is related to the focal length \( f \) and the radius of curvature \( R \) by the formula: \[ \frac{1}{f} = (n - 1) \left( \frac{2}{R} \right). \] Substituting the known values \( f = 30 \, \text{cm} \) and \( R = 60 \, \text{cm} \): \[ \frac{1}{30} = (n - 1) \left( \frac{2}{60} \right). \] Simplifying the equation: \[ \frac{1}{30} = (n - 1) \cdot \frac{1}{30}. \] Thus, we get: \[ n - 1 = 1 \quad \Rightarrow \quad n = 2. \] The refractive index of the lens is 2. Now, since the refractive index is given as \( \frac{\alpha}{10} \), we equate: \[ \frac{\alpha}{10} = 2. \] Thus, the value of \( \alpha \) is: \[ \boxed{20}. \]
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