Ray Optics

JEE Main level, numerical answer, medium

126 of 295

Question 126 of 295

If sunlight is focused on a paper using convex lens, it starts burning the paper in shortest time when the lens is kept at 3030 cm above the paper. If the radius of curvature of the lens is 6060 cm then the refractive index of the lens material is α10\dfrac{\alpha}{10}. The value of α\alpha is _______.

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Answer: 20

Sunlight comes from infinity and is focused at the principal focus of the convex lens. The paper burns in the shortest time when the intensity of light is maximum, which occurs at the focal point.

Therefore, the focal length of the convex lens is f=30f = 30 cm.

Assuming the lens is equiconvex, the radii of curvature are R1=60R_1 = 60 cm and R2=−60R_2 = -60 cm.

Using the Lens Maker's formula:

1f=(μ−1)(1R1−1R2)\dfrac{1}{f} = (\mu - 1) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)

Substituting the given values:

130=(μ−1)(160−(−160))\dfrac{1}{30} = (\mu - 1) \left( \dfrac{1}{60} - \left( -\dfrac{1}{60} \right) \right)

130=(μ−1)(260)\dfrac{1}{30} = (\mu - 1) \left( \dfrac{2}{60} \right)

130=(μ−1)(130)\dfrac{1}{30} = (\mu - 1) \left( \dfrac{1}{30} \right)

μ−1=1⇒μ=2\mu - 1 = 1 \Rightarrow \mu = 2

Given that the refractive index is α10\dfrac{\alpha}{10}:

α10=2⇒α=20\dfrac{\alpha}{10} = 2 \Rightarrow \alpha = 20

Answer: 2020

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