The power of a lens (biconvex) is 1.25 m⁻¹ in particular medium. Refractive index of the lens is 1.5 and radii of curvature are 20 cm and 40 cm respectively. The refractive index of surrounding medium: (a) 1.0 (b) 9/7 (c) 3/2 (d) 4/3

Answer: (b). P = ((μ)/(μm) - 1)(1/R₁ + 1/R₂), radii in m: 1/0.2 + 1/0.4 = 7.5 m⁻¹. 1.25 = ((1.5)/(μm) - 1) × 7.5 ⇒ (1.5)/(μm) = 7/6 μm = (1.5 × 6)/(7) = 9/7 So the answer is (b) 9/7.

The power of a lens (biconvex) is 1.25 m⁻¹ in particular medium… | JEE Main 2022 (28 Jul, Shift 2)

Q64JEE Main 2022 · 28 JUL · SHIFT 2Medium

Spherical Surfaces & Thin Lenses

The power of a lens (biconvex) is $1.25\text{ m}^{-1}$ in particular medium. Refractive index of the lens is 1.5 and radii of curvature are $20\text{ cm}$ and $40\text{ cm}$ respectively. The refractive index of surrounding medium:

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Answer: (b)

$P = \left(\frac{\mu}{\mu_m} - 1\right)\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$ , radii in m: $\frac{1}{0.2} + \frac{1}{0.4} = 7.5\ \text{m}^{-1}$ .
$1.25 = \left(\frac{1.5}{\mu_m} - 1\right) \times 7.5 \Rightarrow \frac{1.5}{\mu_m} = \frac{7}{6}$
$\mu_m = \frac{1.5 \times 6}{7} = \frac{9}{7}$
So the answer is (b) $\frac{9}{7}$ .

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