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JEE Main 2025 (28 Jan, Shift 2) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from JEE Main 2025 (28 Jan, Shift 2)

Q25JEE Main 2025 · 28 JAN · SHIFT 2Medium

Spherical Surfaces & Thin Lenses

In a long glass tube, mixture of two liquids A and B with refractive indices 1.3 and 1.4 respectively, forms a convex refractive meniscus towards A. If an object placed at 13 cm from the vertex of the meniscus in A forms an image with a magnification of −2-2, then the radius of curvature of meniscus is:

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Answer: (a)

  1. Refraction at a spherical surface: μ2v−μ1u=μ2−μ1R\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}, with μ1=1.3\mu_1 = 1.3 (A, object side), μ2=1.4\mu_2 = 1.4 (B), u=−13u = -13 cm.
  2. Magnification m=μ1vμ2u=−2⇒v=−2×1.4×(−13)1.3=28m = \frac{\mu_1 v}{\mu_2 u} = -2 \Rightarrow v = \frac{-2 \times 1.4 \times (-13)}{1.3} = 28 cm.
  3. 1.428+1.313=0.1R⇒0.05+0.1=0.1R\frac{1.4}{28} + \frac{1.3}{13} = \frac{0.1}{R} \Rightarrow 0.05 + 0.1 = \frac{0.1}{R}
  4. ∣R∣=0.10.15=23|R| = \frac{0.1}{0.15} = \frac{2}{3} cm.
  5. So the answer is (a) −23-\frac{2}{3} cm (magnitude 23\frac{2}{3} cm).
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