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JEE Main 2025 (3 Apr, Shift 2) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from JEE Main 2025 (3 Apr, Shift 2)

Q27JEE Main 2025 · 3 APR · SHIFT 2Medium

Spherical Surfaces & Thin Lenses

Light from a point source in air falls on a spherical glass surface (refractive index, μ=1.5\mu = 1.5 and radius of curvature =50 cm= 50\text{ cm} ). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is ___________m.

Numerical answer type. Work it out, then open the answer.

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Answer: 4

  1. μ2v−μ1u=μ2−μ1R\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} with v=+200v = +200 cm, R=+50R = +50 cm.
  2. 1.5200−1u=0.550=0.01\frac{1.5}{200} - \frac{1}{u} = \frac{0.5}{50} = 0.01
  3. 1u=0.0075−0.01=−0.0025⇒u=−400\frac{1}{u} = 0.0075 - 0.01 = -0.0025 \Rightarrow u = -400 cm
  4. So the source is at 4 m.
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