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JEE Main 2025 (23 Jan, Shift 1) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from JEE Main 2025 (23 Jan, Shift 1)

Q26JEE Main 2025 · 23 JAN · SHIFT 1Medium

Spherical Surfaces & Thin Lenses

A spherical surface of radius of curvature RR, separates air from glass (refractive index =1.5= 1.5 ). The centre of curvature is in the glass medium. A point object ' OO ' placed in air on the optic axis of the surface, so that its real image is formed at 'I' inside glass. The line OI intersects the spherical surface at PP and PO=PIPO = PI. The distance POPO equals to

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Answer: (a)

  1. μ2v−μ1u=μ2−μ1R\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} with μ1=1\mu_1 = 1, μ2=1.5\mu_2 = 1.5, R=+RR = +R (centre in glass).
  2. Let PO=PI=xPO = PI = x: u=−xu = -x, v=+xv = +x.
  3. 1.5x+1x=0.5R⇒2.5x=0.5R\frac{1.5}{x} + \frac{1}{x} = \frac{0.5}{R} \Rightarrow \frac{2.5}{x} = \frac{0.5}{R}
  4. So the answer is (a) x=5Rx = 5R.
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