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JEE Main 2025 (2 Apr, Shift 1) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from JEE Main 2025 (2 Apr, Shift 1)

Q24JEE Main 2025 · 2 APR · SHIFT 1Medium

Spherical Surfaces & Thin Lenses

Figure

A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure. Distance of the image of an object ' O ', is:

( C is the center of curvature of the spherical surface and R is the radius of curvature)

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Answer: (c)

  1. μ2v−μ1u=μ2−μ1R\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} with u=−0.2u = -0.2 m, R=+0.4R = +0.4 m.
  2. 1.5v+10.2=0.50.4⇒1.5v=1.25−5=−3.75\frac{1.5}{v} + \frac{1}{0.2} = \frac{0.5}{0.4} \Rightarrow \frac{1.5}{v} = 1.25 - 5 = -3.75
  3. v=−0.4 mv = -0.4\text{ m}: a virtual image on the object side.
  4. So the answer is (c) 0.4 m left to the spherical surface.
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