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JEE Main 2023 (11 Apr, Shift 1) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from JEE Main 2023 (11 Apr, Shift 1)

Q58JEE Main 2023 · 11 APR · SHIFT 1Medium

Spherical Surfaces & Thin Lenses

The radius of curvature of each surface of a convex lens having refractive index 1.8 is 20 cm20\text{ cm}. The lens is now immersed in a liquid of refractive index 1.5. The ratio of power of lens in air to its power in the liquid will be x:1x : 1. The value of xx is _________.

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

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Answer: 4

  1. Lens maker: P∝(μLμm−1)P \propto \left(\frac{\mu_L}{\mu_m} - 1\right) for the same radii.
  2. In air: 1.81−1=0.8\frac{1.8}{1} - 1 = 0.8
  3. In liquid: 1.81.5−1=0.2\frac{1.8}{1.5} - 1 = 0.2
  4. PairPliq=0.80.2=4\frac{P_{air}}{P_{liq}} = \frac{0.8}{0.2} = 4, so x=4x = 4.
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