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MHT CET 2025 (26 Apr, Shift 2) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from MHT CET 2025 (26 Apr, Shift 2)

Q29MHT CET 2025 · 26 APR · SHIFT 2Medium

Spherical Surfaces & Thin Lenses

Refractive index of a glass convex lens is 1.5. The radius of curvature of each of the two surfaces of the lens is 40 cm. The ratio of the power of the lens when immersed in a liquid of refractive index 1.25 to that when placed in air is

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Answer: (c)

  1. PlPa=μμl−1μ−1\frac{P_l}{P_a} = \frac{\frac{\mu}{\mu_l} - 1}{\mu - 1}
  2. 1.51.25−1=0.2\frac{1.5}{1.25} - 1 = 0.2, so PlPa=0.20.5=25\frac{P_l}{P_a} = \frac{0.2}{0.5} = \frac{2}{5}
  3. So the answer is (c) 2:52 : 5.
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