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A lens made of glass with a refractive index of 1.5 has a certain focal length in air. When this lens is completely immersed in an unknown transparent liquid, its focal length becomes 2.5 times its focal length in air. The refractive index of the unknown liquid is

Options

  1. A1.20
  2. B1.80
  3. C1.25
  4. D0.67

Correct answer

C. 1.25

Step-by-step solution

Let the focal length of the lens in air be f_a and in the liquid be f_l . Using the Lens Maker's Formula for the lens in air: 1 f_a = ( _g - 1) ( 1 R₁ - 1 R₂ ) For the lens in the liquid, the relative refractive index is _g _l . The formula becomes: 1 f_l = ( _g _l - 1 ) ( 1 R₁ - 1 R₂ ) Dividing the first equation by the second gives: f_l f_a = _g - 1 _g _l - 1 We are given that f_l = 2.5 f_a and _g = 1.5 . Substituting these values: 2.5 = 1.5 - 1 1.5 _l - 1 2.5 = 0.5 1.5 _l - 1 1.5 _l - 1 = 0.5 2.5 = 0.2 1.5 _l =

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