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A biconvex lens has a focal length f in air. When it is immersed in a transparent liquid of refractive index 1.4 , it behaves as a converging lens of focal length 3f . The refractive index of the material of the lens is:

Options

  1. A1.60
  2. B0.875
  3. C1.75
  4. D4.20

Correct answer

C. 1.75

Step-by-step solution

Let the refractive index of the lens material be _L . The focal length of the lens in air is given by: 1 f = ( _L - 1) ( 1 R₁ - 1 R₂ ) When immersed in a liquid of refractive index _m = 1.4 , the focal length becomes 3f . Using the Lens Maker's formula for the medium: 1 3f = ( _L 1.4 - 1 ) ( 1 R₁ - 1 R₂ ) Dividing the first equation by the second equation, we get: 3f f = _L - 1 _L 1.4 - 1 3 = _L - 1 _L 1.4 - 1 Cross-multiplying yields: 3 ( _L 1.4 - 1 ) = _L - 1 3 _L 1.4 - 3 = _L - 1 _L ( 3 1.4 - 1 ) = 2 _L ( 3 - 1.

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