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A biconvex lens made of a material with refractive index 1.5 has a focal length of +24 cm in air. If this lens is completely immersed in a transparent liquid of refractive index 1.6 , its new focal length will be

Options

  1. A+192 cm
  2. B+180 cm
  3. C+20 cm
  4. D-192 cm

Correct answer

D. -192 cm

Step-by-step solution

Let the geometric factor of the lens be K = ( 1 R₁ - 1 R₂ ) . Using the Lens Maker's Formula for the lens in air: 1 f_a = ( _L - 1)K Substitute the given values ( f_a = +24 cm , _L = 1.5 ): 1 24 = (1.5 - 1)K = 0.5K K = 1 12 cm ⁻¹ When the lens is immersed in the liquid, the relative refractive index of the lens with respect to the liquid is _ rel = _L _m = 1.5 1.6 = 15 16 . The new focal length f_l is given by: 1 f_l = ( _ rel - 1)K 1 f_l = ( 15 16 - 1 ) 1 12 1 f_l = (- 1 16 ) 1 12 = - 1 192 f_l = -192 cm The negat

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