A biconvex lens with focal length f in air and refractive index of 1.5 is floating on the surfaceof a deep pond of water (refractive index 1.33). If an object is placed at a height of 2 f verticallyabove the lens, then the distance between the lens and the image is
Sheela Suresh , 7 Years ago
Grade 12th pass
1 Answers
Vikas TU
Last Activity: 7 Years ago
R. I regarding Air(µa) = 1.5
R.I. regarding water(µw) = 1.33
1/Fa = (µa – 1)(1/R1 – 1/R2)
Central length in water
1/Fw =( µw – 1) (1/R1 – 1/R2)
Accordingly,
Fa/Fw = ( µw – 1)/(µa – 1)
= (1.33-1)/(1.5-1)
= 0.66
Be that as it may, here they are not given the question separate, at that point picture remove cannot be figured .
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