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A biconvex lens with focal length f in air and refractive index of 1.5 is floating on the surface of a deep pond of water (refractive index 1.33). If an object is placed at a height of 2 f vertically above the lens, then the distance between the lens and the image is

A biconvex lens with focal length f in air and refractive index of 1.5 is floating on the surface
of a deep pond of water (refractive index 1.33). If an object is placed at a height of 2 f vertically
above the lens, then the distance between the lens and the image is

Grade:12th pass

1 Answers

Vikas TU
14149 Points
6 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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