Lens-Makers Equation: Finding Focal Length in Different Mediums

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AI Thread Summary
The lens-makers' equation can be adapted for lenses in different mediums by replacing the refractive index (n) with the ratio of the indices of refraction of the lens material (n2) and the surrounding medium (n1). For a lens with a focal length of 92.6 cm in air and a refractive index of 1.55, the focal length in water can be calculated using this modified equation. A user initially struggled with the calculations but later resolved the issue with guidance from others in the discussion. The correct approach involves applying the lens-makers' equation accurately to find the new focal length. Understanding the relationship between the indices of refraction is crucial for solving these problems effectively.
PhyHyped

Homework Statement


The lens-makers' equation applies to a lens immersed in a liquid if n in the equation is replaced by n2/n1. Here n2 refers to the index of refraction of the lens material and n1 is that of the medium surrounding the lens.
(a) A certain lens has focal length 92.6 cm in the air and index of refraction 1.55. Find its focal length in water.
(b) A certain mirror has focal length 92.6 cm in the air. Find its focal length in water.

Homework Equations


1/f = (n-1)(1/R1 - 1/R2)

The Attempt at a Solution


I'm not really sure how to set this up. Think it looks something like this f2/f1 =(n1-1)/(n2-1) index of refraction of water 1.33 plugged in the given variables and got 55.56 which was wrong.

Comments
Not sure what direction to go with this equation.

Any help would be greatly appreciated.
 
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PhyHyped said:
f2/f1 =(n1-1)/(n2-1)
That does not result from replacing n with n1/n2 in 1/f = (n-1)(1/R1 - 1/R2), as instructed.
 
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thanks! I figured it out earlier! Thanks for the reply though
 

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