Focal distance of lens submerged in water

AI Thread Summary
The discussion revolves around calculating the focal length of a biconvex lens submerged in water, initially having a focal length of 15 cm in air. The formula used is 1/f = (n-1)(1/R1 - 1/R2), where the radius of curvature is 15 cm and the refractive index of water is 1.33. An attempt to solve the equation leads to confusion regarding the values used for n, resulting in an incorrect focal length calculation. Participants emphasize the need to refer back to textbooks or notes for clarification on the correct approach. The importance of accurately applying the lens formula in different mediums is highlighted.
Gauss M.D.
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Homework Statement



A biconvex lens with radius of curvature 15 cm has a focal length of 15 cm in air. What is its focal length if it is submerged in water (n=1.33)?

Homework Equations



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

The Attempt at a Solution



1/f = 1/0.15 = (n-1)(2/0.15)

Solving for n gives n = 1.5

So the new equation will be:

1/f = (1.5 - 1.33)(2/0.15) = 2.27

f = 0.44

What am I doing wrong?
 
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Gauss M.D. said:

Homework Statement



A biconvex lens with radius of curvature 15 cm has a focal length of 15 cm in air. What is its focal length if it is submerged in water (n=1.33)?

Homework Equations



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

The Attempt at a Solution



1/f = 1/0.15 = (n-1)(2/0.15)

Solving for n gives n = 1.5

So the new equation will be:

1/f = (1.5 - 1.33)(2/0.15) = 2.27

No. Refer your textbook or notes again.
 
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