Focal length question

Question

A converging glass lens (index of refraction 1.63) has a focal length of 40 cm in air. Find its focal length when it is immersed in a liquid, which has an index of refraction 1.32.

 

The way I set it up was...

 

The original equation: 1/f=(n-1)(1/r1-1/r2)

 

The (1/r1-1/r2) part of the equation stays constant so to "manipulate" the equation to solve for f2 when the lens is in the water...

 

f(liq)/f(lens) = (nlens-1)/(nliq-1)

x/40 cm = (1-1.63/(1-1.32)

x=78.75 cm 

 

I got the question wrong so I was wondering how to solve this problem correctly? 

Answer 1

You are using the lens maker's equation for a lens of index n₁ immersed in air.  However, if the lens is immersed in another medium of index n₂, the lens maker's equation takes on a different form:

 

1/f₂ = (n₁/n₂ - 1)(1/r₁ - 1/r₂)

 

f₂ = focal length in medium of index n₂

 

In air, 

 

1/f₁ = (n₁ - 1)(1/r₁ - 1/r₂)

 

(1/f₁) / (1/f₂) = [(n₁ - 1)(1/r₁ - 1/r₂)] / [(n₁/n₂ - 1)(1/r₁ - 1/r₂)]

 

f₂/f₁ = (n₁ - 1) / (n₁/n₂ - 1)

 

f₂ = [(n₁ - 1) / (n₁/n₂ - 1)] f₁

 

n₁ = 1.63

n₂ = 1.32

f₁ = 40 cm

 

f₂ = [(1.63 - 1) / (1.63/1.32 - 1)] (40 cm) = 107.3 cm