# Focal Points of a Thick lens

1. Feb 8, 2012

### PChar

1. The problem statement, all variables and given/known data

A thick lens has an index of refraction of 1.560, thickness of 3.0cm, and radii of curvature of R1= -4.50cm, R2= -3.60cm.

Calculate the positions of the focal points (relative to the vertices).

2. Relevant equations

$\frac{n}{s}+\frac{n'}{s'}=\frac{n'-n}{R}$

$ffl = \frac{f1(t + f2)}{t-(f1 + f2)}$

3. The attempt at a solution

I assume I need to first focal length for each lens to plug into the front focal length and back focal length formulas, meaning s' = ∞ so for L1:

f1 = $\frac{R}{n'-n}=\frac{-4.50cm}{1.560-1.000}$ = -8.04cm

Do I have the right idea? If so, will I need to swap the indexes of refraction when doing the opposite side of the lens, so that light is coming from the left side?

2. Feb 9, 2012

### Spinnor

These might help?

"For the case of a lens of thickness d in air, and surfaces with radii of curvature R1 and R2, the effective focal length f is given by:"

From half way down the following page,

Worked examples here,