Calculating Focal Points of a Thick Lens

[PChar]

Homework Statement

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).

Homework Equations

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

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

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?

 

[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,

http://www.answers.com/topic/focal-length

Worked examples here,

http://www.drdrbill.com/downloads/optics/geometric-optics/Thick_Lenses.pdf

Found via,

https://www.google.com/search?hl=en...&um=1&ie=UTF-8&tbm=isch&source=og&sa=N&tab=wi

https://www.google.com/webhp?hl=en#...w.,cf.osb&fp=fc8f4a6474bcae3f&biw=917&bih=404

Good luck!