Simple Optics Question : Back focal length

[getcarter]

Homework Statement

[PLAIN]http://img256.imageshack.us/img256/6330/35727193.jpg

Homework Equations

[URL]http://upload.wikimedia.org/math/2/0/a/20aa1e3d192ecb3164ac4f2095c86cd3.png[/URL]

The Attempt at a Solution

i have used triangle from up side of h1 to the b. and by using sinß and sin(tetha) i have found h1 and h2 interms of sin. but i don't know how can i reach the question's answer in terms of R2 because i drawn only one triangle.

and also i have used the equation x^2 + y^2 = R^2
in order to calculate hypotenus of triangle.

 

[revacious]

Assuming I've read your diagram correctly... this is 2 thin lenses of power F1 and F2 separated by a distance d.

Let the back vertex power be denoted by Fv'.

Fv' = (F1 + F2 - dF1F2)/(1-dF1) by definition.

Then the back vertex length fv' is given by 1/Fv'

So tan(theta) = h1/fv' = h1Fv'

Now F1 has primary focal length f1'

Using similar triangles,
f1'/(f1-d) = h2/h1.

 

[getcarter]

thanx but i have to use only quantities given on the question.
and start from snell's law