general treatment of two-lens system (not for examination

We can derive a general expression for the focal length of a two-lens system, made of two thin lenses, using the geometry shown in figure L33.2.

  

Figure L33.2: The geometry of a two-lens system.

Consider the two-lens system in which the incoming parallel light is focused by lens 1 at the point G. This will form the virtual object for lens 2, and so this object is a distance f₁ - d to the right of lens 2. Then the distance s from lens 2 to the focus F₂ will be given by

$${1 \over s}$$ - $${1 \over f_1 - d}$$ = $${1 \over f_2}$$.

If the initial ray enters lens 1 at a height h₁ and exits lens 2 at a height h₂ then the intersection point (giving the second principal point H₂ ) is found from the similar triangles CF₂V₂ and BF₂H₂

$${h_2 \over s}$$ = $${h_1 \over f}$$

and from CGV₂ and AGV₁

$${h_2 \over f_1 - d}$$ = $${h_1 \over f}$$

which gives us two expressions for h₂/h₁

$${h_2 \over h_1}$$ = $${s \over f}$$ = $${f_1 - d \over f_1}$$

giving us a value for s to substitute back

$${f_1 \over f (f_1 - d)}$$ - $${1 \over f_1 - d}$$ = $${1 \over f_2}$$

which may be rearranged to give

$${1 \over f}$$ = $${1 \over f_1}$$ + $${1 \over f_2}$$ - $${d \over f_1 f_2}$$.

Similarly, one may go back to find the positions of the principal points. As

s = $${f \over f_1}$$(f₁ - d )

the position of H₂ relative to the second lens is then

s - f = $${f \over f_1}$$(f₁ - d )- f = - $${f d \over f_1}$$.

Note the sign, showing that if both lenses are converging then the second principal point lies to the left of the second lens.

Similarly, the first principal point is at fd$\over$f₂ relative to the first lens.