Derive the equation for effective focal length for lenses in contact.

Explanation

Consider a two lenses of focal length `f_(1) and f_(2)` arranged coaxially but separated by a distance d can be considered. For a parallel ray that falls on the arrangement, the two lenses produces deviations `delta_(1) and delta_(2)` respectively and The net deviation `delta` is.
`delta = delta_(1) + delta_(2)`
From Angle of deviation in lens equation, `delta = (h)/(f)`
`delta_(1) = (h_(1))/(f_(1)):delta_(2) = (h_(2))/(f_(2)) and delta = (h_(1))/(f)`
The equation (1) becomes,
`(h_(1))/(f) = (h_(1))/(f_(1)) + (h_(2))/(f_(2))`
From the geometry.
`h_(2)-h_(1)=P_(2)G-P_(2)C=CG`
`h_(2) - h_(1) = BG tan delta_(1) = BG delta_(1)`
`h_(2) - h_(1) = d(h_(1))/(f_(1))`
`h_(2) = h_(1) + d(h_(1))/(f_(1))`
Substituting the above equaiton in Equation (3)
`(h_(1))/(f)=(h_(1))/(f_(1))+(h_(1))/(f_2)+(h_(1)d)/(f_(1)f_(2))`
On further simplification,
`(1)/(f) = (1)/(f_(1)) + (1)/(f_(2)) + (d)/(f_(1)f_(2))`
The above equation could be used to find the equivalent focal lenght.