An aligned optical system consists of two thin lenses with focal lengths `f_(1)` and `f_(2)`, the distance between the lenses being equal to `d`. The given system has to be replaced by one thin lens which, at any position of an object, would provide the same transverse magnification as the system. what must the focal length of this lens be equal to and in what position must it be placed with respect to the two-lens system?

Explanation

The optical power of the system of two thin lenses placed in air is given as,
`Phi = Phi_(1) + Phi_(2) - d Phi_(1)Phi_(2)`
or, `(1)/(f) = (1)/(f_(1)) + (1)/(f_(2)) - (d)/(f_(1)f_(2))`, where `f` is the equivalent focal length
So, `(1)/(f) = (f_(2) + f_(1) - d)/(f_(1)f_(2))`
or, `f = (f_(1)f_(2))/(f_(1) + f_(2) - d)......... (1)`
This equivalent focal length of the system of two lenses is measured from the primary principle plane.
As clear from the figure, the distance of the primary principle plane from the optical centre of the first is
`O_(1)H = x =+ (n//Phi) (d//n') Phi_(1)`
`= (d Phi_(1))/(Phi)`, as `n = n' = 1`, for air.
`= (df)/(f_(1))`
`= ((d)/(f_(1))) ((f_(1)f_(2))/(f_(1) + f_(2) - d))`
`= (df_(2))/(f_(1) + f_(2) - d)`
Now, if we place the equivalent lens at the primary principle plane of the lens system, it will provide the same transverse magnification as the system. So, the distance of equivalent lens from the vertex of the first lens is,
`x = (df_(2))/(f_(1) + f_(2) - d)`