= Two equal lenses with a virtual intermediate object
For two thin converging lenses, the first lens may form its would-be image beyond the second lens. The rays entering lens 2 then converge, so its Cartesian real-object distance is negative. Apply $1/f=1/u+1/v$ separately at each lens rather than treating that intermediate focus as an object upstream. In the example $u_1=2f$ with separation $d<f$, $u_2=-(2f-d)$ and $v_2=f(2f-d)/(3f-d)$. Multiplying the two signed <transverse magnifications> determines final inversion; a positive outgoing image distance means a <real image>.
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