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 Incoming links: Image distance for two separated thin lenses

Use the thin lens sign convention in which a real object on the incident side has positive object distance, a real image on the outgoing side has positive image distance, and a converging lens has positive focal length. The first image distance is

v_{1} = \frac{f _{1} d _{1}}{d _{1} - f _{1}} .

u_{2} = L - v_{1}

, which can be negative when its intermediate object is virtual. Applying the thin lens equation again gives

d_{2} = \frac{f _{2} u _{2}}{u _{2} - f _{2}} = \frac{f _{2} [( L - f _{1} ) d _{1} - L f _{1} ]}{( L - f _{1} - f _{2} ) d _{1} + f _{1} f _{2} - L f _{1}} .

The image is

d_{2}

beyond the second lens when

d_{2} > 0

;

d_{2} < 0

denotes a virtual image on its incident side. A vanishing denominator with nonzero numerator denotes an image at infinity. The rational expression also supplies appropriate limits when the first image is at infinity; for example,

d_{1} = f_{1}

gives

d_{2} = f_{2}

for finite nonzero focal lengths.

L \to 0

, the effective focal length is

F = f_{1} f_{2} / (f_{1} + f_{2})

and

d_{2} = \frac{F d _{1}}{d _{1} - F} .

For a real incident object,

d_{1} > 0

, and positive

F

, the real-image condition is exactly

d_{1} > F

, yielding the requested inequality. This includes two converging lenses. The positivity assumption matters: with

f_{1} = f_{2} = - 1 m

and

d_{1} = 1 m

, one has

F = - 0.5 m

and

d_{1} > F

, but

d_{2} = - 1/3 m

is virtual. If

f_{1} + f_{2} = 0

, the touching pair has zero net optical power and produces no real image of this real object in the ideal thin lens model.