Is the equivalent lens of two such that f_1+f_2<h divergent?

[crick]

The focal of the lens equivalent of two thin lens at distance h is
$$1/f=1/f_1+1/f_2+h/(f_1 f_2)$$

Therefore, supposing that $f_1>0$ and $f_2>0$ (both lenses are convergent), if $f_1+f_2 <h$ then the equivalent lens should be divergent.

Nevertheless consider the example in picture

The two lenses have focals such that $f_1+f_2 <h$ but the image is real, i.e. the equivalent lens cannot be divergent. I understood the ray diagram, but how can this hold true?

 

[Daniel Gallimore]

Therefore, supposing that f1>0f1>0f_1>0 and f2>0f2>0f_2>0 (both lenses are convergent), if f1+f2<hf1+f2<hf_1+f_2

Can you quote the source for this claim?

 

[pixel]

Check out Fig. 5.30 here: http://engineering.tufts.edu/bme/people/georgakoudi/EN31Lab4ThinLensCombinations.pdf

It shows two thin lenses separated by a distance greater than the sum of their focal lengths and a real, erect image.

 

[pixel]

Maybe the condition for the equivalent lens to be diverging is $f_1+f_2 >h$ . In that case, f₁ is to the right of f₂ in your diagram. A real image formed by the first lens will be within a focal length of the second lens and the result will be a virtual image.