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

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1. May 6, 2017

### 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?

2. May 7, 2017

### Daniel Gallimore

Can you quote the source for this claim?

3. May 7, 2017

4. May 8, 2017

### 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.