87. ⁠Now in the first place it will be immediately seen that this expression gives the principal focal distance, which we will call ${\displaystyle F}$, by leaving out the last term, which is equivalent to making ${\displaystyle {\frac {1}{\Delta }}=0}$, or ${\displaystyle \Delta }$ infinite: we have thus

${\displaystyle {\frac {1}{F}}=(m-1)\left\{{\frac {1}{r}}-{\frac {1}{r'}}\right\}}$^[1]

and then,

${\displaystyle {\frac {1}{\Delta ^{n}}}={\frac {1}{F}}+{\frac {1}{\Delta }}.}$

It appears from the former of these that ${\displaystyle F}$ is positive or negative according as ${\displaystyle {\frac {1}{r}}-{\frac {1}{r'}}}$ is so: let us examine what sign this is affected with in different cases.

In the concavo-convex lens placed as in Fig. 86, ${\displaystyle r<r'}$ and ${\displaystyle F}$ is positive.

When this lens is turned the contrary way, ${\displaystyle r>r'}$, but they are both negative, we have then

${\displaystyle {\frac {1}{F}}=(m-1)\left\{{\frac {1}{r'}}-{\frac {1}{r}}\right\},}$

and ${\displaystyle F}$ is positive as before.

In the meniscus, either ${\displaystyle r>r',}$ both being positive, and then

${\displaystyle {\frac {1}{F}}=-(m-1)\left\{{\frac {1}{r'}}-{\frac {1}{r}}\right\},}$

or ${\displaystyle r<r'}$, and both are negative: so that

${\displaystyle {\frac {1}{F}}=-(m-1)\left\{{\frac {1}{r}}-{\frac {1}{r'}}\right\}.}$