Page:Optics.djvu/86

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

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

and then,

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

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

In the concavo-convex lens placed as in Fig. 86, $$rr'$$, but they are both negative, we have then $$\frac{1}{F}=(m-1) \left \{ \frac{1}{r'}-\frac{1}{r} \right \},$$

and $$F$$ is positive as before.

In the meniscus, either $$r>r',$$ both being positive, and then

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

or $$r<r'$$, and both are negative: so that

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