Page:Optics.djvu/92

 Now, suppose a second lens be placed close to the first, (Fig. 94.) having for its principal focal length $$F',$$ or $$\frac{m'-1}{\rho'}.$$

In order to find $$\phi',$$ the distance of the focus after the second refraction, we must consider $$\phi$$ and $$\phi'$$ as representing $$\Delta$$ and $$\Delta''$$ in the formula, so that

$$\frac{1}{\phi'} = \frac{1}{F'} + \frac{1}{\phi},$$ or $$\frac{m' - 1}{\rho'} + \frac{1}{\phi};$$

$$\therefore \frac{1}{\phi'} = \frac{1}{F} + \frac{1}{F'}+\frac{1}{\Delta},$$ or $$\frac{m - 1}{\rho} + \frac{m' - 1}{\rho'} +\frac{1}{\phi} +\frac{1}{\Delta}.$$

And in like manner, if there be any number $$n$$ of lenses acting together, we shall have

$$\frac{1}{\phi^{(n)}} = \frac{1}{F}+ \frac{1}{F'} +.... +\frac{1}{F^{(n)}}+ \frac{1}{\Delta},$$

or $$=\frac{m-1}{\rho} + \frac{m'-1}{\rho'}+.... +\frac{m^{(n)}-1}{\rho^{(n)}}+ \frac{1}{\Delta};$$

so that their joint effect is the same as that of a single lens, having, for its principal focal length unity divided by

$$\frac{1}{F}+ \frac{1}{F'}+ \frac{1}{F''} +.... +\frac{1}{F^{(n)}},$$

or $$=\frac{m-1}{\rho} + \frac{m'-1}{\rho'}+\frac{m-1}{\rho}+.... +\frac{m^{(n)}-1}{\rho^{(n)}}.$$

95. Mr. Herschel calls the reciprocal quantity $$\frac{1}{F}$$ the power of a lens, and enounces the last result thus:

"The power of any system of lenses is the sum of the powers of the component lenses."

Of course, regard must be had to the signs: the power of a concave-lens must be considered as positive, that of a convex one, negative.

96. The same method by which we found the focal length of a lens may be easily applied to any number of surfaces, having a common axis.