Page:Optics.djvu/89

 but as the equation gives $$\Delta$$ negative in the latter case, we may take $$\Delta-\Delta$$ as its general value.

Now

This quantity evidently admits of a minimum value. To find this, we will equate to 0 the differential of its logarithm, which gives

The negative sign shows that the incident rays are converging to a point beyond the lens.

91. To return to the original question: if it be not thought proper to neglect $$t,$$ the thickness of the lens, we may make the calculation rather simpler by measuring $$\Delta''$$ from the second surface.

Then, $$\frac{1}{\Delta''}=-\frac{m-1}{r'}+\frac{m}{\Delta' + t}$$

$$ =-\frac{m-1}{r'}+m \left \{\frac{m\Delta r}{mr+(m-1)\Delta} +t\right \}^{-1}   $$

$$ =-\frac{m-1}{r'}+\frac{mr+(m-1)\Delta}{\Delta r} \left \{1+\frac{mr+(m-1)\Delta }{m\Delta r} t\right \}^{-1}.$$

The binomial in the second term may be expanded, and as many terms taken as thought proper.

If we consider only parallel incident rays, the equation becomes of course much simpler;

$$ \frac{1}{F}=-\frac{m-1}{r'}+m \left \{\frac{mr} {m-1}+t\right \}^{-1} $$

$$ =-\frac{m-1}{r'}+ \frac{m-1}{r} \cdot \left \{1+\frac{m-1}{m r} t\right \}^{-1} $$

$$ =-(m-1) \cdot \left \{\frac{1}{r'}-\frac{1}{r}\Bigl(1+\frac{m-1 }{m}\frac{t}{r}\Bigr) ^{-1}\right \}. $$