# Page:BatemanConformal.djvu/19

88
[Nov. 12,
Mr. H. Bateman

The linear magnification in the new system is

 ${\displaystyle m_{1}={\frac {\frac {A_{1}Q'_{1}}{\mu '}}{\frac {A_{1}Q_{1}}{\mu }}}={\frac {\mu x'(z-2x)}{\mu 'x(z-2x')}}=m{\frac {z-2x}{z-2x'}}.}$

Let c be the distance of the second interface from the first; then we associate the length -c with the second interface B, and the formulæ of the transformation become

 ${\displaystyle B_{1}Q'_{1}=-\xi '_{1}={\frac {k^{2}\xi '}{(\zeta '-\xi ')^{2}-(\xi '-c)^{2}}}={\frac {k^{2}\xi '}{(\zeta '-c)\left(\zeta '+c-2\xi '\right)}},}$ ${\displaystyle \zeta '_{1}={\frac {k^{2}\zeta }{\zeta ^{2}-c{}^{2}}},\ c_{1}=-{\frac {k^{2}c}{\zeta ^{2}-c{}^{2}}},}$ ${\displaystyle \zeta '_{1}-\xi '_{1}={\frac {k^{2}(\zeta '-\xi ')}{(\zeta '-\xi ')^{2}-(\xi '-c)^{2}}}={\frac {k^{2}(\zeta '-\xi ')}{(\zeta '-c)\left(\zeta '+c-2\xi '\right)}},}$ ${\displaystyle \xi '_{1}-c_{1}={\frac {k^{2}(c-\xi ')}{(\zeta '-c)\left(\zeta '+c-2\xi '\right)}},}$ ${\displaystyle \zeta '_{1}+c_{1}-2\xi '_{1}={\frac {k^{2}}{\zeta '+c-2\xi '}}.}$

The last of which may be written

 ${\displaystyle \zeta '_{1}-c_{1}-2\left(\xi '_{1}-c_{1}\right)={\frac {k^{2}}{\zeta '-c-2\left(\xi '-c\right)}},}$

i.e.,

 ${\displaystyle z_{1}-2x'_{1}={\frac {k^{2}}{z-2x'}},}$

where z and x' are measured, as before, from the first surface. This is exactly the former relation between two corresponding points; consequently the whole course of a ray in one instrument corresponds in the transformation to that of the corresponding ray in the corresponding instrument.[1]

1. We can verify the equation
 ${\displaystyle {\frac {\mu ''}{B_{1}Q''_{1}}}-{\frac {\mu '}{B_{1}Q'{}_{1}}}={\frac {\mu ''-\mu '}{B_{1}O'{}_{1}}},}$

connecting two conjugate points in the second instrument as follows

 ${\displaystyle {\frac {1}{B_{1}Q''_{1}}}={\frac {(\zeta '-c)\left(\zeta '+c-2\xi ''\right)}{k^{2}\xi ''}},\ {\frac {1}{B_{1}Q'{}_{1}}}={\frac {(\zeta '-c)\left(\zeta '+c-2\xi '\right)}{k^{2}\xi '}},}$ ${\displaystyle B_{1}O'_{1}={\frac {(\zeta '-c)\left(\zeta '+c-2a'\right)}{k^{2}a'}}}$

and

 ${\displaystyle {\frac {\mu ''}{\xi ''}}-{\frac {\mu '}{\xi '}}={\frac {\mu ''-\mu '}{a'}},}$

since Q" and Q' are conjugate points in the first instrument; hence the relation is satisfied.