88
[Nov. 12,
Mr. H. Bateman
The linear magnification in the new system is
$m_{1}={\frac {\frac {A_{1}Q'_{1}}{\mu '}}{\frac {A_{1}Q_{1}}{\mu }}}={\frac {\mu x'(z2x)}{\mu 'x(z2x')}}=m{\frac {z2x}{z2x'}}.$


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
$B_{1}Q'_{1}=\xi '_{1}={\frac {k^{2}\xi '}{(\zeta '\xi ')^{2}(\xi 'c)^{2}}}={\frac {k^{2}\xi '}{(\zeta 'c)\left(\zeta '+c2\xi '\right)}},$
$\zeta '_{1}={\frac {k^{2}\zeta }{\zeta ^{2}c{}^{2}}},\ c_{1}={\frac {k^{2}c}{\zeta ^{2}c{}^{2}}},$
$\zeta '_{1}\xi '_{1}={\frac {k^{2}(\zeta '\xi ')}{(\zeta '\xi ')^{2}(\xi 'c)^{2}}}={\frac {k^{2}(\zeta '\xi ')}{(\zeta 'c)\left(\zeta '+c2\xi '\right)}},$
$\xi '_{1}c_{1}={\frac {k^{2}(c\xi ')}{(\zeta 'c)\left(\zeta '+c2\xi '\right)}},$
$\zeta '_{1}+c_{1}2\xi '_{1}={\frac {k^{2}}{\zeta '+c2\xi '}}.$


The last of which may be written
$\zeta '_{1}c_{1}2\left(\xi '_{1}c_{1}\right)={\frac {k^{2}}{\zeta 'c2\left(\xi 'c\right)}},$


i.e.,
$z_{1}2x'_{1}={\frac {k^{2}}{z2x'}},$


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]}
 ↑ We can verify the equation
${\frac {\mu ''}{B_{1}Q''_{1}}}{\frac {\mu '}{B_{1}Q'{}_{1}}}={\frac {\mu ''\mu '}{B_{1}O'{}_{1}}},$


connecting two conjugate points in the second instrument as follows
${\frac {1}{B_{1}Q''_{1}}}={\frac {(\zeta 'c)\left(\zeta '+c2\xi ''\right)}{k^{2}\xi ''}},\ {\frac {1}{B_{1}Q'{}_{1}}}={\frac {(\zeta 'c)\left(\zeta '+c2\xi '\right)}{k^{2}\xi '}},$
$B_{1}O'_{1}={\frac {(\zeta 'c)\left(\zeta '+c2a'\right)}{k^{2}a'}}$


and
${\frac {\mu ''}{\xi ''}}{\frac {\mu '}{\xi '}}={\frac {\mu ''\mu '}{a'}},$


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