![{\displaystyle \left.{\begin{array}{ll}{\frac {P''-p}{\lambda ''{\sqrt {1+P''^{2}+Q''^{2}}}}}&=\pm {\frac {P'-p}{\lambda '{\sqrt {1+P'^{2}+Q'^{2}}}}},\\\\{\frac {Q''-q}{\lambda ''{\sqrt {1+P''^{2}+Q''^{2}}}}}&=\pm {\frac {Q'-q}{\lambda '{\sqrt {1+P'^{2}+Q'^{2}}}}}.\end{array}}\right\}\quad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/edd1cf738767b827c53b86a9dded80bc00d2af45)
(III)
équations qui pourront remplacer (I) et (II), dans la recherche de
et
en
et
Dans le cas de la réflexion, on prendra les signes inférieurs, en posant
tandis qu’au contraire, dans le cas de la réfraction, ce sera les signes supérieurs qu’il faudra prendre (21).
23. On tire des équations (III), par différentiation,
![{\displaystyle {\begin{array}{ll}{\frac {1}{\lambda ''}}.{\frac {\operatorname {d} .}{\operatorname {d} y}}{\frac {P''-p}{\sqrt {1+P''^{2}+Q''^{2}}}}&=\pm {\frac {1}{\lambda '}}.{\frac {\operatorname {d} .}{\operatorname {d} y}}{\frac {P'-p}{\sqrt {1+P'^{2}+Q'^{2}}}},\\\\{\frac {1}{\lambda ''}}.{\frac {\operatorname {d} .}{\operatorname {d} x}}{\frac {Q''-q}{\sqrt {1+P''^{2}+Q''^{2}}}}&=\pm {\frac {1}{\lambda '}}.{\frac {\operatorname {d} .}{\operatorname {d} x}}{\frac {Q'-q}{\sqrt {1+P'^{2}+Q'^{2}}}}\,;\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06e7a7d92e48aa584cc49d2174b82a00dda0def0)
d’où en retranchant,
![{\displaystyle {\begin{aligned}&{\frac {1}{\lambda ''}}\left\{{\frac {\operatorname {d} .}{\operatorname {d} y}}{\frac {P''-p}{\sqrt {1+P''^{2}+Q''^{2}}}}-{\frac {\operatorname {d} .}{\operatorname {d} x}}{\frac {Q''-q}{\sqrt {1+P''^{2}+Q''^{2}}}}\right\}\\\\=\pm &{\frac {1}{\lambda '}}\left\{{\frac {\operatorname {d} .}{\operatorname {d} y}}{\frac {P'-p}{\sqrt {1+P'^{2}+Q'^{2}}}}-{\frac {\operatorname {d} .}{\operatorname {d} x}}{\frac {Q'-q}{\sqrt {1+P'^{2}+Q'^{2}}}}\right\}\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d65d088ca7f0a2399deba539b2b8c45310826b0e)
donc, suivant qu’on aura ou qu’on n’aura pas
![{\displaystyle {\frac {\operatorname {d} .}{\operatorname {d} y}}{\frac {P'-p}{\sqrt {1+P'^{2}+Q'^{2}}}}={\frac {\operatorname {d} .}{\operatorname {d} x}}{\frac {Q'-q}{\sqrt {1+P'^{2}+Q'^{2}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/824d8fe2625e311f03c8b3da85eb9d92dc68d30d)
on aura ou on n’aura pas
![{\displaystyle {\frac {\operatorname {d} .}{\operatorname {d} y}}{\frac {P''-p}{\sqrt {1+P''^{2}+Q''^{2}}}}={\frac {\operatorname {d} .}{\operatorname {d} x}}{\frac {Q''-q}{\sqrt {1+P''^{2}+Q''^{2}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68d752b0c7c941f6dcbe2d505cf9ceb93d4082fd)