Page:Annales de mathématiques pures et appliquées, 1823-1824, Tome 14.djvu/179

 Les corrections sont expliquées en page de discussion

$\left.{\begin{array}{ll}{\frac {P''-p}{\lambda ''{\sqrt {1+P''^{2}+Q''^{2}}}}}&=\pm {\frac {P'-p}{\lambda '{\sqrt {1+P'^{2}+Q'^{2}}}}},\\\\{\frac {Q''-q}{\lambda ''{\sqrt {1+P''^{2}+Q''^{2}}}}}&=\pm {\frac {Q'-q}{\lambda '{\sqrt {1+P'^{2}+Q'^{2}}}}}.\end{array}}\right\}\quad$(III)

équations qui pourront remplacer (I) et (II), dans la recherche de $P''$ et $Q''$ en $P'$ et $Q'.$ Dans le cas de la réflexion, on prendra les signes inférieurs, en posant $\lambda ''=\lambda ',$ 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,

${\begin{array}{ll}{\frac {1}{\lambda ''}}.{\frac {\operatorname {d} .}{\operatorname {d} y}}{\frac {P''-p}{\sqrt {1+P''^{2}+Q''^{2}}}}&=\pm {\frac {1}{\lambda '}}.{\frac {\operatorname {d} .}{\operatorname {d} y}}{\frac {P'-p}{\sqrt {1+P'^{2}+Q'^{2}}}},\\\\{\frac {1}{\lambda ''}}.{\frac {\operatorname {d} .}{\operatorname {d} x}}{\frac {Q''-q}{\sqrt {1+P''^{2}+Q''^{2}}}}&=\pm {\frac {1}{\lambda '}}.{\frac {\operatorname {d} .}{\operatorname {d} x}}{\frac {Q'-q}{\sqrt {1+P'^{2}+Q'^{2}}}}\,;\end{array}}$

d’où en retranchant,

${\begin{aligned}&{\frac {1}{\lambda ''}}\left\{{\frac {\operatorname {d} .}{\operatorname {d} y}}{\frac {P''-p}{\sqrt {1+P''^{2}+Q''^{2}}}}-{\frac {\operatorname {d} .}{\operatorname {d} x}}{\frac {Q''-q}{\sqrt {1+P''^{2}+Q''^{2}}}}\right\}\\\\=\pm &{\frac {1}{\lambda '}}\left\{{\frac {\operatorname {d} .}{\operatorname {d} y}}{\frac {P'-p}{\sqrt {1+P'^{2}+Q'^{2}}}}-{\frac {\operatorname {d} .}{\operatorname {d} x}}{\frac {Q'-q}{\sqrt {1+P'^{2}+Q'^{2}}}}\right\}\,;\end{aligned}}$

donc, suivant qu’on aura ou qu’on n’aura pas

${\frac {\operatorname {d} .}{\operatorname {d} y}}{\frac {P'-p}{\sqrt {1+P'^{2}+Q'^{2}}}}={\frac {\operatorname {d} .}{\operatorname {d} x}}{\frac {Q'-q}{\sqrt {1+P'^{2}+Q'^{2}}}},$

on aura ou on n’aura pas

${\frac {\operatorname {d} .}{\operatorname {d} y}}{\frac {P''-p}{\sqrt {1+P''^{2}+Q''^{2}}}}={\frac {\operatorname {d} .}{\operatorname {d} x}}{\frac {Q''-q}{\sqrt {1+P''^{2}+Q''^{2}}}},$