The correct relations showing Raoult's law are

According to Raoult's law, Relative lowering in vapour pressure = Mole fraction of solute
                                $\frac{{\mathrm{p}}^{\circ }-\mathrm{p}}{{\mathrm{p}}^{\circ }}={\mathrm{x}}_2\Rightarrow \frac{{\mathrm{p}}^{\circ }-\mathrm{p}}{{\mathrm{p}}^{\circ }}=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}$
or                           $\frac{{\mathrm{p}}^{\circ }}{{\mathrm{p}}^{\circ }-\mathrm{p}}=\frac{{\mathrm{n}}_1+{\mathrm{n}}_2}{{\mathrm{n}}_2}$
or                          $\frac{{\mathrm{p}}^{\circ }}{{\mathrm{p}}^{\circ }-\mathrm{p}}=1+\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}$
                    $\frac{{\mathrm{p}}^{\circ }}{{\mathrm{p}}^{\circ }-\mathrm{p}}-1=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}\Rightarrow \frac{\mathrm{p}}{{\mathrm{p}}^{\circ }-\mathrm{p}}=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}$
or                              $\frac{{\mathrm{p}}^{\circ }-\mathrm{p}}{\mathrm{p}}=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}$