\(n=A+\frac{B}{\lambda^2}\)

\(v_g=v_p(1+\frac{\lambda}{n}\frac{dn}{d \lambda})\)

\(\frac{dn}{d \lambda}=\frac{-2B}{\lambda^3}=\frac{-2(2.5*10^6 \overset\circ{A^2})}{5000\overset\circ{A^3}}=-4*10^{-5} \overset\circ{A^{-1}}\)

\(v_p=\frac{c}{n}=\frac{c}{1.5}\)

\(p\)=1.5

\(v_g=\frac{c}{1.5}(1+\frac{5000\overset\circ{A}}{1.5}(-4*10^{-5}\overset\circ{A^{-1}}))=\frac{c}{1.73}\)

\(g\)=1.73