The refractive index variation for $\mathrm{CaF}_{2}$ (in the visible region of the spectrum) can be written in the form $^{*}$
$$n^{2}=6.09+\frac{6.12 \times 10^{-15}}{\lambda^{2}-8.88 \times 10^{-15}}+\frac{5.10 \times 10^{-9}}{\lambda^{2}-1.26 \times 10^{-9}}$$
where $\lambda$ is in meters
(a) Plot the variation of $n^{2}$ with $\lambda$ in the visible region.
(b) From the values of $A_{1}$ and $A_{2}$ show that $m / M \approx 2.07 \times$ $10^{-5}$ and compare this with the exact value.
(c) Show that using the constants $A_{1}, A_{2}, \lambda_{1}$ and $\lambda_{2}$ we obtain $n^{2} \approx 5.73$ which agrees reasonably well with the experimental value given above.