A1  ^?? With refractometry.py find the value of $n$ for each wavelength $\lambda$ in the table.

Using the program, we can find the following pairs of $n$.

$\lambda,~\text{nm}$	$\Phi=60^\circ$		$\Phi=50^\circ$
	$n_1$	$n_2$	$n_1$	$n_2$
378	1.4732	2.0771	1.0129	1.4726
406	1.4698	2.0833	1.0144	1.4691
434	1.4671	2.0880	1.0153	1.4667
466	1.4646	2.0926	1.0162	1.4645
499	1.4624	2.0966	1.0171	1.4624
535	1.4606	2.0999	1.0179	1.4605
573	1.4590	2.1029	1.0186	1.4589
614	1.4578	2.1051	1.0191	1.4577
658	1.4566	2.1073	1.0197	1.4561
706	1.4549	2.1104	1.0202	1.4551
756	1.4537	2.1127	1.0206	1.4541
810	1.4526	2.1149	1.0210	1.4532

As you can see the $n_1$ for $\Phi=60^\circ$ is approximately equal to the $n_2$ for $\Phi=50^\circ$. Let's use their average as the answer.

A2  ^?? Plot a linear graph $n(\lambda)$ and find the values of $A$ and $B$ for fused glass.

Let's plot $n$ vs. $1/\lambda^2$, so that $A$ is an intercept and $B$ is a slope.