Answer
(a) 3rd order of the first color overlaps with the 2nd order of the second color.
(b) $35^{\circ}$
Work Step by Step
(a) We know that
$\frac{m_1}{m_2}=\frac{\lambda_2}{\lambda_1}$
We plug in the known values to obtain:
$\frac{m_1}{m_2}=\frac{630nm}{420nm}=\frac{3}{2}$
This shows that the 3rd order of the first color overlaps with the 2nd order of the second color.
(b) We know that
$\theta=sin^{-1}[\frac{m_1\lambda_1}{d}]$
We plug in the known values to obtain:
$\theta=sin^{-1}[\frac{2(420\times 10^{-9})}{2.222\times 10^{-6}m}]$
$\theta=35^{\circ}$
