Answer
$528nm$
Work Step by Step
We know that
$\frac{m_b}{m_R}=\frac{\lambda_R}{\lambda_b}$
We plug in the known values to obtain:
$\frac{m_b}{m_R}=\frac{687nm}{458nm}=\frac{3}{2}$
Now we can find the required thickness as follows:
$t=\frac{m_b\lambda_b}{2n}$
We plug in the known values to obtain:
$t=\frac{3(458\times 10^{-9}m)}{2(1.30)}$
$t=528nm$
