Answer
$658nm$
Work Step by Step
We can find the required wavelength as follows:
$tan\theta=\frac{y}{L}$
$\theta=tan^{-1}(\frac{y}{L})$
$\implies \theta=tan^{-1}(\frac{7.15\times 10^{-3}m}{2.50})=0.1638$
Now $\lambda=\frac{dsin\theta}{n}$
We plug in the known values to obtain:
$\lambda=\frac{(0.230\times 10^{-3}m)sin(0.1638)}{(1)}$
$\lambda=657.75\times 10^{-9}m$
$\lambda=658nm$
