Answer
$24.8cm$
Work Step by Step
We can find the required linear distance as follows:
$\theta=sin^{-1}[\frac{(m+\frac{1}{2})\lambda}{w}]$
$\implies \theta=sin^{-1}[\frac{(1+\frac{1}{2})\lambda}{w}]=sin^{-1}(\frac{3\lambda}{2w})$
We plug in the known values to obtain:
$\theta=sin^{-1}[\frac{3}{2}(\frac{676\times 10^{-9}}{7.64\times 10^{-6}})]=7.6269^{\circ}$
Now $y=Ltan\theta$
We plug in the known values to obtain:
$y=(1.85m)tan{7.62^{\circ}}=0.2477m$
$\implies y=24.8cm$
