Chapter 28 - Physical Optics: Interference and Diffraction - Problems and Conceptual Exercises - Page 1006: 48

$51cm$

We know that $\theta=sin^{-1}(\frac{3\lambda}{W})$ $\theta=sin^{-1}[\frac{(3)(676\times 10^{-9}m)}{7.64\times 10^{-6}m}]=15.39^{\circ}$ Now $y=Ltan\theta$ We plug in the known values to obtain: $y=(1.85m)tan(15.39^{\circ})$ $y=50.9cm=51cm$