What is the distance that a beam of light of wavelength 500 nm can travel without significant broadening, if the diffracting aperture is 3 mm wide ?

18 m

The correct answer is Option (4) → 18 m

The angular spread θ for the first diffraction minimum is -

$\sin θ=\frac{λ}{d}$

for small angle $\sin θ≃θ$ (radians)

$θ=\frac{λ}{d}=\frac{500×10^{-9}}{3×10^{-3}}=1.67×10^{-4}$ radians

Distance, $L=\frac{d}{θ}=\frac{3×10^{-3}}{1.67×10^{-4}}≃18m$