Answer:
X = 5.48 mm
Explanation:
for single slit
By Rayleigh criterian
[tex]sin\theta _R = \frac{\lambda}{d}[/tex]
[tex]sin\theta _R = tan\theta _R ≈ \frac{x}{3}[/tex]
where d = slit width =0.5 mm
wavelength [tex]\lambda = 686 nm[/tex]
[tex]\frac{x}{3} =\frac{686*10^{-9}}{0.5*10^{-3}}[/tex]
therefore maximum of value of X can be calculated from above
[tex]X = 5.48*10^{-3} m[/tex]
X = 5.48 mm
