Answer:
[tex]Given:\\d=5mm=0.005\\lambda=600nm=600*10^-^9[/tex]
[tex]y=3m\\L=?[/tex]
applying Rayleigh criterion
[tex]sin\alpha =y/L=1.22(lambda/d)=1.22(600*10^-^9nm/0.005)=1.464*10^-^4\\\alpha =1.464*10^-^4rad[/tex]
[tex]sin\alpha =y/L\\=1.22(lambda/d)\\L=yd/1.22*lambda\\=3m*0.005/1.22*600*10^-^9\\=20 km[/tex]