Question

O retea de difractie are n=1000tr/mm. Care este numarul total de maxime care se pot obtine folosind o radiatie cu lungimea de unda λ =500nm

Answer 1

$\delta=lsin\alpha\\\delta=k\lambda\\n=\frac{1}{l}\\ lsin\alpha=k\lambda=>sin\alpha=\frac{k\lambda}{l}=kn\lambda\\\alpha\leq 90^o=>sin\alpha\leq 1=>nk\lambda\leq 1=>k\leq \frac{1}{n\lambda}\\ k\in Z=>k_{max}=[\frac{1}{n\lambda}]=[\frac{1}{1000*500*10^{-6}} ]=2\\ N=2k_{max}+1=2*2+1=4+1=5 maxime\\N=5 maxime$