When deriving the color resolution of the grating, according to the Rayleigh criterion, the angular width Δ θ = j Δ λ d cos θ \Delta \theta=\frac{j\Delta\lambda}{d\cos \theta}D i=dcosij D land the angular width due to grating diffraction θ 1 = λ N d cos θ \theta_1=\frac{\lambda}{Nd\cos\theta}i1=NdcosilThe color resolution of the grating obtained by being equal R = λ Δ λ = j NR =\frac{\lambda}{\Delta\lambda}=jNR=D ll=j N
buthere θ 1 \theta_1i1How did you get it?
First of all, it is necessary to deduce the position of the dark grating
: if it is a dark grating, then the cancellation condition is satisfied between adjacent slits of the grating. And the phase difference between adjacent slits is: δ = 2 π λ d sin θ \delta =\frac{2\pi}\lambda d\sin\thetad=l2 p.mdsinθ , then by the amplitude vector method, the dark fringe condition satisfied by the common phase difference of N slits is:N δ = ± 2 m π N\delta=\pm2m\piN d=± 2 m π
It is worth noting that, assuming that m here is a multiple of N, it will become the position of the main maximum. 1So there are restrictions on the range of m here:m ≠ k N m\neq kNm=k N
attaches a diagram: (source of wave optics MOOC courseware of East China Normal University)
and the grating equation satisfiesd sin θ = ± k λ d\sin\theta = \pm k\lambdadsini=±kλ , k k k is any integer.
According to the physical meaning, the half-angle width is the distance from a certain bright fringe to an adjacent dark fringe. According to this condition, the column formula is:
Let the half-angle width be Δ θ \Delta \thetaΔ θ , then
Form : d sin θ k = k λ (1) Form:d\sin\theta_k = k\lambda \tag{1}Bright pattern:dsinik=kλ( 1 ) Adjacent dark lines: N d sin ( θ k + Δ θ ) = ( k N + 1 ) λ (2) Adjacent dark lines: Nd\sin(\theta_k+\Delta\theta)=( kN+1)\lambda\tag{2}adjacent shadows _:Ndsin ( ik+D i )=(kN+1 ) l(2)
Use the trigonometric formula to have:
N d ( sin θ k cos Δ θ + cos θ k sin Δ θ ) = ( k N + 1 ) λ (3) Nd(\sin\theta_k\cos\Delta\theta+\cos\theta_k\sin \Delta\theta)=(kN+1)\lambda\tag{3}Nd(sinikcosD i+cosiksinD i )=(kN+1 ) l(3)
And in Δ θ \Delta\thetaΔ θ较小时,有cos Δ θ ≈ 1 , sin Δ θ ≈ Δ θ cos\Delta\theta \approx1,\sin\Delta\theta\approx\Delta\thetac o s D i≈1,sinD i≈Δ θ ; (3) and (1) are subtracted to get the expression of the half-angle width:
Δ θ = λ N d cos θ \Delta\theta = \frac{\lambda}{Nd\cos\theta};D i=Ndcosil
Certificate completed
(As for why both the main maxima and the minima satisfy this positional condition, they become the main maxima in the end. Simply put, when the diffraction factor tends to 0 at the same time, the limit tends to 1) ↩︎