Gratings: Dispersion and Resolving Power — Question 3

Question 3

Rayleigh separation of grating peaks leads to resolving power mN

For N uniformly illuminated slits, the angular distance from a principal maximum to its nearest zero is $\delta\theta=\lambda/(Nd\cos\theta)$. The angular separation of nearby wavelengths in order m is $\Delta\theta\simeq m\Delta\lambda/(d\cos\theta)$.

1. Equate these quantities to derive the ideal resolving power.

2. For N=10000 in order m=2 at $600\,\mathrm{nm}$, calculate the limiting wavelength separation.

Question 3 - Solution

Rayleigh separation of grating peaks leads to resolving power mN

(a) Method and calculation

$$\frac{m\Delta\lambda}{d\cos\theta}=\frac\lambda{Nd\cos\theta}\Longrightarrow\Delta\lambda=\lambda/(mN).$$$$\boxed{\mathcal R=\lambda/\Delta\lambda=mN}.$$

(b) Interpretation and further calculation

$$\boxed{\mathcal R=2(10000)=20000}.$$$$\boxed{\Delta\lambda=600\,\mathrm{nm}/20000=0.0300\,\mathrm{nm}}.$$

Physical check

Large angular dispersion alone does not establish resolution; the diffracted line width must be compared with the separation.