Gratings: Dispersion and Resolving Power — Question 1

Question 1

Differentiating the grating equation gives angular dispersion

A normally illuminated grating obeys $d\sin\theta=m\lambda$. Let $d=2.00\,\mu\mathrm m$, order m=2, and central wavelength $500\,\mathrm{nm}$.

1. Derive and evaluate angular dispersion $\mathcal D=d\theta/d\lambda$ at this wavelength.

2. Estimate the angular separation of two lines differing by $2.00\,\mathrm{nm}$.

Question 1 - Solution

Differentiating the grating equation gives angular dispersion

(a) Method and calculation

$$d\cos\theta\,d\theta=m\,d\lambda\Longrightarrow\boxed{\mathcal D=\frac m{d\cos\theta}}.$$$$\sin\theta=0.500\Longrightarrow\boxed{\theta=30^\circ,\quad\mathcal D=1.15470\times10^6\,\mathrm{rad/m}}.$$

(b) Interpretation and further calculation

$$\mathcal D=1.15470\,\mathrm{mrad/nm}.$$$$\boxed{\Delta\theta\simeq\mathcal D\Delta\lambda=2.30940\,\mathrm{mrad}}.$$

Physical check

The derivative is a local linear approximation; finite wavelength intervals require the inverse-sine grating equation if higher accuracy is needed.