Gratings: Dispersion and Resolving Power — Question 5

Question 5

A fixed illuminated width imposes a geometric ceiling on resolving power

A normally illuminated grating has illuminated width $W=Nd$. In order m, $m\lambda=d\sin\theta$ and ideal resolving power is $\mathcal R=mN$. The collection system requires $|\theta|\leq60^\circ$. Let $W=10.0\,\mathrm{mm}$ and $\lambda=500\,\mathrm{nm}$.

1. Eliminate m, N, and d to express resolving power in terms of W and angle.

2. Find the maximum geometric bound and corresponding wavelength-separation bound.

Question 5 - Solution

A fixed illuminated width imposes a geometric ceiling on resolving power

(a) Method and calculation

$$\mathcal R=mN=\frac{d\sin\theta}{\lambda}\frac Wd.$$$$\boxed{|\mathcal R|=\frac W\lambda|\sin\theta|}.$$

(b) Interpretation and further calculation

$$\boxed{\mathcal R\leq\frac{0.0100}{500\times10^{-9}}\sin60^\circ=17320.5}.$$$$\boxed{\Delta\lambda\geq\lambda/17320.5=0.0288675\,\mathrm{nm}}.$$

Physical check

This is an ideal geometric bound; integer slit counts, available groove spacing, and instrument broadening can make the achievable value lower.