Gratings: Dispersion and Resolving Power — Question 4

Question 4

A higher-order resolution estimate must pass the propagation constraint

A grating has 1200 lines per millimeter and must resolve two lines near $500\,\mathrm{nm}$ separated by $0.0500\,\mathrm{nm}$ at normal incidence. Use ideal resolving power $mN$ and illuminated width $W=Nd$.

1. Find the required number of slits and illuminated width in first order.

2. Compute the formal second-order width and determine whether that order can actually propagate.

Question 4 - Solution

A higher-order resolution estimate must pass the propagation constraint

(a) Method and calculation

$$\mathcal R_{\mathrm{req}}=500/0.0500=10000,\quad d=1/1200\,\mathrm{mm}.$$$$m=1:\quad\boxed{N=10000,\quad W=8.33333\,\mathrm{mm}}.$$

(b) Interpretation and further calculation

$$m=2:\quad N=5000\Longrightarrow\boxed{W_{\mathrm{formal}}=4.16667\,\mathrm{mm}}.$$$$\sin\theta_2=2\lambda/d=1.20>1\Longrightarrow\boxed{\text{second-order design is impossible}}.$$

Physical check

Resolving-power algebra cannot create a grating order outside the propagating angular range.