Gratings: Dispersion and Resolving Power — Question 2

Question 2

Linear dispersion depends on how the detector plane is oriented

A grating order is centered at $\theta=30^\circ$ with angular dispersion $\mathcal D=1.1547005\times10^6\,\mathrm{rad/m}$. Compare a screen using $x=f\tan\theta$ with a camera centered on that order, using local displacement $\xi\simeq f\Delta\theta$. Take $f=0.300\,\mathrm m$.

1. Derive the screen dispersion and evaluate it in millimeters per nanometer.

2. Calculate the centered-camera dispersion and the ratio between the two values.

Question 2 - Solution

Linear dispersion depends on how the detector plane is oriented

(a) Method and calculation

$$\frac{dx}{d\lambda}=f\sec^2\theta\frac{d\theta}{d\lambda}=\boxed{f\mathcal D/\cos^2\theta}.$$$$\boxed{dx/d\lambda=0.461880\,\mathrm{mm/nm}}.$$

(b) Interpretation and further calculation

$$\boxed{d\xi/d\lambda=f\mathcal D=0.346410\,\mathrm{mm/nm}}.$$$$\boxed{(dx/d\lambda)/(d\xi/d\lambda)=\sec^230^\circ=4/3}.$$

Physical check

A local focal-plane displacement and a coordinate measured on a plane normal to the original axis have different geometric projection factors.