Question

A monochromator with a focal length of $0.75\mathrm{m}$was equipped with an echellette grating with $3000$blazes per millimeter.

(a) Calculate the reciprocal linear dispersion of the instrument for first-order spectra.

(b) If localid="1646643497497" $2.0\mathrm{cm}$of the grating were illuminated, what is the first-order resolving power of the monochromator?

(c) At approximatelylocalid="1646643508082" $400\mathrm{nm}$, what minimum wavelength difference could in theory be completely resolved by the instrument?

Short answer

$$\begin{gathered} \mathrm{a})0.44\mathrm{nm}/\mathrm{mm} \\ \mathrm{b})6\times {10}^4 \\ \mathrm{c})200\mathrm{nm} \end{gathered}$$

Step-by-step solution

Given information

$$\begin{gathered} \mathrm{f}=0.75\mathrm{m} \\ 3000\mathrm{blaze}/\mathrm{mm} \\ b)\mathrm{grating}=2\mathrm{cm} \\ c)𝜆=400\mathrm{nm} \end{gathered}$$

Part a) Step 1: Calculation

Formula for e the reciprocal linear dispersion is

$D^{-1}=\frac{d}{nf}$

Putting values

localid="1646643603007" $$\begin{gathered} n=1 \\ {D}^{-1}=\frac{\left({\displaystyle \frac{1\mathrm{mm}}{3000\mathrm{blazes}/\mathrm{mm}}}\right)}{1\times 0.75\mathrm{m}} \\ {D}^{-1}=\frac{\left({\displaystyle \frac{1\mathrm{mm}}{3000\mathrm{blazes}/\mathrm{mm}}\times \frac{{10}^6\mathrm{nm}}{\mathrm{mm}}}\right)}{1\times 0.75\mathrm{m}\times \left({\displaystyle \frac{{10}^3\mathrm{mm}}{\mathrm{m}}}\right)} \\ {D}^{-1}=0.44\mathrm{nm}/\mathrm{mm} \end{gathered}$$

Part b) Step 1: Calculations

Formula for resolving power of the monochromator is

$R=nN$

Putting values in formula

localid="1646643626931" $$\begin{gathered} R=1\times 3000\mathrm{blazes}/\mathrm{mm}\times 2\mathrm{cm}\times 10\mathrm{mm}/\mathrm{cm} \\ R=60000 \\ R=6\times {10}^4 \end{gathered}$$

Part c) Step 1: Calculations

Formula for wavelength difference is

$∆{𝜆}_{diff}=\frac{1}{2}({𝜆}_{max}-{𝜆}_{min})$

Putting values in formula

localid="1646643644855" $$\begin{gathered} ∆{𝜆}_{diff}=\frac{1}{2}(400\mathrm{nm}) \\ ∆{𝜆}_{diff}=200\mathrm{nm} \end{gathered}$$