Question

A monochromator has a focal length of $1.6\mathrm{m}$and a collimating mirror with a diameter of $3.5\mathrm{cm}$. The dispersing device was a grating with $1500\mathrm{lines}/\mathrm{mm}$. For first-order diffraction,

(a) what is the resolving power of the monochromator if a collimated beam illuminated $3.2\mathrm{cm}$of the grating?

(b) what are the first- and second-order reciprocal linear dispersions of the monochromator?

Short answer

$$\begin{gathered} a)4.8\times {10}^4 \\ b)0.42\mathrm{nm}/\mathrm{mm}\mathrm{and}0.21\mathrm{nm}/\mathrm{mm} \end{gathered}$$

Step-by-step solution

Part a) Step 1: Given Information

$$\begin{gathered} f=1.6\mathrm{m} \\ d=3.5\mathrm{cm} \\ \mathrm{lines}/\mathrm{mm}=1500 \\ a)N=3.2\mathrm{cm} \end{gathered}$$

Part a) Step 2: Calculation

Formula for resolving power of the monochromator is

$R=nN$

By putting values

localid="1646640071221" $$\begin{gathered} R=1\times 1500\mathrm{lines}/\mathrm{mm}\times 3.2\mathrm{cm}\times 10\mathrm{mm}/\mathrm{cm} \\ R=48000 \\ R=4.8\times {10}^4 \end{gathered}$$

Part b) Step 1: Formula.

The formula for reciprocal linear dispersions of the monochromator is

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

Where,

$d$is the distance

$f$is the focal length

Part b) Step 2: Calculation for first order

Putting values in formula

$$\begin{gathered} n=1 \\ {D}^{-1}=\frac{\left({\displaystyle \frac{1\mathrm{mm}}{1500\mathrm{lines}/\mathrm{mm}}}\right)}{1\times 1.6\mathrm{m}} \\ {D}^{-1}=\frac{\left({\displaystyle \frac{1\mathrm{mm}}{1500\mathrm{lines}/\mathrm{mm}}\times \frac{{10}^6\mathrm{nm}}{\mathrm{mm}}}\right)}{1\times 1.6\mathrm{m}\times \left({\displaystyle \frac{{10}^3\mathrm{mm}}{\mathrm{m}}}\right)} \\ {D}^{-1}=\frac{{10}^3}{2400} \\ {D}^{-1}=0.42\mathrm{nm}/\mathrm{mm} \end{gathered}$$

Part b) Step 3: Calculation for second order

Putting values in formula

$$\begin{gathered} n=2 \\ {D}^{-1}=\frac{\left({\displaystyle \frac{1\mathrm{mm}}{1500\mathrm{lines}/\mathrm{mm}}}\right)}{2\times 1.6\mathrm{m}} \\ {D}^{-1}=\frac{\left({\displaystyle \frac{1\mathrm{mm}}{1500\mathrm{lines}/\mathrm{mm}}\times \frac{{10}^6\mathrm{nm}}{\mathrm{mm}}}\right)}{2\times 1.6\mathrm{m}\times \left({\displaystyle \frac{{10}^3\mathrm{mm}}{\mathrm{m}}}\right)} \\ {D}^{-1}=\frac{{10}^3}{4800} \\ {D}^{-1}=0.21\mathrm{nm}/\mathrm{mm} \end{gathered}$$