Light at wavelength 589 nm from a sodium lamp is incident perpendicularly on a grating with 40,000 rulings over width 76 mm. What are the first-order (a) dispersion $\mathit{D}$and (b) resolving power $\mathit{R}$, the second-order (c) $\mathit{D}$and (d) $\mathit{R}$,and the third-order (e) $\mathit{D}$and (f) $\mathit{R}$?

It is known the resolving power of a grating is given by$\mathrm{R}=\mathrm{Nm}$ , where is $\mathrm{N}$the number of rulings in the grating and $\mathrm{m}$is the order of the lines.

(a)

Here, the width is and the rulings are 40,000. So, the width of the single grating is:

$$\begin{gathered} d=\frac{76\times {10}^{-3}}{40000} \\ =1900\times {10}^{-9} \text{m} \\ =1900 \text{nm} \end{gathered}$$

For the first-order maxima, we have $\lambda =d\sin \theta$. So, the angle can be obtained as follows:

$$\begin{gathered} \sin \theta =\frac{\lambda }{d} \\ \theta ={\sin }^{-1}\left(\frac{589}{1900}\right) \\ \theta =18° \end{gathered}$$

Now, the dispersion of given by$D=\frac{m}{d\cos \theta }$. So, the first-order dispersion can be obtained as follows:

$$\begin{gathered} D=\frac{m}{d\cos \theta } \\ =\frac{1}{1900\cos \left(18°\right)} \\ =5.5\times {10}^{-4} \text{rad/nm} \\ =0.032°\text{/nm} \end{gathered}$$

(b)

It is known that the resolving power is given by$R=Nm$. Here$N=40000$, and $m=1$. So, the resolving power can be obtained as follows:

$$\begin{gathered} R=40000\times 1 \\ R=4.0\times {10}^4 \end{gathered}$$

Thus, the first-order resolving power is$4\times {10}^4$ .

(c)

Here, the width is$76 \text{mm}$ and the rulings are 40,000. So, the width of the single grating is:

$$\begin{gathered} d=\frac{76\times {10}^{-3}}{40000} \\ =1900\times {10}^{-9} \text{m} \\ =1900 \text{nm} \end{gathered}$$

For the second-order maxima, we have$2\lambda =d\sin \theta$ . So, the angle can be obtained as follows:

$$\begin{gathered} \sin \theta =\frac{2\lambda }{d} \\ \theta ={\sin }^{-1}\left(\frac{2\times 589}{1900}\right) \\ \theta =38° \end{gathered}$$

Now, the dispersion of given by$D=\frac{m}{d\cos \theta }$ . So, the second-order dispersion can be obtained as follows:

$$\begin{gathered} D=\frac{m}{d\cos \theta } \\ =\frac{2}{1900\cos \left(38°\right)} \\ =13.4\times {10}^{-4} \text{rad/nm} \\ =0.076°\text{/nm} \end{gathered}$$

Thus, the second-order dispersion is $0.076°\text{/nm}$.

(d)

It is known that the resolving power is given by$R=Nm$. Here, $N=40000$and$m=2$. So, the resolving power can be obtained as follows:

$$\begin{gathered} R=40000\times 2 \\ R=8.0\times {10}^4 \end{gathered}$$

Thus, the second-order resolving power is$8.0\times {10}^4$ .

(e)

Here, the width is $76mm$and the rulings are 40,000. So, the width of the single grating is:

$$\begin{gathered} d=\frac{76\times {10}^{-3}}{40000} \\ =1900\times {10}^{-9} \text{m} \\ =1900 \text{nm} \end{gathered}$$

For the third-order maxima, we have$3\lambda =d\sin \theta$. So, the angle can be obtained as follows:

$$\begin{gathered} \sin \theta =\frac{3\lambda }{d} \\ \theta ={\sin }^{-1}\left(\frac{3\times 589}{1900}\right) \\ \theta =68° \end{gathered}$$

Now, the dispersion of given by$D=\frac{m}{d\cos \theta }$. So, the third-order dispersion can be obtained as follows:

$$\begin{gathered} D=\frac{m}{d\cos \theta } \\ =\frac{3}{1900\cos \left(68°\right)} \\ =42.1\times {10}^{-4} \text{rad/nm} \\ =0.24°\text{/nm} \end{gathered}$$

Thus, the third-order dispersion is $0.24°\text{/nm}$.

(f)

It is known that the resolving power is given by$R=Nm$. Here, $N=40000$and $m=3$. So, the resolving power can be obtained as follows:

$$\begin{gathered} R=40000\times 3 \\ R=1.2\times {10}^5 \end{gathered}$$

Thus, the third-order resolving power is$1.2\times {10}^5$.