91Ó°ÊÓ

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:

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

For the first-order maxima, we have $\mathrm{λ}=d\sin \mathrm{θ}$. So, the angle can be obtained as follows:

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

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

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

(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:

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

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

(c)

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

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

For the second-order maxima, we have$2\mathrm{λ}=d\sin \mathrm{θ}$ . So, the angle can be obtained as follows:

$$\begin{gathered} \sin \mathrm{θ}=\frac{2\mathrm{λ}}{d} \\ \mathrm{θ}={\sin }^{-1}\left(\frac{2×589}{1900}\right) \\ \mathrm{θ}=38° \end{gathered}$$

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

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

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:

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

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

(e)

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

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

For the third-order maxima, we have$3\mathrm{λ}=d\sin \mathrm{θ}$. So, the angle can be obtained as follows:

$$\begin{gathered} \sin \mathrm{θ}=\frac{3\mathrm{λ}}{d} \\ \mathrm{θ}={\sin }^{-1}\left(\frac{3×589}{1900}\right) \\ \mathrm{θ}=68° \end{gathered}$$

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

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

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:

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

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