Chapter 25, Problem 45P

(a)

To determine

The number of orders of lines that can be seen with the grating.

Answer to Problem 45P

The number of orders of lines that can be seen with the grating is $\underset{\_}{3}$.

Explanation of Solution

Given that the width of the grating is $1.600\text{cm}$, the number of slits is $12000$, and the wavelengths of light are ${\lambda }_{\text{a}}=440.000\text{nm}$ and ${\lambda }_{\text{b}}=440.936\text{nm}$.

Write the expression for the slit width of the grating.

  $d=\frac{\text{width of grating}}{\text{total number of slits}}$                                                                                     (I)

Here, $d$ is the slit width.

Write the expression for the diffraction maxima using grating.

  $d\sin \theta =m\lambda$                                                                                                    (II)

Here, $\theta$ is the angle of diffraction, $\lambda$ is the wavelength of light, and $m$ is the order of diffraction.

Deduce the number of orders that can be seen, from equation (II).

  $m\le \frac{d\sin \theta }{\lambda }$                                                                                                      (III)

The maximum number of orders is obtained for $\theta =90°$ (since $\sin \theta =90°$).

Conclusion:

Substitute $1.600\text{cm}$ for width of grating, and $12000$ for the total number of slits in equation (I) to find $d$.

  $\begin{array}{c}d=\frac{1.600\text{cm}}{12000}\\ =\frac{1.600\text{cm}\times \frac{1\text{m}}{100\text{cm}}}{12000}\\ =1.333\times {10}^{-6}\text{m}\end{array}$

Substitute $1.333\times {10}^{-6}\text{m}$ for $d$, $90°$ for $\theta$, and $440.000\text{nm}$ for $\lambda$ in equation (III) to find $m$.

  $\begin{array}{c}m\le \frac{\left(1.333\times {10}^{-6}\text{m}\right)\sin 90°}{440.000\text{nm}}\\ =3.024\end{array}$

Since $m$ must be an integer, it can be taken as $m=3$.

Therefore, the number of orders of lines that can be seen with the grating is $\underset{\_}{3}$.

(b)

To determine

The angular separation ${\theta }_{\text{b}}-{\theta }_{\text{a}}$ between the lines in each order.

Answer to Problem 45P

The angular separation ${\theta }_{\text{b}}-{\theta }_{\text{a}}$ between the lines in each order are $\underset{\_}{{\theta }_{\text{b1}}-{\theta }_{\text{a1}}=0.04°}$, $\underset{\_}{{\theta }_{\text{b2}}-{\theta }_{\text{a2}}=0.11°}$, and $\underset{\_}{{\theta }_{\text{b3}}-{\theta }_{\text{a3}}=0.91°}$.

Explanation of Solution

Given that the width of the grating is $1.600\text{cm}$, the number of slits is $12000$, and the wavelengths of light are ${\lambda }_{\text{a}}=440.000\text{nm}$ and ${\lambda }_{\text{b}}=440.936\text{nm}$.

Solve equation (II) for $\theta$.

  $\theta ={\sin }^{-1}\left(\frac{m\lambda }{d}\right)$                                                                                          (IV)

The angles corresponding to each wavelengths for each orders can be computed and their difference gives the angular separation.

Conclusion:

The value of $1/d$ can be computed from equation (I) as,

  $\begin{array}{c}\frac{1}{d}=\frac{12000}{1.600\text{cm}}\\ =\frac{12000}{1.600\text{cm}\times \frac{1\text{m}}{100\text{cm}}}\\ =7.500\times {10}^5\text{/m}\end{array}$

Substitute $1$ for $m$, $440.000\text{nm}$ for $\lambda$, and $7.500\times {10}^5\text{/m}$ for $1/d$ in equation (IV) to find $\theta$ corresponding to order 1 of ${\lambda }_{\text{a}}$, (${\theta }_{\text{a1}}$).

  $\begin{array}{c}{\theta }_{\text{a1}}={\sin }^{-1}\left[\left(1\right)\left(440.000\text{nm}\right)\left(7.500\times {10}^5\text{/m}\right)\right]\\ ={\sin }^{-1}\left[\left(1\right)\left(440.000\text{nm}\times \frac{1\text{m}}{1\times {10}^9\text{m}}\right)\left(7.500\times {10}^5\text{/m}\right)\right]\\ =19.27°\end{array}$

Substitute $2$ for $m$, $440.000\text{nm}$ for $\lambda$, and $7.500\times {10}^5\text{/m}$ for $1/d$ in equation (IV) to find $\theta$ corresponding to order 2 of ${\lambda }_{\text{a}}$, (${\theta }_{\text{a2}}$)

  $\begin{array}{c}{\theta }_{\text{a2}}={\sin }^{-1}\left[\left(2\right)\left(440.000\text{nm}\right)\left(7.500\times {10}^5\text{/m}\right)\right]\\ ={\sin }^{-1}\left[\left(2\right)\left(440.000\text{nm}\times \frac{1\text{m}}{1\times {10}^9\text{m}}\right)\left(7.500\times {10}^5\text{/m}\right)\right]\\ =41.30°\end{array}$

Substitute $3$ for $m$, $440.000\text{nm}$ for $\lambda$, and $7.500\times {10}^5\text{/m}$ for $1/d$ in equation (IV) to find $\theta$ corresponding to order 3 of ${\lambda }_{\text{a}}$, (${\theta }_{\text{a3}}$)

  $\begin{array}{c}{\theta }_{\text{a3}}={\sin }^{-1}\left[\left(3\right)\left(440.000\text{nm}\right)\left(7.500\times {10}^5\text{/m}\right)\right]\\ ={\sin }^{-1}\left[\left(3\right)\left(440.000\text{nm}\times \frac{1\text{m}}{1\times {10}^9\text{m}}\right)\left(7.500\times {10}^5\text{/m}\right)\right]\\ =81.89°\end{array}$

Substitute $1$ for $m$, $440.936\text{nm}$ for $\lambda$, and $7.500\times {10}^5\text{/m}$ for $1/d$ in equation (IV) to find $\theta$ corresponding to order 1 of ${\lambda }_{\text{b}}$, (${\theta }_{\text{b1}}$).

  $\begin{array}{c}{\theta }_{\text{b1}}={\sin }^{-1}\left[\left(1\right)\left(440.936\text{nm}\right)\left(7.500\times {10}^5\text{/m}\right)\right]\\ ={\sin }^{-1}\left[\left(1\right)\left(440.936\text{nm}\times \frac{1\text{m}}{1\times {10}^9\text{m}}\right)\left(7.500\times {10}^5\text{/m}\right)\right]\\ =19.31°\end{array}$

Substitute $2$ for $m$, $440.936\text{nm}$ for $\lambda$, and $7.500\times {10}^5\text{/m}$ for $1/d$ in equation (IV) to find $\theta$ corresponding to order 2 of ${\lambda }_{\text{b}}$, (${\theta }_{\text{b2}}$)

  $\begin{array}{c}{\theta }_{\text{b2}}={\sin }^{-1}\left[\left(2\right)\left(440.936\text{nm}\right)\left(7.500\times {10}^5\text{/m}\right)\right]\\ ={\sin }^{-1}\left[\left(2\right)\left(440.936\text{nm}\times \frac{1\text{m}}{1\times {10}^9\text{m}}\right)\left(7.500\times {10}^5\text{/m}\right)\right]\\ =41.41°\end{array}$

Substitute $3$ for $m$, $440.936\text{nm}$ for $\lambda$, and $7.500\times {10}^5\text{/m}$ for $1/d$ in equation (IV) to find $\theta$ corresponding to order 3 of ${\lambda }_{\text{b}}$, (${\theta }_{\text{b3}}$)

  $\begin{array}{c}{\theta }_{\text{b3}}={\sin }^{-1}\left[\left(3\right)\left(440.936\text{nm}\right)\left(7.500\times {10}^5\text{/m}\right)\right]\\ ={\sin }^{-1}\left[\left(3\right)\left(440.936\text{nm}\times \frac{1\text{m}}{1\times {10}^9\text{m}}\right)\left(7.500\times {10}^5\text{/m}\right)\right]\\ =82.80°\end{array}$

Compute the angular separation for each orders.

  $\begin{array}{c}{\theta }_{\text{b1}}-{\theta }_{\text{a1}}=19.31°-19.27°\\ =0.04°\end{array}$

  $\begin{array}{c}{\theta }_{\text{b2}}-{\theta }_{\text{a2}}=41.41°-41.30°\\ =0.11°\end{array}$

  $\begin{array}{c}{\theta }_{\text{b3}}-{\theta }_{\text{b3}}=82.80°-81.89°\\ =0.91°\end{array}$

Therefore, the angular separation ${\theta }_{\text{b}}-{\theta }_{\text{a}}$ between the lines in each order are $\underset{\_}{{\theta }_{\text{b1}}-{\theta }_{\text{a1}}=0.04°}$, $\underset{\_}{{\theta }_{\text{b2}}-{\theta }_{\text{a2}}=0.11°}$, and $\underset{\_}{{\theta }_{\text{b3}}-{\theta }_{\text{a3}}=0.91°}$.

(c)

To determine

The order of the spectrum which best resolves the two lines.

Answer to Problem 45P

The third order spectrum best resolves the two lines.

Explanation of Solution

The angular separation for different orders of the spectrum are obtained as ${\theta }_{\text{b1}}-{\theta }_{\text{a1}}=0.04°$, ${\theta }_{\text{b2}}-{\theta }_{\text{a2}}=0.11°$, and ${\theta }_{\text{b3}}-{\theta }_{\text{a3}}=0.91°$.

The higher resolution corresponds to larger angular separation in the diffraction process. Among the obtained values of the angular separations, the third order spectrum has the largest angular separation. Thus, the two lines are best resolved in the third-order spectrum.

Conclusion:

Therefore, the third order spectrum best resolves the two lines.