Chapter 36, Problem 30PQ

(a)

To determine

The half width of the spectral line.

Answer to Problem 30PQ

The half width of the spectral line is $9.0\times {10}^{-6}\text{rad}$.

Explanation of Solution

Write the expression for the grating spacing.

    $d=\frac{1}{n}$                     (I)

Here, $d$ is the grating spacing, $n$ is the density of ruling.

Write the expression for the grating spacing in terms of number of ruling on the grating.

    $d=\frac{l}{N}$                     (II)

Here, $N$ is the number of ruling on the grating and $l$ is the length of the grating.

Rewrite the equation (II) to determine $N$.

    $N=\frac{l}{d}$                     (III)

Write the expression for the half width of the spectral line.

    $\Delta {\theta }_{\text{hw}}=\frac{\lambda }{Nd\cos \theta }$                            (IV)

Here, $\Delta {\theta }_{\text{hw}}$ is the half width of the spectral line $\lambda$ is the wavelength of light used, $d$ is the separation between rulings of grating and $\theta$ is the angle of diffraction corresponding to particular order maxima.

Conclusion:

Substitute $\text{515}\text{rulings}/\text{mm}$ for $n$ in equation (I) to calculate $d$.

    $\begin{array}{c}d=\frac{1}{\text{515}\text{rulings}/\text{mm}}\\ =1.94\times {10}^{-3}\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\\ =1.94\times {10}^{-6}\text{m}\end{array}$

Substitute $\text{5}\text{.00 cm}$ for $l$ and $1.94\times {10}^{-6}\text{ m}$ for $d$ in equation (III) to calculate $N$.

    $\begin{array}{c}N=\frac{\text{5}\text{.00}\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{ cm}}\right)}{1.94\times {10}^{-6}\text{m}}\\ =25733\text{rulings}\end{array}$

Substitute $25733$ for $N$, $0°$ for $\theta$, $450.0\text{nm}$ for $\lambda$ and $1.94\times {10}^{-6}\text{ m}$ for $d$ in equation (IV) to calculate $\Delta {\theta }_{\text{hw}}$.

    $\begin{array}{c}\Delta {\theta }_{\text{hw}}=\frac{450.0\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{\left(25733\right)\left(1.94\times {10}^{-6}\text{m}\right)\left(\cos 0°\right)}\\ =9\times {10}^{-6}\text{rad}\end{array}$

Therefore, the half width of the spectral line is $9.0\times {10}^{-6}\text{rad}$.

(b)

To determine

The half width of the spectral line.

Answer to Problem 30PQ

The half width of the spectral line is $1.30\times {10}^{-5}\text{rad}$.

Explanation of Solution

Conclusion:

Substitute $25733$ for $N$, $0°$ for $\theta$, $650.0\text{nm}$ for $\lambda$ and $1.94\times {10}^{-6}\text{m}$ for $d$ in equation (IV) to calculate $\Delta {\theta }_{\text{hw}}$.

    $\begin{array}{c}\Delta {\theta }_{\text{hw}}=\frac{650.0\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{ nm}}\right)}{\left(25733\right)\left(1.94\times {10}^{-6}\text{m}\right)\left(\cos 0°\right)}\\ =1.30\times {10}^{-5}\text{rad}\end{array}$

Therefore, the half width of the spectral line is $1.30\times {10}^{-5}\text{rad}$.