Chapter 36, Problem 44PQ

To determine

The dispersion of the first three maxima that fall on the screen.

Answer to Problem 44PQ

The dispersion of the first order is $15.4\times {10}^4\text{rad}/\text{m}$, second order is $31\times {10}^4\text{rad}/\text{m}$ and the third order is $46.7\times {10}^4\text{rad}/\text{m}$.

Explanation of Solution

Write the expression for $m^{\text{th}}$ maxima of diffraction grating.

    $\begin{array}{l}d\sin \theta =m\lambda \\ \sin \theta =\frac{m\lambda }{d}\\ \theta ={\sin }^{-1}\left(\frac{m\lambda }{d}\right)\end{array}$                                                                                                         (I)

Here, $d$ is the spacing grating, $\lambda$ is the wavelength and $m$ is the order.

Write the expression for diffraction grating in the first order.

    $D=\frac{m}{d\cos \theta }$                                                                                                            (II)

Here, $D$ is the diffraction grating.

Conclusion:

Calculate grating spacing as follows.

    $\begin{array}{c}d=\frac{1}{1.54\times {10}^3\text{rulings}}\\ =6.49\times {10}^{-4}\text{cm}\times \frac{1\text{m}}{100\text{cm}}\\ =6.49\times {10}^{-6}\text{m}\end{array}$

Substitute $1$ for $m$, $310\text{nm}$ for $\lambda$ and $6.49\times {10}^{-6}\text{m}$ for $d$ in equation (I) to calculate ${\theta }_1$.

    $\begin{array}{c}{\theta }_1={\sin }^{-1}\left(\frac{1\times \left(310\text{nm}\times \frac{1\text{m}}{{10}^9\text{nm}}\right)}{6.49\times {10}^{-6}\text{m}}\right)\\ ={\sin }^{-1}\left(0.0477\right)\\ =2.74°\end{array}$

Substitute $1$ for $m$, $6.49\times {10}^{-6}\text{m}$ for $d$ and $2.74°$ for ${\theta }_1$ in equation (II) to calculate $D_1$.

    $\begin{array}{c}{D}_1=\frac{1}{6.49\times {10}^{-6}\text{m}\times \cos \left(2.74°\right)}\\ =15.4\times {10}^4\text{rad}/\text{m}\end{array}$

Substitute $2$ for $m$, $310\text{nm}$ for $\lambda$ and $6.49\times {10}^{-6}\text{m}$ for $d$ in equation (I) to calculate ${\theta }_2$.

    $\begin{array}{c}{\theta }_2={\sin }^{-1}\left(\frac{2\times \left(310\text{nm}\times \frac{1\text{m}}{{10}^9\text{nm}}\right)}{6.49\times {10}^{-6}\text{m}}\right)\\ ={\sin }^{-1}\left(0.0954\right)\\ =5.47°\end{array}$

Substitute $2$ for $m$, $6.49\times {10}^{-6}\text{m}$ for $d$ and $5.47°$ for ${\theta }_2$ in equation (II) to calculate $D_2$.

    $\begin{array}{c}{D}_2=\frac{2}{6.49\times {10}^{-6}\text{m}\times \cos \left(5.47°\right)}\\ =31\times {10}^4\text{rad}/\text{m}\end{array}$

Substitute $3$ for $m$, $310\text{nm}$ for $\lambda$ and $6.49\times {10}^{-6}\text{m}$ for $d$ in equation (I) to calculate ${\theta }_3$.

    $\begin{array}{c}{\theta }_3={\sin }^{-1}\left(\frac{3\times \left(310\text{nm}\times \frac{1\text{m}}{{10}^9\text{nm}}\right)}{6.49\times {10}^{-6}\text{m}}\right)\\ ={\sin }^{-1}\left(0.1431\right)\\ =8.23°\end{array}$

Substitute $3$ for $m$, $6.49\times {10}^{-6}\text{m}$ for $d$ and $8.23°$ for ${\theta }_3$ in equation (II) to calculate $D_3$.

    $\begin{array}{c}{D}_3=\frac{3}{6.49\times {10}^{-6}\text{m}\times \cos \left(8.23°\right)}\\ =46.7\times {10}^4\text{rad}/\text{m}\end{array}$

Therefore, the dispersion of the first order is $15.4\times {10}^4\text{rad}/\text{m}$, second order is $31\times {10}^4\text{rad}/\text{m}$ and the third order is $46.7\times {10}^4\text{rad}/\text{m}$.