Chapter 27, Problem 42P

To determine

The angular separation between spectral lines in the visible spectrum.

Answer to Problem 42P

The angular separation between spectral lines of first, second and third order is $5.91°,13.2°\text{and}\text{26}\text{.5}°$.

Explanation of Solution

Draw the diagram of spectral lines as per the given data.

Write the equation to find the grating spacing.

    $d=\frac{1\text{cm}}{n_{\text{grooves}}}$        (I)

Here, $d$ is the grating spacing and $n_{\text{grooves}}$ is the number of grooves per cm.

Write the equation to find the diffraction angle.

    $\sin \theta =\frac{m\lambda }{d}$        (II)

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

Write the equation to find the angular separation between red and blue lines.

    $\Delta \theta ={\theta }_{\text{red}}-{\theta }_{\text{blue}}$        (III)

Here, $\Delta \theta$ is the angular separation, ${\theta }_{\text{red}}$ is the diffraction angle of red wavelength and ${\theta }_{\text{blue}}$ is the diffraction angle of blue wavelength.

Conclusion:

Substitute $4500\text{grooves}/\text{cm}$ for $n_{\text{grooves}}$ in equation (I).

    $\begin{array}{c}d=\frac{1\text{cm}}{4500\text{grooves}/\text{cm}}\\ =2.22\times {10}^{-6}\text{m}\end{array}$

Substitute $1$ for $m$, $656\text{nm}$ for $\lambda$ and $2.22\times {10}^{-6}\text{m}$ for $d$ in equation (II) to find $\theta$ for red light.

  $\begin{array}{c}\sin {\theta }_{\text{red}}=\frac{\left(1\right)\left(656\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}{2.22\times {10}^{-6}\text{m}}\\ =0.295\end{array}$

Rewrite the expression for ${\theta }_{\text{red}}$.

  $\begin{array}{c}{\theta }_{\text{red}}={\sin }^{-1}\left(0.295\right)\\ =17.17°\end{array}$

Substitute $1$ for $m$, $434\text{nm}$ for $\lambda$ and $2.22\times {10}^{-6}\text{m}$ for $d$ in equation (II) to find $\theta$ for blue light.

  $\begin{array}{c}\sin {\theta }_{\text{blue}}=\frac{\left(1\right)\left(434\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}{2.22\times {10}^{-6}\text{m}}\\ =0.195\end{array}$

Rewrite the expression for ${\theta }_{\text{blue}}$.

  $\begin{array}{c}{\theta }_{\text{blue}}={\sin }^{-1}\left(0.195\right)\\ =11.26°\end{array}$

Substitute $17.17°$ for ${\theta }_{\text{red}}$ and $11.26°$ for ${\theta }_{\text{blue}}$ in equation (III) to find $\Delta \theta$ for first order.

    $\begin{array}{c}\Delta \theta =17.17°-11.26°\\ =5.91°\end{array}$

Substitute $2$ for $m$, $656\text{nm}$ for $\lambda$ and $2.22\times {10}^{-6}\text{m}$ for $d$ in equation (II) to find $\theta$ for red light.

  $\begin{array}{c}\sin {\theta }_{\text{red}}=\frac{\left(2\right)\left(656\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}{2.22\times {10}^{-6}\text{m}}\\ =0.59\end{array}$

Rewrite the expression for ${\theta }_{\text{red}}$.

  $\begin{array}{c}{\theta }_{\text{red}}={\sin }^{-1}\left(0.59\right)\\ =36.16°\end{array}$

Substitute $2$ for $m$, $434\text{nm}$ for $\lambda$ and $2.22\times {10}^{-6}\text{m}$ for $d$ in equation (II) to find $\theta$ for blue light.

  $\begin{array}{c}\sin {\theta }_{\text{blue}}=\frac{\left(2\right)\left(434\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}{2.22\times {10}^{-6}\text{m}}\\ =0.39\end{array}$

Rewrite the expression for ${\theta }_{\text{blue}}$.

  $\begin{array}{c}{\theta }_{\text{blue}}={\sin }^{-1}\left(0.39\right)\\ =22.95°\end{array}$

Substitute $36.16°$ for ${\theta }_{\text{red}}$ and $22.95°$ for ${\theta }_{\text{blue}}$ in equation (III) to find $\Delta \theta$ for second order.

    $\begin{array}{c}\Delta \theta =36.16°-22.95°\\ =13.2°\end{array}$

Substitute $3$ for $m$, $656\text{nm}$ for $\lambda$ and $2.22\times {10}^{-6}\text{m}$ for $d$ in equation (II) to find $\theta$ for red light.

  $\begin{array}{c}\sin {\theta }_{\text{red}}=\frac{\left(3\right)\left(656\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}{2.22\times {10}^{-6}\text{m}}\\ =0.885\end{array}$

Rewrite the expression for ${\theta }_{\text{red}}$.

  $\begin{array}{c}{\theta }_{\text{red}}={\sin }^{-1}\left(0.885\right)\\ =62.25°\end{array}$

Substitute $3$ for $m$, $434\text{nm}$ for $\lambda$ and $2.22\times {10}^{-6}\text{m}$ for $d$ in equation (II) to find $\theta$ for blue light.

  $\begin{array}{c}\sin {\theta }_{\text{blue}}=\frac{\left(3\right)\left(434\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}{2.22\times {10}^{-6}\text{m}}\\ =0.585\end{array}$

Rewrite the expression for ${\theta }_{\text{blue}}$.

  $\begin{array}{c}{\theta }_{\text{blue}}={\sin }^{-1}\left(0.585\right)\\ =35.8°\end{array}$

Substitute $62.25°$ for ${\theta }_{\text{red}}$ and $35.8°$ for ${\theta }_{\text{blue}}$ in equation (III) to find $\Delta \theta$ for third order.

    $\begin{array}{c}\Delta \theta =62.25°-35.8°\\ =26.5°\end{array}$

Thus, the angular separation between spectral lines of first, second and third order is $5.91°,13.2°\text{and}\text{26}\text{.5}°$.