Chapter 25, Problem 43P

(a)

To determine

The number of wavelengths present in the spectrum of the light source.

Answer to Problem 43P

There are two spectral lines with wavelengths $449.2\text{nm}$ and $665.1\text{nm}$.

Explanation of Solution

Write the expression to calculate the slit width.

  $d=\frac{1}{n}$

Here, d is the slit width and n is the number of slits.

Substitute $5000.0{\text{cm}}^{-1}$ for n in the above equation to calculate d.

  $\begin{array}{c}d=\frac{1}{5000.0{\text{cm}}^{-1}}\\ =2.00\times {10}^{-4}\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =2.00\times {10}^{-6}\text{m}\end{array}$

Write the expression to calculate the first order wavelength for the smallest angle.

  $\lambda =\frac{d\sin \theta }{m}$

Here, $\lambda$ is the wavelength.

Substitute $2.00\times {10}^{-6}\text{m}$ for d, $12.98°$ for $\theta$ and 1 for m in the above equation to calculate $\lambda$.

  $\begin{array}{c}\lambda =\frac{\left(2.00\times {10}^{-6}\text{m}\right)\sin 12.98°}{1}\\ =0.4492\times {10}^{-6}\text{m}\\ =\text{449}\text{.2}\times {10}^{-9}\text{m}\left(\frac{1\text{nm}}{{10}^{-9}\text{m}}\right)\\ =449.2\text{nm}\end{array}$

Write the expression for the angle of second order lines for the above wavelength.

  $\theta ={\sin }^{-1}\left(\frac{2\lambda }{d}\right)$

Substitute $449.2\text{nm}$ for $\lambda$ and $2.00\times {10}^{-6}\text{m}$ for d in the above equation to calculate $\theta$.

  $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{2\left(449.2\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}{2.00\times {10}^{-6}\text{m}}\right)\\ =26.69°\end{array}$

Write the expression for the angle of third order lines for the above wavelength.

  $\theta ={\sin }^{-1}\left(\frac{3\lambda }{d}\right)$

Substitute $449.2\text{nm}$ for $\lambda$ and $2.00\times {10}^{-6}\text{m}$ for d in the above equation to calculate $\theta$.

  $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{3\left(449.2\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}{2.00\times {10}^{-6}\text{m}}\right)\\ =42.36°\end{array}$

Write the expression for the angle of fourth order lines for the above wavelength.

  $\theta ={\sin }^{-1}\left(\frac{4\lambda }{d}\right)$

Substitute $449.2\text{nm}$ for $\lambda$ and $2.00\times {10}^{-6}\text{m}$ for d in the above equation to calculate $\theta$.

  $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{4\left(449.2\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}{2.00\times {10}^{-6}\text{m}}\right)\\ =63.95°\end{array}$

Thus, these angles belong to the second order wavelengths and these angles could be omitted.

Write the expression to calculate the first order wavelength for the angle $19.0°$.

  $\lambda =\frac{d\sin \theta }{m}$

Substitute $2.00\times {10}^{-6}\text{m}$ for d, $19.0°$ for $\theta$ and 1 for m in the above equation to calculate $\lambda$.

  $\begin{array}{c}\lambda =\frac{\left(2.00\times {10}^{-6}\text{m}\right)\sin 19.0°}{1}\\ =0.6510\times {10}^{-6}\text{m}\\ =\text{665}\text{.1}\times {10}^{-9}\text{m}\left(\frac{1\text{nm}}{{10}^{-9}\text{m}}\right)\\ =665.1\text{nm}\end{array}$

Write the expression for the angle of second order lines for the above wavelength.

  $\theta ={\sin }^{-1}\left(\frac{2\lambda }{d}\right)$

Substitute $665.1\text{nm}$ for $\lambda$ and $2.00\times {10}^{-6}\text{m}$ for d in the above equation to calculate $\theta$.

  $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{2\left(665.1\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}{2.00\times {10}^{-6}\text{m}}\right)\\ =40.6°\end{array}$

Write the expression for the angle of third order lines for the above wavelength.

  $\theta ={\sin }^{-1}\left(\frac{3\lambda }{d}\right)$

Substitute $665.1\text{nm}$ for $\lambda$ and $2.00\times {10}^{-6}\text{m}$ for d in the above equation to calculate $\theta$.

  $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{3\left(665.1\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}{2.00\times {10}^{-6}\text{m}}\right)\\ =77.6°\end{array}$

Thus, the angle corresponding the wavelength could be omitted.

Conclusion:

Therefore, there are two spectral lines with wavelengths $449.2\text{nm}$ and $665.1\text{nm}$.

(b)

To determine

The number of spectral lines on one side of the central maximum.

Answer to Problem 43P

The number of spectral lines is $19$.

Explanation of Solution

Write the expression to calculate the slit width.

  $d=\frac{1}{n}$

Here, d is the slit width and n is the number of slits.

Substitute $2000.0{\text{cm}}^{-1}$ for n in the above equation to calculate d.

  $\begin{array}{c}d=\frac{1}{2000.0{\text{cm}}^{-1}}\\ =5.00\times {10}^{-4}\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =5.00\times {10}^{-6}\text{m}\end{array}$

Write the expression to calculate order corresponding to the wavelength $449.2\text{nm}$.

  $m=\frac{d\sin \theta }{\lambda }$

Substitute $90.0°$ for $\theta$, $5.00\times {10}^{-6}\text{m}$ for d and $449.2\text{nm}$ for $\lambda$ in the above equation to calculate m.

  $\begin{array}{c}m=\frac{\left(5.00\times {10}^{-6}\text{m}\right)\sin 90.0°}{449.2\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\\ =11.1\\ \approx 11\end{array}$

Write the expression to calculate order corresponding to the wavelength $665.1\text{nm}$.

  $m=\frac{d\sin \theta }{\lambda }$

Substitute $90.0°$ for $\theta$, $5.00\times {10}^{-6}\text{m}$ for d and $665.1\text{nm}$ for $\lambda$ in the above equation to calculate m.

  $\begin{array}{c}m=\frac{\left(5.00\times {10}^{-6}\text{m}\right)\sin 90.0°}{665.1\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\\ =7.5\\ \approx 8\end{array}$

Thus, the total number of spectral lines is $11+8=19$.

Conclusion:

Therefore, the number of spectral lines is $19$.