Chapter 38, Problem 38.28P

Three discrete spectral lines occur at angles of 10.1°, 13.7°, and 14.8° in the first-order spectrum of a grating spectrometer. (a) If the grating has 3 660 slits/cm, what are the wavelengths of the light? (b) At what angles are these lines found in the second-order spectrum?

To determine

The wavelengths of the light.

Answer to Problem 38.28P

The wavelengths of light are $487.52\text{nm}$, $658.41\text{nm}$ and $710.14\text{nm}$ at spectral angles of $10.1°$, $13.7°$ and $14.8°$ respectively.

Explanation of Solution

Given info: Angles of spectral lines are $10.1°$, $13.7°$ and $14.8°$, Slits on the grating are $3600\text{slits}/\text{cm}$.

The width of slit can be given as,

$d=\frac{1}{N}$

Here,

$N$ is the number of slits per length.

$d$ is the width of the slit.

Substitute $3600\text{slits}/\text{cm}$ for $N$ in the above equation,

$\begin{array}{c}d=\frac{1}{3600\text{slits}/\text{cm}}\\ =\left(2.78\times {10}^{-4}\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =2.78\times {10}^{-6}\text{m}\end{array}$

The condition for first order diffraction grating can be given as,

$\lambda =d\sin \theta$ (1)

Here,

$\theta$ is the angle of spectral line.

$\lambda$ is the wavelength of light.

For the angle of $10.1°$:

Substitute $2.78\times {10}^{-6}\text{m}$ for $d$, ${\lambda }_1$ for $\lambda$ and $10.1°$ for $\theta$ in the equation (1),

$\begin{array}{c}{\lambda }_1=\left(2.78\times {10}^{-6}\text{m}\right)\left(\sin 10.1°\right)\\ =487.52\times {\text{10}}^{-9}\text{m}\\ =487.52\text{nm}\end{array}$

Thus, the wavelength of the light is $487.52\text{nm}$ at angle of $10.1°$.

For the angle of $13.7°$:

Substitute $2.78\times {10}^{-6}\text{m}$ for $d$, ${\lambda }_2$ for $\lambda$ and $13.7°$ for $\theta$ in the equation (1),

$\begin{array}{c}{\lambda }_2=\left(2.78\times {10}^{-6}\text{m}\right)\left(\sin 13.7°\right)\\ =658.41\times {\text{10}}^{-9}\text{m}\\ =658.41\text{nm}\end{array}$

Thus, the wavelength of the light is $658.41\text{nm}$ at angle of $13.7°$.

For the angle of $14.8°$:

Substitute $2.78\times {10}^{-6}\text{m}$ for $d$, ${\lambda }_3$ for $\lambda$ and $14.8°$ for $\theta$ in the equation (1),

$\begin{array}{c}{\lambda }_3=\left(2.78\times {10}^{-6}\text{m}\right)\left(\sin 14.8°\right)\\ =710.14\times {\text{10}}^{-9}\text{m}\\ =710.14\text{nm}\end{array}$

Thus, the wavelength of the light is $710.14\text{nm}$ at angle of $14.8°$.

Conclusion:

Therefore, the wavelengths of light are $487.52\text{nm}$, $658.41\text{nm}$ and $710.14\text{nm}$ at spectral angles of $10.1°$, $13.7°$ and $14.8°$ respectively.

To determine

The angles for the lines in the second order spectrum.

Answer to Problem 38.28P

The angles for the second order lines are $20.53°$, $28.27°$ and $30.72°$ for the wavelengths of $487.52\text{nm}$, $658.41\text{nm}$ and $710.14\text{nm}$ respectively.

Explanation of Solution

Given info: Angles of spectral lines are $10.1°$, $13.7°$ and $14.8°$, Slits on the grating are $3600\text{slits}/\text{cm}$.

The condition for first order diffraction grating can be given as,

$d\sin \theta =2\lambda$ (1)

$\theta$ is the angle of spectral line.

$\lambda$ is the wavelength of light.

Rearrange the above equation for $\theta$,

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

For the wavelength $487.52\text{nm}$:

Substitute $2.78\times {10}^{-6}\text{m}$ for $d$, $487.52\times {10}^{-9}\text{m}$ for $\lambda$ and ${\theta }_1$ for $\theta$ in the equation (2),

$\begin{array}{c}{\theta }_1={\sin }^{-1}\left(\frac{2\left(487.52\times {10}^{-9}\text{m}\right)}{\left(2.78\times {10}^{-6}\text{m}\right)}\right)\\ =20.53°\end{array}$

Thus, the angle of the spectral line is $20.53°$ for wavelength of $487.52\text{nm}$.

For the wavelength $658.41\text{nm}$:

Substitute $2.78\times {10}^{-6}\text{m}$ for $d$, $658.41\times {10}^{-9}\text{m}$ for $\lambda$ and ${\theta }_2$ for $\theta$ in the equation (2),

$\begin{array}{c}{\theta }_2={\sin }^{-1}\left(\frac{2\left(658.41\times {10}^{-9}\text{m}\right)}{\left(2.78\times {10}^{-6}\text{m}\right)}\right)\\ =28.27°\end{array}$

Thus, the angle of the spectral line is $28.27°$ for wavelength of $658.41\text{nm}$.

For the wavelength $710.14\text{nm}$:

Substitute $2.78\times {10}^{-6}\text{m}$ for $d$, $710.14\times {10}^{-9}\text{m}$ for $\lambda$ and ${\theta }_3$ for $\theta$ in the equation (2),

$\begin{array}{c}{\theta }_3={\sin }^{-1}\left(\frac{2\left(710.14\times {10}^{-9}\text{m}\right)}{\left(2.78\times {10}^{-6}\text{m}\right)}\right)\\ =30.72°\end{array}$

Thus, the angle of the spectral line is $30.72°$ for wavelength of $710.14\text{nm}$.

Conclusion:

Therefore, the angles of lines are $20.53°$, $28.27°$ and $30.72°$ for the wavelengths of $487.52\text{nm}$, $658.41\text{nm}$ and $710.14\text{nm}$ respectively.