Chapter 24, Problem 39P

Three discrete spectral lines occur at angles of 10.1°, 13.7°, and 14.8°, respectively, in the first-order spectrum of a diffraction 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 spectra?

To determine

The wavelengths of the incident light.

Answer to Problem 39P

The wavelengths of the incident light are $479\text{nm, 647}\text{nm and 698}\text{nm}$ for the angles $10.1°,13.7°\text{and}\text{14}\text{.8}°$ respectively.

Explanation of Solution

Given Info: The slit width is $1\text{cm}/3660$, the angles are $10.1°,13.7°\text{and}\text{14}\text{.8}°$ the order is 1.

Formula to calculate the wavelength is,

$\lambda =\frac{d\sin \theta }{m}$

• $\lambda$ is the wavelength.
• $m$ is the order.
• $d$ is the grid separation.
• $\theta$ is the angle.

Substitute $1\text{cm}/3660$ for $d$, $10.1°$ for $\theta$ and $1$ for $m$ in equation I to find $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(\left(1\text{cm}/3660\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)\sin \left(10.1°\right)}{1}\\ =479\times {10}^{-9}\text{m}\\ \text{=}\left(479\times {10}^{-9}\text{m}\right)\left(\frac{1\text{nm}}{{10}^{-9}\text{m}}\right)\\ =479\text{nm}\end{array}$

Substitute $1\text{m}/3660$ for $d$, $13.7°$ for $\theta$ and $1$ for $m$ in equation I to find $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(\left(1\text{cm}/3660\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)\sin \left(13.7°\right)}{1}\\ =647\times {10}^{-9}\text{m}\\ \text{=}\left(647\times {10}^{-9}\text{m}\right)\left(\frac{1\text{nm}}{{10}^{-9}\text{m}}\right)\\ =647\text{nm}\end{array}$

Substitute $1\text{m}/3660$ for $d$, $14.8°$ for $\theta$ and $1$ for $m$ in equation I to find $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(\left(1\text{cm}/3660\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)\sin \left(14.8°\right)}{1}\\ =698\times {10}^{-9}\text{m}\\ \text{=}\left(698\times {10}^{-9}\text{m}\right)\left(\frac{1\text{nm}}{{10}^{-9}\text{m}}\right)\\ =698\text{nm}\end{array}$

Thus, the wavelengths of the incident light are $479\text{nm, 647}\text{nm and 698}\text{nm}$ for the angles $10.1°,13.7°\text{and}\text{14}\text{.8}°$ respectively.

Conclusion:

The wavelengths of the incident light are $479\text{nm, 647}\text{nm and 698}\text{nm}$ for the angles $10.1°,13.7°\text{and}\text{14}\text{.8}°$ respectively.

To determine

The angles at which the lines found second order spectra.

Answer to Problem 39P

The angles at which the lines found second order spectra are $20.5°,28.3°\text{and 30}\text{.7}°$ for the wavelengths $479\text{nm, 647}\text{nm and 698}\text{nm}$ respectively.

Explanation of Solution

Given Info: The wavelengths of the incident light are $479\text{nm, 647}\text{nm and 698}\text{nm}$ and the slit width is $1\text{cm}/3660$.

Formula to calculate the angle is,

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

• $\lambda$ is the wavelength.
• $m$ is the order.
• $d$ is the grid separation.
• $\theta$ is the angle.

Substitute $1\text{cm}/3660$ for $d$, $479\text{nm}$ for $\lambda$ and $2$ for $m$ in equation II to find $\theta$.

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(2\right)\left(479\text{nm}\right)\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}{\left(\left(1\text{cm}/3660\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}\right)\\ =20.5°\end{array}$

Substitute $1\text{cm}/3660$ for $d$, $647\text{nm}$ for $\lambda$ and $2$ for $m$ in equation II to find $\theta$.

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(2\right)\left(647\text{nm}\right)\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}{\left(\left(1\text{cm}/3660\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}\right)\\ =28.3°\end{array}$

Substitute $1\text{cm}/3660$ for $d$, $698\text{nm}$ for $\lambda$ and $2$ for $m$ in equation II to find $\theta$.

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(2\right)\left(698\text{nm}\right)\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}{\left(\left(1\text{cm}/3660\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}\right)\\ =30.7°\end{array}$

Thus, the angles at which the lines found second order spectra are $20.5°,28.3°\text{and 30}\text{.7}°$ for the wavelengths $479\text{nm, 647}\text{nm and 698}\text{nm}$ respectively.

Conclusion:

The angles at which the lines found second order spectra are $20.5°,28.3°\text{and 30}\text{.7}°$ for the wavelengths $479\text{nm, 647}\text{nm and 698}\text{nm}$ respectively.