Chapter 36, Problem 62PQ

White light is incident on a diffraction grating that has 2.85 × 10³ rulings per centimeter, producing first-order maxima at 6.55°, 8.75°, and 11.2°.

a. What are the wavelengths responsible for these three first-order maxima?

b. At which angles do these wavelengths produce second-order maxima?

To determine

The wavelength responsible for the three first-order maxima.

Answer to Problem 62PQ

The wavelength for $1^{\text{st}}$ first-order maxima is $4\times {10}^{-7}\text{m}$, the wavelength for $2^{\text{nd}}$ first-order maxima is $5.34\times {10}^{-7}\text{m}$ and the wavelength for $3^{\text{rd}}$ first-order maxima is $6.82\times {10}^{-7}\text{m}$.

Explanation of Solution

A set of equally spaced parallel lines, producing a spectrum by diffraction and interference of light that are used to separate an incident wave front into its component wavelengths is known as diffraction grating.

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

    $\begin{array}{c}d\sin \theta =m\lambda \\ \lambda =\frac{d\sin \theta }{m}\end{array}$                                                                                           (I)

Here, $d$ is the separation between the rulings of the grating, $\lambda$ is the wavelength of the light used and $m$ is an integer which varies from $0,\pm 1,\pm 2,\pm \mathrm{3....}$.

Write the formula for grating spacing.

    $d=\frac{1}{N}$                                                                                                        (II)

Here, $N$ is the number of rulings.

Conclusion:

Substitute $2.85\times {10}^3\text{rulings}$ for $N$ in equation (II) to find $d$.

    $\begin{array}{c}d=\frac{1}{2.85\times {10}^3\text{rulings/cm}}\\ =3.51\times {10}^{-4}c\text{m}\left(\frac{{10}^{-2}\text{m}}{1\text{mm}}\right)\\ =3.51\times {10}^{-6}\text{m}\end{array}$

Calculate the wavelength for $1^{\text{st}}$ first-order maxima.

Substitute $1$ for $m$, $6.55°$ for $\theta$ and $3.51\times {10}^{-6}\text{m}$ for $d$ in equation (I) to find $\lambda$.

    $\begin{array}{c}\lambda =\frac{\left(3.51\times {10}^{-6}\text{m}\right)\left(\sin 6.55°\right)}{1}\\ =4\times {10}^{-7}\text{m}\end{array}$

Calculate the wavelength for $2^{\text{nd}}$ first-order maxima.

Substitute $1$ for $m$, $9.75°$ for $\theta$ and $3.51\times {10}^{-6}\text{m}$ for $d$ in equation (I) to find $\lambda$.

    $\begin{array}{c}\lambda =\frac{\left(3.51\times {10}^{-6}\text{m}\right)\left(\sin 8.75°\right)}{1}\\ =5.34\times {10}^{-7}\text{m}\end{array}$

Calculate the wavelength for $3^{\text{rd}}$ first-order maxima.

Substitute $1$ for $m$, $11.2°$ for $\theta$ and $3.51\times {10}^{-6}\text{m}$ for $d$ in equation (I) to find $\lambda$.

    $\begin{array}{c}\lambda =\frac{\left(3.51\times {10}^{-6}\text{m}\right)\left(\sin 11.2°\right)}{1}\\ =6.82\times {10}^{-7}\text{m}\end{array}$

Therefore, the wavelength for $1^{\text{st}}$ first-order maxima is $4\times {10}^{-7}\text{m}$, the wavelength for $2^{\text{nd}}$ first-order maxima is $5.34\times {10}^{-7}\text{m}$ and wavelength for $3^{\text{rd}}$ first-order maxima is $6.82\times {10}^{-7}\text{m}$.

To determine

The angles at which the three wavelengths produce second-order maxima.

Answer to Problem 62PQ

The angle at which the second-order maxima occurs due to $1^{\text{st}}$ wavelength is $13.17°$, the angle at which the second-order maxima occurs due to $2^{\text{nd}}$ wavelength is $17.71°$ and the angle at which the second-order maxima occurs due to $3^{\text{rd}}$ wavelength is $22.87°$.

Explanation of Solution

Rearrange the equation (I) to find $\theta$.

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

Conclusion:

Calculate the angle at which the second-order maxima occurs due to $1^{\text{st}}$ wavelength.

Substitute $4\times {10}^{-7}\text{m}$ for $\lambda$, $2$ for $m$ and $3.51\times {10}^{-6}\text{m}$ for $d$ in equation (III) to find $\theta$.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{2\left(4\times {10}^{-7}\text{m}\right)}{3.51\times {10}^{-6}\text{m}}\right)\\ =13.17°\end{array}$

Calculate the angle at which the second-order maxima occurs due to $2^{\text{nd}}$ wavelength.

Substitute $5.34\times {10}^{-7}\text{m}$ for $\lambda$, $2$ for $m$ and $3.51\times {10}^{-6}\text{m}$ for $d$ in equation (III) to find $\theta$.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{2\left(5.34\times {10}^{-7}\text{m}\right)}{3.51\times {10}^{-6}\text{m}}\right)\\ =17.71°\end{array}$

Calculate the angle at which the second-order maxima occurs due to $3^{\text{rd}}$ wavelength.

Substitute $6.82\times {10}^{-7}\text{m}$ for $\lambda$, $2$ for $m$ and $3.51\times {10}^{-6}\text{m}$ for $d$ in equation (III) to find $\theta$.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{2\left(6.82\times {10}^{-7}\text{m}\right)}{3.51\times {10}^{-6}\text{m}}\right)\\ =22.87°\end{array}$

Therefore, the angle at which the second-order maxima occurs due to $1^{\text{st}}$ wavelength is $13.17°$, the angle at which the second-order maxima occurs due to $2^{\text{nd}}$ wavelength is $17.71°$ and the angle at which the second-order maxima occurs due to $3^{\text{rd}}$ wavelength is $22.87°$.