Chapter 38, Problem 73AP

(a)

To determine

The angular dispersion of the grating is $\frac{d\theta }{d\lambda }=\frac{m}{d\cos \theta }$.

Answer to Problem 73AP

 The angular dispersion of the grating is $\frac{d\theta }{d\lambda }=\frac{m}{d\cos \theta }$.

Explanation of Solution

Formula to calculate the angles of bright beams diffracted from the grafting is,

    $d\sin \theta =m\lambda$                                                                                                (I)

Here, $d$ is the spacing between adjacent slits, $m$ is order number of intensity maximum, $\lambda$ is wavelength of light, and $\theta$ is the angle by which ray is diffracted.

Differentiate equation (I) with respect to $\theta$.

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

Conclusion:

Therefore, the angular dispersion of the grating is $\frac{d\theta }{d\lambda }=\frac{m}{d\cos \theta }$.

(b)

To determine

The angular separation of the two wavelengths in the second order spectrum.

Answer to Problem 73AP

The angular separation of the two wavelengths in the second order spectrum is $0.109°$.

Explanation of Solution

Number of equally spaced slit to analyze the spectrum of mercury is 8,000. The square grating on each side is $2\text{cm}$.

The wavelengths of the incident lights are $579.065\text{nm}$ and $576.959\text{nm}$.

Write the expression of the average wavelength.

    $\begin{array}{c}{\lambda }_{\text{avg}}=\frac{579.065\text{nm}+576.959\text{nm}}{2}\\ =578.012\text{nm}\\ =578.012\times {10}^{-9}\text{m}\end{array}$

Formula to calculate the spacing between the adjacent slit is,

    $d=\frac{\text{Square}\text{grating}}{\text{Number}\text{of}\text{equally}\text{spaced}\text{slits}}$                                                            (II)

Substitute 0.02 for square grating and 8000 for number of equally spaced slits in equation (II).

    $\begin{array}{c}d=\frac{\text{0}\text{.02}\text{m}}{8000}\\ =2.5\times {10}^{-6}\end{array}$

Substitute $2.5\times {10}^{-6}$ for $d$, 2 for $m$, $578.012$ for $\lambda$ in equation (1) to calculate $\theta$.

    $\begin{array}{c}\frac{0.02}{8000}\sin \theta =2\left(578.012\times {10}^{-9}\text{m}\right)\\ \theta ={\sin }^{-1}\frac{2\times 578\times {10}^{-9}\text{m}}{2.5\times {10}^{-6}\text{m}}\\ =27.54°\end{array}$

Formula to calculate the separation angle between the lines is,

    $\Delta \theta =\frac{m}{d\cos \theta }\cdot \Delta \lambda$                                                                                 (III)

Here, $\Delta \theta$ is angular separation, $m$ is order number of intensity maximum, and

$\Delta \lambda$ is difference of wavelength of lights.

The difference in wavelength of lights is,

    $\begin{array}{c}\Delta \lambda =579.065\text{nm}-576.959\text{nm}\\ =\text{2}\text{.106}\text{nm}\\ =\text{2}\text{.106}\times {\text{10}}^{-9}\text{m}\end{array}$

Substitute 2 for $m$, $27.54$ for $\theta$ and $\text{2}\text{.106}\times {\text{10}}^{-9}$ for $\Delta \lambda$ in equation (3) to calculate $\Delta \theta$.

    $\begin{array}{c}\Delta \theta =\frac{2}{\left(2.5\times {10}^{-6}\text{m}\right)\text{cos}\left(\text{27}\text{.5}°\right)}\left(2.106\times {10}^{-9}\text{m}\right)\\ =\frac{4.212\times {10}^{-9}}{2.217\times {10}^{-6}}\\ =0.00190\text{rad}\left(\frac{360°}{2\pi \text{rad}}\right)\\ =0.109°\end{array}$ \

Conclusion:

Therefore, the angular separation of the two wavelengths in the second order spectrum is $0.109°$.