Chapter 25, Problem 86P

To determine

The sketch of the grating spectra with its labels.

Answer to Problem 86P

The diffraction spectra is given in figure 1.

Explanation of Solution

The wavelength of the red light is $690\text{nm}$, wavelength of the blue light is $460\text{nm}$, the grating-screen distance is $2.0\text{m}$ and the slit density is $10000.0\text{slits/cm}$.

Write the expression for slit separation

    $d=\frac{1}{d_{\text{n}}}$                                                                                                    (I)

Here, $d_{\text{n}}$ is the slit density and $d$ is the slit separation.

Substitute $10000.0\text{slits/cm}$ for $d_{\text{n}}$ in (I) to find $d$

  $\begin{array}{c}d=\frac{1}{10000.0\text{slits/cm}}\\ =\frac{1}{10000.0\text{slits/cm}}\cdot \frac{{10}^{-2}\text{m }}{1\text{cm}}\\ =1.00000\times {10}^{-6}\text{m}\end{array}$

Write the expression for diffraction maxima

  $d\sin \theta =m\lambda$

Here, $\lambda$ is the wavelength, $m$ is the order of the fringe and $\theta$ is the angular spacing for fringes.

Rearrange for $\theta$

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

Write the expression for $\theta$

  $\tan \theta =\frac{\Delta x}{D}$

Here, $\Delta x$ is the distance from the central fringe and $D$ is the grating-screen distance.

Rearrange for $\Delta x$

  $\Delta x=D\tan \theta$                                                                                          (III)

Substitute (II) in (III)

  $\Delta x=D\tan \left({\sin }^{-1}\left(\frac{m\lambda }{d}\right)\right)$                                                                       (IV)

Substitute $0$ for $m$, $2.0\text{m}$ for $D$, $690\text{nm}$ for $\lambda$ and $1.00000\times {10}^{-6}\text{m}$ for $d$ in (IV) to find $\Delta x$ for central red fringe.

  $\begin{array}{c}\Delta x=690\text{nm}\tan \left({\sin }^{-1}\frac{0\times 690\text{nm}}{1.00000\times {10}^{-6}\text{m}}\right)\\ =0\end{array}$

Substitute $\pm 1$ for $m$, $2.0\text{m}$ for $D$, $690\text{nm}$ for $\lambda$ and $1.00000\times {10}^{-6}\text{m}$ for $d$ in (IV) to find $\Delta x$ for first order red light.

  $\begin{array}{c}\Delta x=690\text{nm}\tan \left({\sin }^{-1}\frac{\left(\pm 1\right)690\text{nm}}{1.00000\times {10}^{-6}\text{m}}\right)\\ =690\text{nm}\tan \left({\sin }^{-1}\frac{\left(\pm 1\right)690\times {10}^{-9}\text{m}}{1.00000\times {10}^{-6}\text{m}}\right)\\ =\pm 1.9\text{nm}\end{array}$

Substitute $0$ for $m$, $2.0\text{m}$ for $D$, $460\text{nm}$ for $\lambda$ and $1.00000\times {10}^{-6}\text{m}$ for $d$ in (IV) to find $\Delta x$ for central blue fringe.

  $\begin{array}{c}\Delta x=460\text{nm}\tan \left({\sin }^{-1}\frac{0\times 690\text{nm}}{1.00000\times {10}^{-6}\text{m}}\right)\\ =0\end{array}$

Substitute $\pm 1$ for $m$, $2.0\text{m}$ for $D$, $460\text{nm}$ for $\lambda$ and $1.00000\times {10}^{-6}\text{m}$ for $d$ in (IV) to find $\Delta x$ for first order blue light.

  $\begin{array}{c}\Delta x=460\text{nm}\tan \left({\sin }^{-1}\frac{\left(\pm 1\right)690\text{nm}}{1.00000\times {10}^{-6}\text{m}}\right)\\ =460\text{nm}\tan \left({\sin }^{-1}\frac{\left(\pm 1\right)690\times {10}^{-9}\text{m}}{1.00000\times {10}^{-6}\text{m}}\right)\\ =\pm 1.0\text{m}\end{array}$

Substitute $\pm 2$ for $m$, $2.0\text{m}$ for $D$, $460\text{nm}$ for $\lambda$ and $1.00000\times {10}^{-6}\text{m}$ for $d$ in (IV) to find $\Delta x$ for second order blue light.

  $\begin{array}{c}\Delta x=460\text{nm}\tan \left({\sin }^{-1}\frac{\left(\pm 2\right)690\text{nm}}{1.00000\times {10}^{-6}\text{m}}\right)\\ =460\text{nm}\tan \left({\sin }^{-1}\frac{\left(\pm 2\right)690\times {10}^{-9}\text{m}}{1.00000\times {10}^{-6}\text{m}}\right)\\ =\pm 4.7\text{m}\end{array}$

The blue light and red light overlap for the central fringe and hence it appears purple.

The grating spectra drawn using the above information is given below: