Chapter 33, Problem 63P

(a)

To determine

The angles in the first order spectrum can be expected in two lines having wavelengths of $434\text{nm}$ and $410\text{nm}$ .

Explanation of Solution

Given:

The wavelength of first violet line is $434\text{nm}$ .

The wavelength of second violet line is $410\text{nm}$ .

The diffraction grating has $2000$ slits per centimeter.

The order of spectrum is $1$ .

Formula used:

Write the expression for slit separation.

  $d=\frac{1}{N}$   ...... (1)

Here, $d$ is the distance between two slits and $N$ is the number of slits per centimeter.

Write the expression for condition of maximum interference.

  $d\sin \theta =m\lambda$

Here, $d$ is the slit separation, $\theta$ is angle at which first spectrum expected, $m$ is the order of spectrum and $\lambda$ is the wavelength of line.

Rearrange the above equation.

  $\sin \theta =\frac{m\lambda }{d}$

Rearrange the above expression for $\theta$ .

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

Calculation:

Substitute $2000$ per centimeter for $N$ in equation (1).

  $\begin{array}{c}d=\frac{1\text{cm}}{2000}\\ =0.0005\text{cm}\end{array}$

Substitute $1$ for $m$ , $0.0005\text{cm}$ for $d$ and $434\text{nm}$ for $\lambda$ to find first anglein equation (2).

  $\begin{array}{c}{\theta }_1={\sin }^{-1}\left(\frac{\left(1\right)\left(434\text{nm}\right)}{\left(0.0005\text{cm}\right)}\right)\\ ={\sin }^{-1}\left(\frac{\left(434\text{nm}\right)\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}{\left(0.0005\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\right)\\ ={\sin }^{-1}\left(868\times {10}^{-4}\right)\\ =4.979°\end{array}$

Substitute $1$ for $m$ , $\frac{1\text{cm}}{2000}$ for $d$ and $410\text{nm}$ for $\lambda$ to find second anglein equation (2).

  $\begin{array}{c}{\theta }_2={\sin }^{-1}\left(\frac{\left(1\right)\left(410\text{nm}\right)}{\left(0.0005\text{cm}\right)}\right)\\ ={\sin }^{-1}\left(\frac{\left(410\text{nm}\right)\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}{\left(0.0005\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\right)\\ ={\sin }^{-1}\left(820\times {10}^{-4}\right)\\ =4.703°\end{array}$

Conclusion:

Thus, the angles at which first order spectrum is expected are $4.979°$ and $4.703°$ .

(b)

To determine

Angles of first order spectrum can be expected in two lines having wavelengths of $434\text{nm}$ , $410\text{nm}$ for which grating has $15000$ slits per centimeter.

Explanation of Solution

Given:

The wavelength of first violet line is $434\text{nm}$ .

The wavelength of second violet line is $410\text{nm}$ .

The diffraction grating has $2000$ slits per centimeter.

The order of spectrum is $1$ .

Formula used:

Write the expression for slit separation.

  $d=\frac{1}{N}$   ...... (1)

Here, $d$ is the distance between two slits and $N$ is the number of slits per centimeter.

Write the expression for condition of maximum interference.

  $d\sin \theta =m\lambda$

Here, $d$ is the slit separation, $\theta$ is angle at which first spectrum expected, $m$ is the order of spectrum and $\lambda$ is the wavelength of line.

Rearrange the above equation.

  $\sin \theta =\frac{m\lambda }{d}$

Rearrange the above expression for $\theta$ .

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

Calculation:

Substitute $2000$ per centimeter for $N$ in equation (1).

  $\begin{array}{c}d=\frac{1\text{cm}}{2000}\\ =0.0005\text{cm}\end{array}$

Substitute $1$ for $m$ , $0.0005\text{cm}$ for $d$ and $434\text{nm}$ for $\lambda$ to find first angle in equation (2).

  $\begin{array}{c}{\theta }_1={\sin }^{-1}\left(\frac{\left(1\right)\left(434\text{nm}\right)}{\left(0.0005\text{cm}\right)}\right)\\ ={\sin }^{-1}\left(\frac{\left(434\text{nm}\right)\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}{\left(0.0005\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\right)\\ ={\sin }^{-1}\left(6510\times {10}^{-4}\right)\\ =40.617°\end{array}$

Substitute $1$ for $m$ , $0.0005\text{cm}$ for $d$ and $410\text{nm}$ for $\lambda$ to find second anglein equation (2).

  $\begin{array}{c}{\theta }_2={\sin }^{-1}\left(\frac{\left(1\right)\left(410\text{nm}\right)}{\left(0.0005\text{cm}\right)}\right)\\ ={\sin }^{-1}\left(\frac{\left(410\text{nm}\right)\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}{\left(0.0005\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\right)\\ ={\sin }^{-1}\left(820\times {10}^{-4}\right)\\ =37.952°\end{array}$

Conclusion:

Thus, angle at which first order spectrum for wavelength $434\text{nm}$ is ${40.617}^{\circ }$ and for wavelength is ${37.952}^{\circ }$ .