Chapter 35, Problem 79PQ

(a)

To determine

The angular separation between the central maximum and an adjacent maximum.

Answer to Problem 79PQ

The angular separation between the central maximum and an adjacent maximum is $29.2°$.

Explanation of Solution

Write the expression for the speed of the wave.

    $v=f\lambda$                                                                                                                 (I)

Here, $v$ is the speed of the wave, $\lambda$ is the wavelength of the wave and $f$ is the frequency of the wave.

Rearrange the above equation.

  $\lambda =\frac{v}{f}$                                                                                                                 (II)

Write the expression for the path difference for bright fringes in Young’s double slit experiment.

    $d\sin {\theta }_{\text{n}}=n\lambda$                                                                                                     (III)

Here, $\lambda$ is the wavelength of light, $d$ is the separation between two slit, $\theta$ is the angle of diffraction and $n$ is an integer.

Rearrange the above equation.

  $\begin{array}{c}d\sin {\theta }_{\text{n}}=n\lambda \\ \sin {\theta }_{\text{n}}=\frac{n\lambda }{d}\\ {\theta }_{\text{n}}={\sin }^{-1}\left(\frac{n\lambda }{d}\right)\end{array}$                                                                                        (IV)

Write the expression for the angle made by the central maximum $\left(n=0\right)$ from the above equation.

    $\begin{array}{l}{\theta }_0={\sin }^{-1}\left(\frac{0\lambda }{d}\right)\\ {\theta }_0=0°\end{array}$

Write the expression for the angle made by the adjacent maximum next to the central maximum $\left(n=1\right)$ from the equation-(IV).

    ${\theta }_1={\sin }^{-1}\left(\frac{\lambda }{d}\right)$

Write the expression for the angular separation between the central maximum and an adjacent maximum.

    $\Delta \theta ={\theta }_1-{\theta }_0$

Substitute ${\sin }^{-1}\left(\frac{\lambda }{d}\right)$ for ${\theta }_1$ and $0°$ for ${\theta }_0$ in the above equation.

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

Conclusion:

Substitute $343\text{m}/\text{s}$ for $v$ and $1850\text{Hz}$ for $f$ in the equation (II) to find $\lambda$.

  $\begin{array}{c}\lambda =\frac{343\text{m}/\text{s}}{1850\text{Hz}}\\ =0.1854\text{m}\end{array}$

Substitute $38.0\text{cm}$ for $d$ and $0.1854\text{m}$ for $\lambda$ in the equation (V) to find $\Delta \theta$.

  $\begin{array}{c}\Delta \theta ={\sin }^{-1}\left[\frac{0.1854\text{m}}{38.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\right]\\ =29.2°\end{array}$

Therefore, the angular separation between the central maximum and an adjacent maximum is $29.2°$.

(b)

To determine

The separation between the slits for the same angular separation between the central maximum and an adjacent maximum.

Answer to Problem 79PQ

The separation between the slits is $5.43\times {10}^{-2}\text{m}$.

Explanation of Solution

Write the expression for the separation between the slits from equation-(V).

    $\begin{array}{c}\Delta \theta ={\sin }^{-1}\left(\frac{\lambda }{d}\right)\\ \sin \Delta \theta =\frac{\lambda }{d}\\ d=\frac{\lambda }{\sin \Delta \theta }\end{array}$                                                                                            (VI)

Conclusion:

Substitute $2.65\text{cm}$ for $\lambda$ and $29.2°$ for $\Delta \theta$ in the above equation to find $d$.

    $\begin{array}{c}d=\frac{2.65\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}{\sin {29.2}^{\circ }}\\ =5.43\times {10}^{-2}\text{m}\end{array}$

Therefore, the separation between the slits is $5.43\times {10}^{-2}\text{m}$.