Chapter 35, Problem 76PQ

(a)

To determine

The angles corresponding to the locations of the first three orders of fringes away from the central bright fringe.

Answer to Problem 76PQ

The angle made by the first bright fringe is $0.090°$. The angle made by the second bright fringe is $0.180°$. The angle made by the third bright fringe is $0.270°$.

Explanation of Solution

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

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

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.

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

Write the expression for the angle made by the first bright fringe from the above equation.

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

Write the expression for the angle made by the second bright fringe from the equation (II).

    ${\theta }_2={\sin }^{-1}\left(\frac{2\lambda }{d}\right)$                                                                         (IV)

Write the expression for the angle made by the third bright fringe from the equation (II).

    ${\theta }_3={\sin }^{-1}\left(\frac{3\lambda }{d}\right)$                                                                        (V)

Conclusion:

Substitute $0.350\text{mm}$ for $d$ and $550\text{nm}$ for $\lambda$ in the equation (III) to find ${\theta }_1$.

  $\begin{array}{c}{\theta }_1={\sin }^{-1}\left[\frac{1\times 550\text{nm}\times \left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{0.350\text{mm}\times \left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}\right]\\ =0.090°\end{array}$

Substitute $0.350\text{mm}$ for $d$ and $550\text{nm}$ for $\lambda$ in the equation (IV) to find ${\theta }_2$.

  $\begin{array}{c}{\theta }_2={\sin }^{-1}\left[\frac{2\times 550\text{nm}\times \left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{0.350\text{mm}\times \left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}\right]\\ =0.180°\end{array}$

Substitute $0.350\text{mm}$ for $d$ and $550\text{nm}$ for $\lambda$ in the equation-(V) to find ${\theta }_3$.

  $\begin{array}{c}{\theta }_3={\sin }^{-1}\left[\frac{3\times 550\text{nm}\times \left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{0.350\text{mm}\times \left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}\right]\\ =0.270°\end{array}$

Therefore, the angle made by the first bright fringe is $0.090°$. The angle made by the second bright fringe is $0.180°$. The angle made by the third bright fringe is $0.270°$.

(b)

To determine

The distance between the first-order and second order bright fringe.

Answer to Problem 76PQ

The distance between the first-order and second order bright fringe is $3.14\times {10}^{-3}\text{m}$.

Explanation of Solution

Write the expression for the angle made by a particular interference fringe on the screen from the central maximum.

    $\tan {\theta }_{\text{n}}=\frac{y_{\text{n}}}{x}$

Here, $y_{\text{n}}$ is the distance of the particular fringe from the central maximum on the screen and $x$ is the distance between slit and the screen.

Here, for the small angle approximation.

    $\tan \theta \approx \sin \theta$

Substitute $\frac{n\lambda }{d}$ for $\tan \theta$ in the above equation to find the expression for $y_{\text{n}}$.

    $\begin{array}{c}\frac{n\lambda }{d}=\frac{y_{\text{n}}}{x}\\ y_{\text{n}}=\frac{n\lambda x}{d}\end{array}$                                                                                (VI)

Write the expression for the distance between two successive maximum fringes on the screen from the above equation.

    $\begin{array}{c}\Delta y=y_{\text{n+1}}-y_{\text{n}}\\ =\frac{\left(n+1\right)\lambda x}{d}-\frac{n\lambda x}{d}\\ =\frac{\lambda x}{d}\end{array}$                                                             (VII)

Conclusion:

Substitute $0.350\text{mm}$ for $d$, $2.00\text{m}$ for $x$ and $550\text{nm}$ for $\lambda$ in the equation (IV) to find $\Delta y$.

  $\begin{array}{c}\Delta y=\frac{550\text{nm}\times \left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\times 2.00\text{m}}{0.350\text{mm}\times \left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}\\ =3.14\times {10}^{-3}\text{m}\end{array}$

Therefore, the distance between the first-order and second order bright fringe is $3.14\times {10}^{-3}\text{m}$.