Chapter 35, Problem 87PQ

(a)

To determine

The distance between the central maxima and adjacent maxima in the interference pattern in the screen.

Answer to Problem 87PQ

The distance between the central maxima and an adjacent maxima in the interference pattern in the screen is $6.03\times {10}^{-3}\text{m}$.

Explanation of Solution

Write the relation of the path difference for bright fringe in young’s double slit experiment.

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

Here, $\lambda$ is wavelength, the distance between the two consecutive slits is $d$ and $n$ is integer number.

Write the equation of the angle made by the particular fringe from the central maxima.

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

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

For small angle $\theta$, consider $\sin \theta =\tan \theta$.

Substitute $\tan \theta$ for $\sin \theta$ in the equation (I).

    $\begin{array}{c}n\lambda =d\tan {\theta }_{\text{n}}\\ \tan {\theta }_{\text{n}}=\frac{n\lambda }{d}\end{array}$

Now, substitute $\frac{n\lambda }{d}$ for $\tan {\theta }_{\text{n}}$ in the equation (II).

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

Substitute, $n=1$ in the above equation to get the distance between the central maxima and the next consecutive maxima.

    $\Delta y=\frac{\lambda x}{d}$

Conclusion:

Substitute, $485\text{nm}$ for $\lambda$, $2.3\text{m}$ for $x$ and $0.185\text{mm}$ for $d$ in the above equation to find $\Delta y$.

    $\begin{array}{c}\Delta y=\frac{\left(485\text{nm}\right)\left(2.3\text{m}\right)}{\left(0.185\text{mm}\right)}\\ =\frac{\left(485\times {10}^{-9}\text{m}\right)\left(2.3\text{m}\right)}{\left(0.185\times {10}^{-6}\text{m}\right)}\\ =6.03\times {10}^{-3}\text{m}\end{array}$

Therefore, the distance between the central maxima and an adjacent maxima in the interference pattern in the screen is $6.03\times {10}^{-3}\text{m}$.

(b)

To determine

The distance between the first and the third dark band of the interference pattern.

Answer to Problem 87PQ

The distance between the first and the third dark band of the interference pattern is $1.2\times {10}^{-2}\text{m}$.

Explanation of Solution

Write the relation of the path difference for dark fringe in young’s double slit experiment.

    $\left(n+\frac{1}{2}\right)\lambda =d\sin {\theta }_{\text{n}}$

Here, $\lambda$ is wavelength, the distance between the two consecutive slits is $d$. $n$ is integer number and ${\theta }_{\text{n}}$ is the angle between the dark band and the central fringe.

Substitute, $\tan \theta$ for $\sin \theta$ in the above equation for small angle $\theta$.

    $\begin{array}{l}\left(n+\frac{1}{2}\right)\lambda =d\tan {\theta }_{\text{n}}\\ \tan {\theta }_{\text{n}}=\frac{\left(n+\frac{1}{2}\right)\lambda }{d}\end{array}$

Consider the equation (2).

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

Substitute $\frac{\left(n+\frac{1}{2}\right)\lambda }{d}$ for $\tan {\theta }_{\text{n}}$ in the above equation.

    $\begin{array}{l}\frac{\left(n+\frac{1}{2}\right)\lambda }{d}=\frac{y_{\text{n}}}{x}\\ y_{\text{n}}=\frac{\left(n+\frac{1}{2}\right)\lambda x}{d}\end{array}$

Calculate the distance between the first and the third fringe.

    $\begin{array}{c}{y}_3-y_1=\frac{\left(3+\frac{1}{2}\right)\lambda x}{d}-\frac{\left(1+\frac{1}{2}\right)\lambda x}{d}\\ =\frac{\left(\frac{7}{2}\right)\lambda x}{d}-\frac{\left(\frac{3}{2}\right)\lambda x}{d}\\ =\frac{\left(7-3\right)\lambda x}{2d}\\ =\frac{4\lambda x}{2d}\end{array}$

Simplify the above equation.

    $y_3-y_1=\frac{2\lambda x}{d}$

Conclusion:

Substitute $485\text{nm}$ for $\lambda$, $2.3\text{m}$ for $x$ and $0.185\text{mm}$ for $d$ in the above equation to find $y_3-y_1$.

    $\begin{array}{c}{y}_3-y_1=\frac{2\times \left(485\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\left(2.3\text{m}\right)}{\left(0.185\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}\\ =\frac{2\times \left(485\times {10}^{-9}\text{m}\right)\left(2.3\text{m}\right)}{\left(0.185\times {10}^{-6}\text{m}\right)}\\ =1.2\times {10}^{-2}\text{m}\end{array}$

Therefore, the distance between the first and the third dark band of the interference pattern is $1.2\times {10}^{-2}\text{m}$.