Chapter 35, Problem 82PQ

To determine

The shortest distance from the central maximum for which the bright fringes for the two wavelengths coincide.

Answer to Problem 82PQ

The shortest distance from the central maximum for which the bright fringes for the two wavelengths coincide is $\text{0}\text{.111}\text{m}$.

Explanation of Solution

Write the for the phase difference for the wave of light $580\text{nm}$.

    $d\sin {\theta }_{\text{n}}=n{\lambda }_1$                                                                                                         (I)

Here, ${\theta }_{\text{n}}$ is the phase difference, $d$ is the separation between the two slits, ${\lambda }_1$ is the wavelength of light wave $580\text{nm}$, and $n$ is an integer.

Write the for the phase difference for the wave of light $620\text{nm}$.

    $d\sin {\theta }_{\text{n}}=n^{\prime }{\lambda }_2$                                                                                                     (II)

Here, ${\lambda }_{620}$ is the wavelength of light wave $620\text{nm}$ and $n^{\prime }$ is an integer.

Divide equation (I) by (II) to get the value of $\frac{n}{n^{\prime }}$.

    $\begin{array}{c}\frac{d\sin {\theta }_{\text{n}}}{d\sin {\theta }_{\text{n}}}=\frac{n{\lambda }_1}{n^{\prime }{\lambda }_2}\\ n^{\prime }{\lambda }_2=n{\lambda }_1\\ \frac{n^{\prime }}{n}=\frac{{\lambda }_1}{{\lambda }_2}\end{array}$                                                                                                 (III)

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

    $\tan {\theta }_{\text{n}}=\frac{y_n}{x}$                                                                                                       (IV)

Here, $x$ is the distance from the slit to the screen, $y_n$ is the distance of the particular time fringe from the central maximum on the screen and ${\theta }_{\text{n}}$ is the angle made by a particular interference fringe from the central maximum of the screen.

Conclusion:

Substitute $580\text{nm}$ for ${\lambda }_1$ and $620\text{nm}$ for ${\lambda }_2$ in equation (III) to calculate the value of $\frac{n^{\prime }}{n}$.

    $\begin{array}{c}\frac{n^{\prime }}{n}=\frac{580\text{nm}}{620\text{nm}}\\ =\frac{29}{31}\end{array}$                                                                                                          (V)

On comparison of values of equation (V),

    $n^{\prime }=29$

    $n=31$

Substitute $31$ for $n$, $0.340\text{mm}$ for $d$ and $580\text{nm}$ for ${\lambda }_1$ in equation (I) to calculate the value of ${\theta }_{\text{n}}$.

    $\begin{array}{c}\left(0.340\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\sin {\theta }_{\text{n}}=\left(31\right)\left(580\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\\ \sin {\theta }_{\text{n}}=\frac{\left(31\right)\left(580\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{0.34\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}}\\ {\theta }_{\text{n}}={\sin }^{-1}\left(0.05288\right)\\ {\theta }_{\text{n}}=3.031°\end{array}$

Substitute $3.031°$ for ${\theta }_{\text{n}}$, $2.10\text{m}$ for $x$ in equation (IV) to calculate the value of $y_{31}$.

    $\begin{array}{c}\tan \left(3.031°\right)=\frac{y_{31}}{2.10\text{m}}\\ y_{31}=\tan \left(3.031°\right)\left(2.10\text{m}\right)\\ =\text{0}\text{.111}\text{m}\end{array}$

Therefore, the shortest distance from the central maximum for which the bright fringes for the two wavelengths coincide is $\text{0}\text{.111}\text{m}$.