Chapter 35, Problem 83PQ

To determine

The distance between the first two adjacent dark fringes on the screen.

Answer to Problem 83PQ

The distance between the first two adjacent dark fringes on the screen is $2.80\times {10}^{-2}\text{m}$.

Explanation of Solution

Write the expression for the phase difference for the dark fringes for wavelength ${\lambda }_1$.

    $d\sin \theta =\left(2n_1+1\right)\frac{{\lambda }_1}{2}$                                                                                     (I)

Here, $\theta$ is the angle, $d$ is the slit width, ${\lambda }_1$ is the first monochromatic wavelength and $n_1$ is an integer.

Write the expression for the phase difference for the dark fringes for wavelength ${\lambda }_2$.

    $d\sin \theta =\left(2n_2+1\right)\frac{{\lambda }_2}{2}$                                                                                  (II)

Here, ${\lambda }_2$ is the second monochromatic wavelength and $n_2$ is an integer.

Divide equation (I) by (II)

$\begin{array}{c}\frac{d\sin \theta }{d\sin \theta }=\frac{\left(2n_1+1\right)\frac{{\lambda }_1}{2}}{\left(2n_2+1\right)\frac{{\lambda }_2}{2}}\\ \left(2n_1+1\right)\frac{{\lambda }_1}{2}=\left(2n_2+1\right)\frac{{\lambda }_2}{2}\\ \left(2n_1+1\right){\lambda }_1=\left(2n_2+1\right){\lambda }_2\\ \frac{\left(2n_1+1\right)}{\left(2n_2+1\right)}=\frac{{\lambda }_2}{{\lambda }_1}\end{array}$ (III)

Write the expression to calculate the value of $n_1{}^{\prime }$.

    $n_1{}^{\prime }=\left(2n_1+1\right)-\left(-n_1\right)$                                                                                 (IV)

Here, $n_1{}^{\prime }$ is an integer.

Write the expression to calculate the value of $n_1{}^{\prime }$.

    $n_2{}^{\prime }=\left(2n_2+1\right)-\left(-n_2\right)$                                                                               (V)

Write the expression for the distance between the two adjacent dark fringes on the screen

    $\Delta s=\frac{{\lambda }_{1}D}{2d}\left(\left(2n_1{}^{\prime }+1\right)-\left(2n_1+1\right)\right)$                                                                (VI)

Here,$D$ is the distance between the slit and the screen.

Conclusion:

Substitute $560.0\text{nm}$ for ${\lambda }_2$ and $400.0\text{nm}$ for ${\lambda }_1$ in equation (III) to calculate the value of $\frac{\left(2n_1+1\right)}{\left(2n_2+1\right)}$

    $\begin{array}{c}\frac{\left(2n_1+1\right)}{\left(2n_2+1\right)}=\frac{560.0\text{nm}}{400.0\text{nm}}\\ \frac{\left(2n_1+1\right)}{\left(2n_2+1\right)}=\frac{7}{5}\end{array}$                                                                              (VII)

Solve equation (VII) for the values of $n_1$ and $n_2$.

    $\begin{array}{c}2n_1+1=7\\ n_1=\left(\frac{7-1}{2}\right)\\ n_1=3\end{array}$

Similarly,

    $\begin{array}{c}2n_2+1=5\\ n_2=\frac{\left(5-1\right)}{2}\\ n_2=2\end{array}$

Substitute $3$ for $n_1$ in equation (V) to calculate the value of $n_1{}^{\prime }$.

    $\begin{array}{c}{n}_1{}^{\prime }=\left(\left(2\times 3\right)+1\right)-\left(-3\right)\\ =10\end{array}$

Substitute $2$ for $n_1$ in equation (VI) to calculate the value of $n_2{}^{\prime }$.

    $\begin{array}{c}{n}_2{}^{\prime }=\left(\left(2\times 2\right)+1\right)-\left(-2\right)\\ =7\end{array}$

Substitute $1.00\text{m}$ for $D$, $0.100\text{mm}$ for $d$, $400.0\text{nm}$ for ${\lambda }_1$, $10$ for $n_1{}^{\prime }$ and $3$ for $n_1$ in equation (VI) to calculate the value of $\Delta s$

    $\begin{array}{c}\Delta s=\frac{\left(400.0\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\left(1.00\text{m}\right)}{\left(2\right)\left(0.100\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}\left(\left(\left(2\times 10\right)+1\right)-\left(\left(2\times 3\right)+1\right)\right)\\ =\left(14.0\right)\left(2.00\times {10}^{-3}\text{m}\right)\\ =2.80\times {10}^{-2}\text{m}\end{array}$

Therefore, the distance between the first two adjacent dark fringes on the screen is $2.80\times {10}^{-2}\text{m}$