In a double-slit experiment, the distance between slits is$\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mm}$ and the slits are $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$ from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength $\mathbf{480}\mathbf{}\mathbf{nm}$, and the other due to light of wavelength $\mathbf{600}\mathbf{}\mathbf{nm}$. What is the separation on the screen between the third-order $\left(m=3\right)$ bright fringes of the two interference patterns?

The distance between the slit, $d=5\text{mm}$

The distance of slit from the screen, $D=1\text{m}$

The wavelength of one light, ${\lambda }_1=480\text{nm}$

The wavelength of other light, ${\lambda }_2=600\text{nm}$

The order of the bright fringe, $m=3$

Young's double-slit experiment. When monochromatic light passing through two narrow slits illuminates a distant screen, a characteristic pattern of bright and dark stripes is observed. This interference pattern is caused by the superposition of overlapping light waves originating from the two slits.

The condition for the maxima in Young’s experiment is given as follows.

$d\sin \theta =m\lambda$ …… (1)

Here, $d$ is the distance between the slits, $\lambda$ is the wavelength, $m$ is the order and $\theta$is the angular separation.

The maximum vertical distance from the centre of the pattern is given by,

$\tan \theta \approx \sin \theta =\frac{y_m}{D}$

Substitute $\frac{y_m}{D}$ for$\sin \theta$ into equation (1).

$\begin{array}{rcl}d\left(\frac{y_m}{D}\right)& =& m\lambda \\ y_m& =& \frac{m\lambda D}{d}\end{array}$

Since there are two interference patterns, therefore the wavelength, $\lambda ={\lambda }_2-{\lambda }_1$

$\begin{array}{rcl}d\left(\frac{y_m}{D}\right)& =& m\lambda \\ y_m& =& \frac{m\left({\lambda }_2-{\lambda }_1\right)D}{d}\end{array}$

Substitute 3 for m , 1 m for D, 600 nm for${\lambda }_2$ , 480 nm for ${\lambda }_1$, and 5 nm for d into above equation.

$\begin{array}{rcl}{y}_m& =& \frac{3\times \left(600-480\right)\times {10}^{-9}\times 1}{5\times {10}^{-3}}\\ & =& \frac{3\times 120\times {10}^{-6}}{5}\\ & =& 3\times 24\times {10}^{-6}\\ & =& 7.2\times {10}^{-5}\text{m}\end{array}$

Hence, the distance between the third-order bright fringes of the two interference patterns is $7.2\times {10}^{-5}\text{m}$.