91Ó°ÊÓ

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, ${\mathrm{λ}}_1=480\text{nm}$

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

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.

$d\sin \mathrm{θ}=m\mathrm{λ}$ …… (1)

Here, $d$ is the distance between the slits, $\mathrm{λ}$ is the wavelength, $m$ is the order and $\mathrm{θ}$is the angular separation.

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

$\tan \mathrm{θ}≈\sin \mathrm{θ}=\frac{y_m}{D}$

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

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

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

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

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

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

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