Question

In a Young's double-slit experimental set-up with slit separation \(0.6\ \text{mm}\), a beam of light consisting of two wavelengths \(440\ \text{nm}\) and \(660\ \text{nm}\) is used to obtain an interference pattern on a screen kept \(1.5\ \text{m}\) in front of the slits. Find the least distance of the point from the central maximum where the bright fringes due to both the wavelengths coincide.

Hint:

For coincidence of bright fringes in YDSE, always use \[ n_1\lambda_1=n_2\lambda_2. \] First find the smallest integral values of \(n_1\) and \(n_2\), and then substitute into \[ y=\frac{nD\lambda}{d} \] to determine the position of the coincident bright fringe.

Answer 1

Concept: In Young's Double-Slit Experiment (YDSE), the condition for the formation of bright fringes (constructive interference) is \[ \Delta = n\lambda, \qquad n=0,1,2,3,\dots \] where \(n\) is the order of the bright fringe. The position of the \(n^{\text{th}}\) bright fringe from the central maximum is given by \[ y_n=\frac{nD\lambda}{d}, \] where
• \(D\) is the distance of the screen from the slits,
• \(d\) is the separation between the slits,
• \(\lambda\) is the wavelength of light. When two different wavelengths are used simultaneously, their bright fringes will coincide only if the positions of the corresponding bright fringes are the same. Thus, for coincidence of bright fringes, \[ \frac{n_1D\lambda_1}{d} = \frac{n_2D\lambda_2}{d} \] or \[ n_1\lambda_1=n_2\lambda_2. \]

Step 1: Write the condition for coincidence of bright fringes.
Given, \[ \lambda_1=440\ \text{nm}, \qquad \lambda_2=660\ \text{nm}. \] For the bright fringes to coincide, \[ n_1(440)=n_2(660). \] Dividing throughout by \(220\), \[ 2n_1=3n_2. \] The smallest integral values satisfying this relation are \[ n_1=3, \qquad n_2=2. \] Thus, the third bright fringe of \(440\ \text{nm}\) light coincides with the second bright fringe of \(660\ \text{nm}\) light.

Step 2: Calculate the position of the coincident bright fringe.
Using \[ y=\frac{nD\lambda}{d}, \] and substituting \[ n=3, \qquad \lambda=440\times10^{-9}\ \text{m}, \] \[ D=1.5\ \text{m}, \qquad d=0.6\ \text{mm}=0.6\times10^{-3}\ \text{m}, \] we get \[ y= \frac{3\times1.5\times440\times10^{-9}} {0.6\times10^{-3}}. \] First, calculate the numerator: \[ 3\times1.5\times440 = 1980. \] Hence, \[ y= \frac{1980\times10^{-9}} {0.6\times10^{-3}}. \] \[ y= \frac{1980}{0.6}\times10^{-6}. \] \[ y= 3300\times10^{-6}\ \text{m}. \] Therefore, \[ y=3.3\times10^{-3}\ \text{m}. \] Thus, \[ \boxed{y=3.3\times10^{-3}\ \text{m}} \] or \[ \boxed{y=3.3\ \text{mm}}. \] Hence, the least distance of the point from the central maximum where the bright fringes due to both wavelengths coincide is \[ \boxed{3.3\ \text{mm}}. \]