Question

A beam of light consisting of two wavelengths \(400\ \text{nm}\) and \(600\ \text{nm}\) is used to illuminate a single slit of width \(1\ \text{mm}\). Find the least distance of the point from the central maximum where the dark fringes due to both wavelengths coincide on the screen placed \(1.5\ \text{m}\) from the slit.

Hint:

For coincidence of dark fringes in single-slit diffraction, 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 \[ y=\frac{nD\lambda}{a} \] to obtain the position of the coincident minimum.

Answer 1

Concept: The diffraction pattern produced by a single slit consists of a central bright maximum and a series of secondary maxima and minima on both sides. For a single slit of width \(a\), the condition for obtaining the dark fringes (minima) is \[ a\sin\theta=n\lambda, \qquad n=1,2,3,\dots \] where
• \(a\) is the width of the slit,
• \(\lambda\) is the wavelength of light,
• \(n\) is the order of the minimum. For small diffraction angles, \[ \sin\theta \approx \tan\theta \approx \theta. \] Hence, if the screen is placed at a distance \(D\) from the slit, the position of the \(n^{\text{th}}\) dark fringe from the central maximum is \[ y_n=\frac{nD\lambda}{a}. \] Since two different wavelengths are used, the dark fringes corresponding to the two wavelengths will coincide only when their positions on the screen are equal.

Step 1: Write the positions of dark fringes for the two wavelengths.
For wavelength \[ \lambda_1=400\ \text{nm}=400\times10^{-9}\ \text{m}, \] the position of the \(n_1^{\text{th}}\) minimum is \[ y_1=\frac{n_1D\lambda_1}{a}. \] For wavelength \[ \lambda_2=600\ \text{nm}=600\times10^{-9}\ \text{m}, \] the position of the \(n_2^{\text{th}}\) minimum is \[ y_2=\frac{n_2D\lambda_2}{a}. \] For coincidence of dark fringes, \[ y_1=y_2. \] Therefore, \[ \frac{n_1D\lambda_1}{a} = \frac{n_2D\lambda_2}{a}. \] After cancelling \(D\) and \(a\), \[ n_1\lambda_1=n_2\lambda_2. \]

Step 2: Determine the least orders for which the minima coincide.
Substituting the wavelengths, \[ n_1(400)=n_2(600). \] Dividing by \(200\), \[ 2n_1=3n_2. \] The smallest integral values satisfying this relation are \[ n_1=3, \qquad n_2=2. \] Thus, the third minimum of \(400\ \text{nm}\) light coincides with the second minimum of \(600\ \text{nm}\) light.

Step 3: Calculate the distance of the coincident dark fringe from the central maximum.
Using \[ y=\frac{nD\lambda}{a}, \] and taking \[ n=3,\qquad \lambda=400\times10^{-9}\ \text{m}, \] \[ D=1.5\ \text{m}, \qquad a=1\ \text{mm}=10^{-3}\ \text{m}, \] we get \[ y= \frac{3\times1.5\times400\times10^{-9}} {10^{-3}}. \] \[ y= \frac{1800\times10^{-9}} {10^{-3}} \] \[ y=1.8\times10^{-3}\ \text{m}. \] Therefore, \[ \boxed{y=1.8\times10^{-3}\ \text{m}} \] or \[ \boxed{y=1.8\ \text{mm}}. \] Hence, the least distance of the point from the central maximum where the dark fringes due to both wavelengths coincide is \[ \boxed{1.8\ \text{mm}}. \]