Question

in a single slit diffraction experiment, the width of the slit through which light passes is reduced. what happens to the width of the central bright fringe in the resulting diffraction pattern

Answer 1

Concepts and reason

The concept required to solve the given problem is width of fringes in single slit diffraction. Initially, write the expression for width of bright fringes in single slit diffraction. Then, obtain the expression for width of central bright fringe by using this expression. Finally, use this expression for width of central bright fringe to determine the width of the central bright fringe when slit width is reduced.

Fundamentals

The width of bright fringes \(y_{\mathrm{m}}\) in a single slit experiment is given by following expression. \(y_{\mathrm{m}}=\frac{m \lambda D}{a}\)

Here, \(a\) is the slit width, \(m\) is the order of diffraction, \(D\) is the distance to the screen, and \(\lambda\) is the wavelength.

The width of dark fringes \(y_{\mathrm{m}}\) in a single slit experiment is given by following expression. \(y_{\mathrm{m}}=\left(m+\frac{1}{2}\right) \frac{\lambda D}{a}\)

The width of bright fringes \(y_{\mathrm{m}}\) in single slit experiment is given by following expression. \(y_{\mathrm{m}}=\frac{m \lambda D}{a}\)

Substitute 1 for \(\mathrm{m}\) in the above equation to solve for width of central bright fringe. \(\begin{aligned} y_{1} &=\frac{(1) \lambda D}{a} \\ &=\frac{\lambda D}{a} \end{aligned}\)

For central fringe, the value of order of diffraction \(\mathrm{m}\) is equal to \(1 .\)

The expression for width of central bright fringe is,

\(y_{1}=\frac{\lambda D}{a}\)

From the above expression, the width of central bright fringe will become wider when slit width is reduced.

The central bright fringe become wider.

The width of central bright fringe is inversely proportional to slit width. Thus, the width of central bright fringe will become wider when slit width reduced.