Question

Using the geometry of the double slit experiment, derive the expression for the fringe width of interference bands.

Hint:

The fringe width is directly proportional to the wavelength of the light and the distance from the slits to the screen, and inversely proportional to the slit separation.

Answer 1

In the double-slit experiment, an interference pattern is formed when light from two slits interacts. The fringe width \( \beta \) is the distance between two consecutive maxima (or minima) observed on the screen. The path difference for two waves reaching a point on the screen is given by: \[ \Delta {Path} = d \sin \theta \] where \( d \) is the separation between the slits and \( \theta \) is the angle at which the maximum or minimum occurs. For constructive interference (bright fringes), the path difference is an integer multiple of the wavelength \( \lambda \): \[ d \sin \theta = m \lambda \] For small angles \( \theta \), \( \sin \theta \approx \tan \theta \), and the position of the maxima can be written as: \[ y_m = m \frac{\lambda L}{d} \] where \( L \) is the distance from the slits to the screen. The fringe width \( \beta \), which is the distance between two successive maxima, is given by: \[ \beta = \frac{\lambda L}{d} \] \bigskip