Question

What is interference of light? Find an expression for fringe width in Young's double slit experiment.

Hint:

To get clearer fringes, increase \( D \) or decrease \( d \). Also, shorter wavelengths produce narrower fringes.

Answer 1

Step 1: When light from two slits separated by distance \( d \) falls on a screen at distance \( D \), the path difference between the two waves at point \( P \) is: \[ \Delta = d \sin \theta. \]
Step 2: For constructive interference (bright fringes), the path difference must be an integer multiple of wavelength: \[ \Delta = n \lambda, \quad n = 0, 1, 2, \ldots \]
Step 3: For small angles, \( \sin \theta \approx \tan \theta = \frac{y}{D} \), where \( y \) is the fringe position on the screen.
Step 4: Substituting, we get: \[ d \frac{y}{D} = n \lambda \implies y = \frac{n \lambda D}{d}. \]
Step 5: The fringe width \( \beta \) is the distance between two consecutive bright fringes: \[ \beta = y_{n+1} - y_n = \frac{\lambda D}{d}. \]