Question

Light consisting of two wavelengths 600 nm and 480 nm is used to obtain interference fringes in a double slit experiment. The screen is placed 1.0 m away from slits which are 1.0 mm apart. (i) Calculate the distance of the third bright fringe on the screen from the central maximum for wavelength 600 nm.

Hint:

The distance between bright fringes in a double slit experiment is directly proportional to the wavelength and the distance from the slits to the screen.

Answer 1

The fringe width \( \beta \) in a double slit experiment is given by the formula: \[ \beta = \frac{\lambda D}{d} \] Where: - \( \lambda \) is the wavelength of the light, - \( D \) is the distance between the slits and the screen, - \( d \) is the distance between the slits. For the wavelength \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \), \( D = 1.0 \, \text{m} \), and \( d = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} \), we can calculate the fringe width: \[ \beta = \frac{600 \times 10^{-9} \times 1.0}{1.0 \times 10^{-3}} = 6.0 \times 10^{-4} \, \text{m} \] The distance of the third bright fringe from the central maximum is: \[ y_3 = 3 \times \beta = 3 \times 6.0 \times 10^{-4} = 1.8 \times 10^{-3} \, \text{m} = 1.8 \, \text{mm} \] Thus, the distance of the third bright fringe is 1.8 mm.