Question

Two slits 0.1 mm apart are arranged 1.20 m from a screen. Light of wavelength 600 nm from a distant source is incident on the slits.

Hint:

In a double-slit experiment, the fringe separation is determined by the wavelength of the light, the distance between the slits, and the distance to the screen.

Answer 1

The distance between adjacent bright fringes in an interference pattern is given by: \[ y = \frac{\lambda L}{d} \] where: - \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \) is the wavelength, - \( L = 1.20 \, \text{m} \) is the distance from the slits to the screen, - \( d = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} \) is the distance between the slits. Substituting the values: \[ y = \frac{600 \times 10^{-9} \times 1.20}{0.1 \times 10^{-3}} = 7.2 \, \text{mm} \] Thus, the distance between adjacent bright interference fringes is 7.2 mm. The angular width \( \theta \) of the first bright fringe (from the center to the first fringe) is given by: \[ \theta = \frac{\lambda}{d} \] Substitute the values: \[ \theta = \frac{600 \times 10^{-9}}{0.1 \times 10^{-3}} = 6 \times 10^{-3} \, \text{radians} \] To convert radians to degrees, multiply by \( \frac{180}{\pi} \): \[ \theta = 6 \times 10^{-3} \times \frac{180}{\pi} \approx 0.344 \, \text{degrees} \] Thus, the angular width of the first bright fringe is approximately 0.344 degrees.