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 fringe spacing \( y \) in an interference pattern is calculated using \( y = \frac{\lambda L}{d} \). Given values are: wavelength \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \), slit-to-screen distance \( L = 1.20 \, \text{m} \), and slit separation \( d = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} \). Substituting these values yields \( y = \frac{600 \times 10^{-9} \times 1.20}{0.1 \times 10^{-3}} = 7.2 \, \text{mm} \). Therefore, adjacent bright fringes are 7.2 mm apart. The angular width \( \theta \) of the first bright fringe from the center is determined by \( \theta = \frac{\lambda}{d} \). Using the provided values, \( \theta = \frac{600 \times 10^{-9}}{0.1 \times 10^{-3}} = 6 \times 10^{-3} \, \text{radians} \). To convert this 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.