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. How far apart will adjacent bright interference fringes be on the screen?

Hint:

Fringe width increases with wavelength and distance to the screen but decreases with increasing slit separation.

Answer 1

The fringe width in Young’s double-slit experiment is given by: \[ y = \frac{\lambda D}{d} \] where:
\( \lambda = 600 \) nm \( = 600 \times 10^{-9} \) m (wavelength),
\( D = 1.2 \) m (distance to the screen),
\( d = 0.1 \) mm \( = 1.0 \times 10^{-4} \) m (slit separation).
Substituting values: \[ y = \frac{(600 \times 10^{-9}) (1.2)}{1.0 \times 10^{-4}} \] \[ y = \frac{7.2 \times 10^{-4}}{10^{-4}} = 7.2 \times 10^{-3} \text{ m} = 7.2 \text{ mm} \] Thus, the fringe width is 7.2 mm.