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

In Young's double-slit experiment, the fringe width is calculated using the formula: \[y = \frac{\lambda D}{d}\] Given the following values: \( \lambda = 600 \) 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 these values into the formula: \[y = \frac{(600 \times 10^{-9}) (1.2)}{1.0 \times 10^{-4}}\] The calculation proceeds as: \[y = \frac{7.2 \times 10^{-4}}{10^{-4}} = 7.2 \times 10^{-3} \text{ m} = 7.2 \text{ mm}\] Therefore, the fringe width is 7.2 mm.