Question

Plane wavefront of light of wavelength 6000 Å is incident on two slits on a screen perpendicular to the direction of light rays. If the total separation of 10 bright fringes on a screen 2 m away is 2 cm, find the distance between the slits.

Hint:

Fringe width in Young’s experiment is \( \beta = \frac{\lambda D}{d} \); total width of \( n \) fringes is \( n \beta \).

Answer 1

Wavelength \( \lambda = 6000 \, \text{Å} = 6 \times 10^{-7} \, \text{m} \).
Fringe width \( \beta = \frac{\lambda D}{d} \), where \( D = 2 \, \text{m} \) (screen distance), \( d \) is slit separation.
Total separation of 10 bright fringes = \( 10 \beta = 2 \, \text{cm} = 0.02 \, \text{m} \).
\[ \beta = \frac{0.02}{10} = 0.002 \, \text{m}. \] \[ \beta = \frac{\lambda D}{d} \quad \Rightarrow \quad 0.002 = \frac{(6 \times 10^{-7}) \times 2}{d}. \] \[ d = \frac{12 \times 10^{-7}}{0.002} = 6 \times 10^{-4} \, \text{m} = 0.6 \, \text{mm}. \] Answer: 0.6 mm.