Question

In Young's double slit experiment, if the distance between the slits is doubled while keeping the wavelength and the distance to the screen constant, the fringe spacing will:

Hint:

Fringe spacing in double slit experiments is inversely proportional to the slit separation.

Answer 1

The fringe spacing (\( \beta \)) in Young's double slit experiment is given by:

\[ \beta = \frac{\lambda L}{d} \]

where:

• \( \lambda \) is the wavelength of light
• \( L \) is the distance from the slits to the screen
• \( d \) is the distance between the slits

If \( d \) is doubled (\( d' = 2d \)), while \( \lambda \) and \( L \) remain constant:

\[ \beta' = \frac{\lambda L}{2d} = \frac{\beta}{2} \]

Thus, the fringe spacing is halved.