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: \begin{itemize} \( \lambda \) is the wavelength of light \( L \) is the distance from the slits to the screen \( d \) is the distance between the slits \end{itemize} 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.