In Young's double slit experiment, the fringe width of the interference pattern produced on the screen is \(2.4\) \(\mu\)m. If the experiment is carried out in another medium having refractive index \(1.2\), the fringe width will be _____ \(\mu\)m.

The fringe width in Young's double slit experiment is given by \(\beta = \dfrac{\lambda D}{d}\).

When the experiment is performed in a medium of refractive index \(\mu\), the wavelength of light becomes \(\lambda' = \dfrac{\lambda}{\mu}\).

The new fringe width is \(\beta' = \dfrac{\lambda' D}{d} = \dfrac{\lambda D}{\mu d} = \dfrac{\beta}{\mu}\).

Substituting the given values, \(\beta = 2.4\) \(\mu\)m and \(\mu = 1.2\):

\(\beta' = \dfrac{2.4}{1.2} = 2\) \(\mu\)m.

Answer: \(2\)