In Young's double slit experiment, the fringe width of the...

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: $2$