In Young's double slit experiment, the distance between screen and aperture is 1 m . The slit width is 2 mm . Light of $6000 Å$ is used. If a thin glass plate ( $\mu=1.5$ ) of thickness 0.04 mm is placed over one of the slits, then there will be a lateral displacement of the fringes by

The lateral displacement of the fringes in Young's double-slit experiment due to a thin glass plate over one slit is given by:

$\Delta x = \frac{D}{d} (\mu - 1)t$

Substituting the given values $D = 1\text{ m}$, $d = 2 \times 10^{-3}\text{ m}$, $\mu = 1.5$, and $t = 4 \times 10^{-5}\text{ m}$, we compute:

$\Delta x = \frac{1}{2 \times 10^{-3}} (1.5 - 1) (4 \times 10^{-5}) = \frac{0.5 \times 4 \times 10^{-5}}{2 \times 10^{-3}} = \frac{2 \times 10^{-5}}{2 \times 10^{-3}} = 1 \times 10^{-2}\text{ m}$

This corresponds to a displacement of $1\text{ cm}$.

Final answer: $\boxed{1\text{ cm}}$