As shown in the figure, in Young's double slit experiment, a thin...

Fringe shift due to slab in Young's double slit experiment is given by $∆ x = \dfrac{D ( \mu - 1 ) t}{d}$

Since, fringe width ${\beta}_{0} = \dfrac{D}{d}\lambda$

Then, $∆ x = \dfrac{{\beta}_{0}\left(\mu - 1\right)t}{\lambda}$

Putting the values, we have

$= \dfrac{10 \times {10}^{- 6} ( 1 . 2 - 1 )}{5 \times {10}^{- 7}}{\beta}_{0}$

$= \dfrac{{10}^{- 5} \times 0 . 2}{5 \times {10}^{- 7}}{\beta}_{0} = 4{\beta}_{0}$

Hence, the value of $x = 4$.