In Young’s double slit experiment, two slits S _ 1 and S _ 2...

The change in path difference due to the two slabs at $P$ is $\Deltax = \left({\mu}_{2} - {\mu}_{1}\right)t$

$= ( 1 . 55 - 1 . 51 ) \times 0 . 1$$= 0 . 04 \times {10}^{- 4}$

$⇒ \Deltax = 4 \times {10}^{- 6} = 4\mu m$

Now, the distance of central maxima from geometrical center is $y = \dfrac{\DeltaxD}{d} = 4 \times {10}^{- 6}\dfrac{D}{d}$

Now, fringe width, $n\beta = 4 \times {10}^{- 6}\dfrac{D}{d}$

Or, $\dfrac{n\lambdaD}{d} = 4 \times {10}^{- 6}\dfrac{D}{d}$

Number of shift, $n = \dfrac{4 \times {10}^{- 6}}{4 \times {10}^{- 7}} = 10$.