1. Idea of shift of central maximum

In YDSE, if an अतिरिक्त optical path difference $\Delta$ is introduced between the two slits, then the central maximum shifts by

$x_0=\frac{D}{d}\,\Delta$

where $D$ is slit-screen distance and $d$ is slit separation.

2. Optical path added by the wedge combination

For a thin plate of refractive index $\mu$ and thickness $t$, the extra optical path introduced is

$(\mu-1)t$

Here, in front of the two slits, wedges $A$ and $B$ are complementary. From the figure description, one slit sees wedge $A$ and the other sees wedge $B$ over the same total thickness $t=12\,\mu\text{m}$.

Hence the net optical path difference produced between the two slits is

$\Delta=(\mu_A-\mu_B)t$

with

$\mu_A=1.7,\qquad \mu_B=1.5$

So,

$\Delta=(1.7-1.5)(12\times 10^{-6})$ $\Delta=0.2\times 12\times 10^{-6}$ $\Delta=2.4\times 10^{-6}\,\text{m}$

3. Shift of central maximum

Given:

$D=2\,\text{m},\qquad d=2\,\text{mm}=2\times 10^{-3}\,\text{m}$

Then

$x_0=\frac{D}{d}\Delta$

$x_0=\frac{2}{2\times 10^{-3}}\times 2.4\times 10^{-6}$

$x_0=10^3\times 2.4\times 10^{-6}$

$x_0=2.4\times 10^{-3}\,\text{m}$

$x_0=2.4\,\text{mm}$

4. Role of $l=1\,\text{mm}$

Since the question asks the shift of the central maximum due to the wedge combination, and refraction at the slanted interface is neglected, only the optical path difference matters. Thus $l$ does not affect the result.

5. Final answer

The central maximum shifts by

$\boxed{2.4\ \text{mm}}$

6. Comparison with stored answer

Stored correct answer is $1.2$. My derived answer is $2.4$ mm, so I disagree. The stored answer appears to be smaller by a factor of $2$. That could arise from incorrectly taking effective thickness difference as $t/2$, but with the given wording, the net OPD is

$(\mu_A-\mu_B)t$

leading to $2.4$ mm.