$\text{In a Young's double slit experiment, a combination of two glass wedges}A\text{and}B,\text{having refractive indices }1.7\text{and}\mathrm{ 1.5},$

$\text{respectively, are placed in front of the slits, as shown in the figure.}\text{The separation between the slits is}d=2\mathrm{mm}$

$\text{and the shortest distance between the slits and the screen is}D=2m.\text{Thickness of the combination of the wedges}$

$\text{is}\mathrm{ t}=12\mu m.\text{The value of }\mathrm{\ell }\text{shown in the figure is}1\mathrm{mm}.\text{Neglect any refraction effect at the slanted interface of}$

$\text{the wedges.}\text{ Due to the combination of the wedges, the central maximum shifts (in mm) with respect to}\mathrm{ O}\text{ by}\text{\_\_\_\_\_}$

The correct answer is 1.2.

Step-by-Step Explanation:

First, let us analyze the geometry of the double-slit setup and the glass wedge combination shown in the figure.

The total height of the rectangular combination of the wedges is:
$H=d+2l$

Given values:
$d=2\text{mm}$
$l=1\text{mm}$

Therefore, the total vertical height of the wedge block is:
$H=2\text{mm}+2\left(1\text{mm}\right)=4\text{mm}$

Let the bottom of the wedge combination be at $y=0$ and the top be at $y=H=4\text{mm}$.

The upper slit, $S_1$, is located at a distance $l=1\text{mm}$ from the top of the block. Thus, its position from the bottom is:
$y_1=H-l=4\text{mm}-1\text{mm}=3\text{mm}$

The lower slit, $S_2$, is located at a distance $l=1\text{mm}$ from the bottom of the block, so:
$y_2=1\text{mm}$

Let us determine the thickness of each wedge at the positions of the two slits. The thickness of wedge A (having refractive index ${\mu }_A=1.7$) decreases linearly from $t$ at the top ($y=H$) to $0$ at the bottom ($y=0$). The thickness of wedge B (having refractive index ${\mu }_B=1.5$) increases linearly from $0$ at the top ($y=H$) to $t$ at the bottom ($y=0$).

Thus, the thickness of wedge A at height $y$ is given by:
$t_A=t·\frac{y}{H}$

And the thickness of wedge B at height $y$ is given by:
$t_B=t·\left(1-\frac{y}{H}\right)$

At the upper slit $S_1$ ($y=3\text{mm}$, where $H=4\text{mm}$):
$t_{A1}=\frac{3}{4}t$
$t_{B1}=\frac{1}{4}t$

At the lower slit $S_2$ ($y=1\text{mm}$):
$t_{A2}=\frac{1}{4}t$
$t_{B2}=\frac{3}{4}t$

The optical path length for the ray passing through $S_1$ in the glass block is:
$x_1={\mu }_A{t}_{A1}+{\mu }_B{t}_{B1}=1.7\left(\frac{3}{4}t\right)+1.5\left(\frac{1}{4}t\right)=1.65t$

The optical path length for the ray passing through $S_2$ in the glass block is:
$x_2={\mu }_A{t}_{A2}+{\mu }_B{t}_{B2}=1.7\left(\frac{1}{4}t\right)+1.5\left(\frac{3}{4}t\right)=1.55t$

Therefore, the optical path difference introduced between the two rays by the wedges is:
$\mathrm{\Delta x}=x_1-x_2=1.65t-1.55t=0.1t$

Since the optical path is larger for the ray passing through the upper slit $S_1$, the central maximum shifts upward towards $S_1$.

The relation for the shift of the central maximum $y$ is given by:
$y=\frac{\mathrm{\Delta x}·D}{d}$

Substituting the given values:
$t=12\mathrm{\mu m}=12\times {10}^{-6}\text{m}$
$\mathrm{\Delta x}=0.1\times 12\times {10}^{-6}\text{m}=1.2\times {10}^{-6}\text{m}$
$D=2\text{m}$
$d=2\text{mm}=2\times {10}^{-3}\text{m}$

Now, calculate the shift:
$y=\frac{(1.2\times {10}^{-6}\text{m})·\left(2\text{m}\right)}{2\times {10}^{-3}\text{m}}=1.2\times {10}^{-3}\text{m}=1.2\text{mm}$

Thus, the shift of the central maximum with respect to point $O$ is 1.2 mm.