Central Maximum Shift in Youngs Double Slit with Glass Wedges

To find the shift of the central maximum on the screen, we first determine the optical path length introduced by the combination of wedges at the positions of the two slits.

1. Geometry of the Wedges

Let the central axis passing through point $O$ on the screen be $y = 0$. The two slits, $S_1$ and $S_2$, are separated by $d = 2\text{ mm}$. Thus, their position coordinates are:

• Upper slit $S_1$: $y_1 = +\frac{d}{2} = +1\text{ mm}$
• Lower slit $S_2$: $y_2 = -\frac{d}{2} = -1\text{ mm}$

The combination of two glass wedges forms a rectangular block of total thickness $t = 12\ \mu\text{m}$. The total height $H$ of this rectangular block is: $H = d + 2l = 2\text{ mm} + 2(1\text{ mm}) = 4\text{ mm}$

The top edge of the block is at $y_{\text{top}} = +2\text{ mm}$ and the bottom edge is at $y_{\text{bottom}} = -2\text{ mm}$.

The thickness of wedge $A$, $t_A(y)$, varies linearly from $t$ at the top edge ($y = +2\text{ mm}$) to $0$ at the bottom edge ($y = -2\text{ mm}$): $t_A(y) = t \left( \frac{y - (-2)}{2 - (-2)} \right) = t \left( \frac{y + 2}{4} \right)$

The thickness of wedge $B$ at height $y$ is: $t_B(y) = t - t_A(y) = t \left( \frac{2 - y}{4} \right)$

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2. Optical Path Length (OPL) through the Slits

• At Slit $S_1$ ($y_1 = +1\text{ mm}$): $t_A(y_1) = t \left( \frac{1 + 2}{4} \right) = \frac{3}{4}t$ $t_B(y_1) = t \left( \frac{2 - 1}{4} \right) = \frac{1}{4}t$

  The optical path length for the ray passing through $S_1$ within the wedge combination is: $\text{OPL}_1 = \mu_A t_A(y_1) + \mu_B t_B(y_1) = \mu_A \left(\frac{3}{4}t\right) + \mu_B \left(\frac{1}{4}t\right) = \frac{t}{4}(3\mu_A + \mu_B)$

• At Slit $S_2$ ($y_2 = -1\text{ mm}$): $t_A(y_2) = t \left( \frac{-1 + 2}{4} \right) = \frac{1}{4}t$ $t_B(y_2) = t \left( \frac{2 - (-1)}{4} \right) = \frac{3}{4}t$

  The optical path length for the ray passing through $S_2$ within the wedge combination is: $\text{OPL}_2 = \mu_A t_A(y_2) + \mu_B t_B(y_2) = \mu_A \left(\frac{1}{4}t\right) + \mu_B \left(\frac{3}{4}t\right) = \frac{t}{4}(\mu_A + 3\mu_B)$

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3. Optical Path Difference ($\Delta \text{OPL}$)

The path difference introduced by the wedge combination between the two rays is: $\Delta \text{OPL} = \text{OPL}_1 - \text{OPL}_2 = \frac{t}{4}\Big((3\mu_A + \mu_B) - (\mu_A + 3\mu_B)\Big) = \frac{t}{2}(\mu_A - \mu_B)$

Given:

• $\mu_A = 1.7$
• $\mu_B = 1.5$
• $t = 12\ \mu\text{m} = 12 \times 10^{-6}\text{ m}$

$\Delta \text{OPL} = \frac{12 \times 10^{-6}\text{ m}}{2} (1.7 - 1.5) = 6 \times 10^{-6} \times 0.2 = 1.2 \times 10^{-6}\text{ m}$

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4. Shift of Central Maximum

For the central maximum, the net optical path difference between the rays arriving at a point $y$ on the screen must be zero: $\Delta x_{\text{geometric}} = \Delta \text{OPL}$ $\frac{d \cdot y}{D} = \Delta \text{OPL}$

Solving for $y$: $y = \frac{D}{d} \cdot \Delta \text{OPL}$

Substitute the values $D = 2\text{ m}$ and $d = 2\text{ mm} = 2 \times 10^{-3}\text{ m}$: $y = \frac{2}{2 \times 10^{-3}} \times (1.2 \times 10^{-6}) = 10^3 \times 1.2 \times 10^{-6}\text{ m} = 1.2 \times 10^{-3}\text{ m} = 1.2\text{ mm}$

Thus, the central maximum shifts with respect to $O$ by 1.2 mm.