Question 14

The shift of the central maximum in YDSE is caused by the extra optical path introduced by the transparent mediums in front of the slits. The shift ($y_0$) is given by $y_0 = \frac{D}{d} \Delta p$, where $\Delta p$ is the optical path difference at the center.

Step 1: Determine the geometry of the wedge combination. Let the center between the two slits be the origin ($y = 0$). Since the slit separation is $d = 2$ mm, the positions of the slits are:

• Top Slit ($S_1$): $y = +1$ mm
• Bottom Slit ($S_2$): $y = -1$ mm

From the figure, the wedge assembly extends a distance $l = 1$ mm above $S_1$ and $l = 1$ mm below $S_2$.

• Top edge of the wedge is at $y = 1 + 1 = +2$ mm.
• Bottom edge of the wedge is at $y = -1 - 1 = -2$ mm. The total height of the wedge assembly is $4$ mm.

Step 2: Calculate the thickness of wedges at each slit. The thickness of wedge A ($t_A$) decreases linearly from $t$ at the top edge ($y = 2$ mm) to $0$ at the bottom edge ($y = -2$ mm). The thickness of wedge B ($t_B$) increases linearly from $0$ to $t$. At any height $y$, the thicknesses are proportional to the distance from the zero-thickness edge:

• At Slit 1 ($y = +1$ mm): Wedge A is $\frac{3}{4}$ of the way from the bottom, so $t_{A1} = \frac{3}{4}t$. Wedge B is $\frac{1}{4}$ of the way from the top, so $t_{B1} = \frac{1}{4}t$.
• At Slit 2 ($y = -1$ mm): By symmetry, $t_{A2} = \frac{1}{4}t$ and $t_{B2} = \frac{3}{4}t$.

Step 3: Calculate the Optical Path Difference ($\Delta p$). The extra optical path introduced at $S_1$ is $\Delta x_1 = (\mu_A - 1)t_{A1} + (\mu_B - 1)t_{B1}$ The extra optical path introduced at $S_2$ is $\Delta x_2 = (\mu_A - 1)t_{A2} + (\mu_B - 1)t_{B2}$

The net path difference $\Delta p$ between the two waves arriving at the central point O is: $\Delta p = \Delta x_1 - \Delta x_2$

$\Delta p = [(\mu_A - 1)\frac{3}{4}t + (\mu_B - 1)\frac{1}{4}t] - [(\mu_A - 1)\frac{1}{4}t + (\mu_B - 1)\frac{3}{4}t]$

$\Delta p = (\mu_A - 1)(\frac{3}{4}t - \frac{1}{4}t) + (\mu_B - 1)(\frac{1}{4}t - \frac{3}{4}t)$

$\Delta p = (\mu_A - 1)\frac{t}{2} - (\mu_B - 1)\frac{t}{2}$

$\Delta p = (\mu_A - \mu_B)\frac{t}{2}$

Step 4: Calculate the numerical shift. Given $\mu_A = 1.7$, $\mu_B = 1.5$, and $t = 12 \times 10^{-6}$ m:

$\Delta p = (1.7 - 1.5) \times \frac{12 \times 10^{-6}}{2}$

$\Delta p = 0.2 \times 6 \times 10^{-6} = 1.2 \times 10^{-6} \text{ m}$

The shift of the central maximum is:

$y_0 = \frac{D}{d} \Delta p$

$y_0 = \frac{2}{2 \times 10^{-3}} \times 1.2 \times 10^{-6} = 10^3 \times 1.2 \times 10^{-6} = 1.2 \times 10^{-3} \text{ m}$

$y_0 = 1.2 \text{ mm}$

Final Answer: The central maximum shifts by 1.2 mm.