Oscillating Slits Fringe Shift in Young Double Slit Experiment

To find the separation between the two extreme positions of the $8^{\text{th}}$ bright fringe, we analyze the position of the fringe as a function of time.

1. Expression for the Position of the $8^{\text{th}}$ Bright Fringe

In Young's double slit experiment, the distance of the $n^{\text{th}}$ bright fringe from the central maximum (point O) is given by: $y_n(t) = \frac{n \lambda D}{d(t)}$

Given parameters:

• Order of fringe, $n = 8$
• Wavelength of light, $\lambda = 6000 \text{ \AA} = 6 \times 10^{-7} \text{ m}$
• Distance between slits and screen, $D = 1 \text{ m}$
• Distance between slits, $d(t) = (0.8 + 0.04 \sin \omega t) \text{ mm}$

Thus, the position of the $8^{\text{th}}$ bright fringe above point O at any time $t$ is: $y_8(t) = \frac{8 \lambda D}{d(t)}$

---

2. Extreme Values of Slit Separation $d$

Since the function $\sin \omega t$ oscillates between $-1$ and $+1$, the slit separation $d(t)$ oscillates between:

• Minimum separation: $d_{\text{min}} = 0.8 - 0.04 = 0.76 \text{ mm} = 0.76 \times 10^{-3} \text{ m}$
• Maximum separation: $d_{\text{max}} = 0.8 + 0.04 = 0.84 \text{ mm} = 0.84 \times 10^{-3} \text{ m}$

---

3. Extreme Positions of the $8^{\text{th}}$ Bright Fringe

The extreme positions of the fringe correspond to $d_{\text{min}}$ and $d_{\text{max}}$:

1. Maximum distance from O ($y_{8, \text{max}}$): $y_{8, \text{max}} = \frac{8 \lambda D}{d_{\text{min}}} = \frac{8 \times (6 \times 10^{-7} \text{ m}) \times 1 \text{ m}}{0.76 \times 10^{-3} \text{ m}} = \frac{4.8 \times 10^{-3}}{0.76} \text{ m}$

2. Minimum distance from O ($y_{8, \text{min}}$): $y_{8, \text{min}} = \frac{8 \lambda D}{d_{\text{max}}} = \frac{8 \times (6 \times 10^{-7} \text{ m}) \times 1 \text{ m}}{0.84 \times 10^{-3} \text{ m}} = \frac{4.8 \times 10^{-3}}{0.84} \text{ m}$

---

4. Separation Between the Extreme Positions

The separation $\Delta y$ between these two extreme positions is: $\Delta y = y_{8, \text{max}} - y_{8, \text{min}} = 8 \lambda D \left( \frac{1}{d_{\text{min}}} - \frac{1}{d_{\text{max}}} \right)$

Substitute the numerical values: $\Delta y = 4.8 \times 10^{-3} \left( \frac{1}{0.76 \times 10^{-3}} - \frac{1}{0.84 \times 10^{-3}} \right) \text{ m}$

$\Delta y = 4.8 \left( \frac{0.84 - 0.76}{0.76 \times 0.84} \right) \text{ mm}$

$\Delta y = 4.8 \times \frac{0.08}{0.6384} \text{ mm} = \frac{0.384}{0.6384} \text{ mm}$

$\Delta y \approx 0.6015037 \text{ mm} = 601.5037 \text{ }\mu\text{m}$

Rounding off to two decimal places: $\Delta y \approx 601.50 \text{ }\mu\text{m}$

(Note: Using first-order differential approximation $\Delta y \approx \left|\frac{dy_8}{dd}\right| \Delta d = \frac{8 \lambda D}{d_0^2} (2 \Delta d) = \frac{4.8 \times 10^{-3}}{(0.8)^2} \times 0.08 = 600 \text{ }\mu\text{m}$.)