In a Young's double slit experiment, each of the two slits A and B , as shown in the figure, are oscillating about their fixed center and with a mean separation of 0.8mm . The distance between the slits at time t is given by d = (0.8 + 0.04 sin ωt)mm , where ω = 0.08 rads⁻¹ . The distance of the screen from the slits is 1m and the wavelength of the light used to illuminate the slits is 6000 ⁰ A . The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point O.

Question : The maximum speed in μm/s at which the 8 ^th bright fringe will move is

The correct answer is 24.

Step-by-Step Explanation:

1. Understanding the given parameters:
From the problem text and the diagram showing two slits A and B separated by a distance $d$ with screen distance $D=1\text{ m}$ and central point $\text{O}$ on the screen:
Distance between the slits at time $t$:
$d\left(t\right)=(0.8+0.04\sin \omega t)\text{ mm}=(0.8+0.04\sin \omega t)\times {10}^{-<!--;</mo--><mn>3</mn>}\text{ m}$
Angular frequency: $\omega =0.08{\text{ rad s}}^{-<!--;</mo--><mn>1</mn>}$
Distance to the screen: $D=1\text{ m}$
Wavelength of light: $\lambda =6000\stackrel{o}{\text{A}}=6000\times {10}^{-<!--;</mo--><mn>10</mn>}\text{ m}=6\times {10}^{-<!--;</mo--><mn>7</mn>}\text{ m}$

2. Position of the 8^th bright fringe:
In Young's double slit experiment, the position $y_n$ of the $n$^th bright fringe from the central zeroth fringe is given by:
$y_n=\frac{n\lambda D}{d}$

For the $8$^th bright fringe ($n=8$):
${\mathrm{y;}}_8=\frac{8\lambda D}{d}$

3. Speed of the 8^th bright fringe:
Differentiating $y_8$ with respect to time $t$ gives the instantaneous speed $v$:

From the given equation for $d$:
$\frac{dd}{dt}=0.04\omega \cos \omega t\text{ mm/s}=0.04\omega \cos \omega t\times {10}^{-<!--;</mo--><mn>3</mn>}\text{ m/s}$

Thus, the magnitude of speed is:
$\left|v\right|=\frac{8\lambda D}{d^2}\times 0.04\omega |\cos \omega t|\times {10}^{-<!--;</mo--><mn>3</mn>}$

4. Calculating the maximum speed:
The maximum speed occurs when $|\cos \omega t|=1$ and $d$ is minimum () or approximated around mean distance $d\approx 0.8\text{ mm}$ for small oscillations. Using $d=0.8\text{ mm}=0.8\times {10}^{-<!--;</mo--><mn>3</mn>}\text{ m}$:
$v_{\text{max}}=\frac{8\times (6\times {10}^{-<!--;</mo--><mn>7</mn>})\times 1}{{(0.8\times {10}^{-<!--;</mo--><mn>3</mn>})}^2}\times (0.04\times {10}^{-<!--;</mo--><mn>3</mn>})\times 0.08\times 1$

Simplifying step by step:
$v_{\text{max}}=\frac{4.8\times {10}^{-<!--;</mo--><mn>6</mn>}}{0.64\times {10}^{-<!--;</mo--><mn>6</mn>}}\times 3.2\times {10}^{-<!--;</mo--><mn>5</mn>}$
$v_{\text{max}}=7.5\times 3.2\times {10}^{-<!--;</mo--><mn>5</mn>}\text{ m/s}=24\times {10}^{-<!--;</mo--><mn>5</mn>}\text{ m/s}=2.4\times {10}^{-<!--;</mo--><mn>4</mn>}\text{ m/s}$

Converting to $\mu \text{m/s}$:
$v_{\text{max}}=24\times {10}^{-<!--;</mo--><mn>6</mn>}\text{ m/s}=24\mu \text{m/s}$

Thus, the maximum speed of the 8^th bright fringe is 24 μm/s.