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 8 ^th bright fringe above the point O oscillates with time between two extreme positions. The separation between these two extreme positions, in micrometer (μm), is

The correct answer is 601.50.

Step-by-step Explanation:

1. Given Data:

As shown in the setup diagram:

Distance of screen from the double-slit system, $D=1\text{ m}$
Wavelength of light used, $\lambda =6000{\text{Å}}^0=6000\times {10}^{-10}\text{ m}=6\times {10}^{-7}\text{ m}$
Time-dependent separation between the two slits A and B: $d\left(t\right)=(0.8+0.04\sin \omega t)\text{ mm}$

2. Position of the 8^th Bright Fringe:

In a Young's double slit experiment, the distance $y_n$ of the $n$^th bright fringe from the central bright fringe at point O is given by:

$y_n=\frac{n\lambda D}{d}$

For the 8^th bright fringe ($n=8$):

$y_8=\frac{8\lambda D}{d}$

3. Maximum and Minimum Positions:

Since $d\left(t\right)=0.8+0.04\sin \omega t\text{ mm}$, the maximum and minimum values of the slit separation $d$ are:

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

Because $y_8$ is inversely proportional to $d$:

$y_{\text{8, min}}=\frac{8\lambda D}{d_{\text{max}}}$
$y_{\text{8, max}}=\frac{8\lambda D}{d_{\text{min}}}$

4. Calculating the Separation Between the Two Extreme Positions:

The total separation $\Delta y$ between the extreme positions of the 8^th bright fringe is:

$\Delta y=y_{\text{8, max}}-y_{\text{8, min}}=8\lambda D(\frac{1}{d_{\text{min}}}-\frac{1}{d_{\text{max}}})$

Substitute the given values into the formula:

$8\lambda D=8\times (6\times {10}^{-7})\times 1=48\times {10}^{-7}\text{ m}$

$\Delta y=48\times {10}^{-7}(\frac{1}{0.76\times {10}^{-3}}-\frac{1}{0.84\times {10}^{-3}})$

$\Delta y=\frac{48\times {10}^{-7}}{{10}^{-3}}\left(\frac{0.84-0.76}{0.76\times 0.84}\right)$

$\Delta y=48\times {10}^{-4}\left(\frac{0.08}{0.6384}\right)$

$\Delta y=\frac{3.84\times {10}^{-4}}{0.6384}\text{ m}\approx 6.015037\times {10}^{-4}\text{ m}$

Converting into micrometers ($\mu \text{m}$):

$\Delta y=6.015037\times {10}^{-4}\times {10}^6\mu \text{m}=601.5037\mu \text{m}\approx 601.50\mu \text{m}$