Animated Solution for Physics - Optics: Comprehension Passage

Imagine a standard Young's double slit setup, but with a thrilling twist! Instead of being rigidly fixed, the slits A and B are oscillating. This means the distance between them, $d$, is constantly changing with time according to the given sine function:

Because the slit separation $d$ dictates the spread of the interference pattern, this entire pattern will "breathe"—expanding and contracting dynamically on the screen. Our goal is to track the motion of a specific fringe: the $8^{\text{th}}$ bright fringe.

We know from wave optics that the position of the $n^{\text{th}}$ bright fringe on the screen, measured from the central maximum, is given by:

Since $d$ is a function of time, the position of our eighth bright fringe, $y_8$, will also be a function of time. As the slits oscillate, this fringe will move up and down along the screen.

To find the separation between the extreme positions of this fringe, we need to determine its maximum and minimum heights. Because $d(t)$ is in the denominator, the fringe reaches its highest point ($y_{\max }$) when the slit separation $d$ is at its minimum. Conversely, it reaches its lowest point ($y_{\min }$) when $d$ is at its maximum.

The sine function oscillates between $-1$ and $+1$, giving us our extreme values for $d$:

The total separation $\Delta y$ is simply the difference between these two extreme positions:

Substituting the given values (and being extremely careful to convert millimeters and Angstroms into standard SI units of meters), we get:

After simplifying the fractions, we arrive at the final separation:

Moving on to the second part of the problem, we need to find the maximum speed of this $8^{\text{th}}$ bright fringe. Speed is the rate of change of position, so we must differentiate our position function $y_8(t)$ with respect to time. Let's write down the exact expression for $y_8(t)$ first:

To find the velocity $v(t)=\frac{d{y}_8}{dt}$, we apply the chain rule. The derivative of $\frac{1}{x}$ is $-\frac{1}{x^2}$, and then we multiply by the derivative of the inside function:

We want the maximum magnitude of this velocity. Look closely at the expression: the numerator contains a $\cos \omega t$ term, and the denominator contains a $(0.8+0.04\sin \omega t{)}^2$ term.

The speed is maximized when the numerator is as large as possible and the denominator is as small as possible. This simultaneous optimization occurs exactly when the fringe passes through its mean position, which corresponds to:

Substituting these conditions and the given value of $\omega =0.08\text{ rad/s}$ into our velocity magnitude equation:

This elegant result shows how beautifully wave optics and kinematics can intertwine in advanced physics problems!