Question 15

The position of the fringe is $y = \frac{n \lambda D}{d}$. The speed $v$ is the magnitude of the rate of change of position: $v = \left| \frac{dy}{dt} \right| = \left| \frac{d}{dt} \left( \frac{n \lambda D}{d} \right) \right| = \frac{n \lambda D}{d^2} \left| \frac{dd}{dt} \right|$

Given $d = d_0 + A \sin \omega t$, where $d_0 = 0.8$ mm, $A = 0.04$ mm, and $\omega = 0.08$ rad/s. Then $\frac{dd}{dt} = A \omega \cos \omega t$. $v(t) = \frac{n \lambda D (A \omega \cos \omega t)}{(d_0 + A \sin \omega t)^2}$

To maximize $v(t)$, we analyze the function $f(\theta) = \frac{\cos \theta}{(d_0 + A \sin \theta)^2}$. Setting the derivative to zero: $- \sin \theta (d_0 + A \sin \theta)^2 - \cos \theta \cdot 2(d_0 + A \sin \theta) A \cos \theta = 0$ $- \sin \theta (d_0 + A \sin \theta) - 2 A \cos^2 \theta = 0$ $- d_0 \sin \theta - A \sin^2 \theta - 2A(1 - \sin^2 \theta) = 0$ $A \sin^2 \theta - d_0 \sin \theta - 2A = 0$

Substituting $d_0 = 0.8$ and $A = 0.04$: $0.04 \sin^2 \theta - 0.8 \sin \theta - 0.08 = 0 \implies \sin^2 \theta - 20 \sin \theta - 2 = 0$ $\sin \theta = \frac{20 \pm \sqrt{400 + 8}}{2} \approx -0.0995 \text{ (since } |\sin \theta| \le 1)$

At $\sin \theta \approx -0.1$, $\cos \theta = \sqrt{1 - (-0.1)^2} \approx 0.995$ and $d = 0.8 + 0.04(-0.1) = 0.796$ mm. $v_{max} = \frac{8 \times 6 \times 10^{-7} \times 1 \times (0.04 \times 10^{-3} \times 0.08 \times 0.995)}{(0.796 \times 10^{-3})^2}$ $v_{max} = \frac{15.36 \times 10^{-12} \times 0.995}{0.6336 \times 10^{-6}} \approx 24.118 \times 10^{-6} \text{ m/s} = 24.12 \text{ } \mu\text{m/s}$

Rounding to two decimal places, we get $24.12$.