The maximum speed in $$\mu$$ m/s at which the $$8^{th}$$ bright fringe will move is ________.

For the 8th bright fringe in Young’s double-slit experiment, the instantaneous position on the screen is

$$y_m(t)=\frac{m\lambda D}{d(t)}$$

with $$m=8,\;D=1\text{ m},\;\lambda =6000\text{ \AA}=6\times10^{-7}\text{ m}$$ and

$$d(t)=\bigl(0.8+0.04\sin\omega t\bigr)\text{ mm} =(8\times10^{-4}+4\times10^{-5}\sin\omega t)\text{ m},\qquad \omega =0.08\text{ rad s}^{-1}$$

The speed of this fringe is obtained by differentiating:

$$v(t)=\frac{dy_m}{dt} =-\frac{m\lambda D}{\bigl[d(t)\bigr]^2}\,\frac{dd}{dt}$$

First derivative of the slit separation:

$$\frac{dd}{dt}=4\times10^{-5}\;\omega\cos\omega t =4\times10^{-5}\times0.08\cos\omega t =3.2\times10^{-6}\cos\omega t\;(\text{m s}^{-1})$$

Hence

$$v(t)=\frac{8\,(6\times10^{-7})(3.2\times10^{-6})\, |\cos\omega t|}{\bigl(8\times10^{-4}+4\times10^{-5}\sin\omega t\bigr)^2}$$

To find the maximum value, put $$\theta=\omega t$$ and maximise

$$f(\theta)=\frac{|\cos\theta|} {\bigl(a+b\sin\theta\bigr)^2},\qquad a=8\times10^{-4},\;b=4\times10^{-5}$$

Within $$0\le\theta\lt2\pi$$, choose the region $$\cos\theta\gt0$$ (same magnitude repeats elsewhere).
Set $$\displaystyle\frac{d}{d\theta}\ln f(\theta)=0$$:

$$-\tan\theta-\frac{2b\cos\theta}{a+b\sin\theta}=0$$

After simplification, using $$x=\sin\theta$$:

$$b x^{2}-a x-2b=0$$

Substituting $$a=8\times10^{-4},\;b=4\times10^{-5}$$ gives

$$x^{2}-20x-2=0\; \Longrightarrow\; x=\sin\theta\approx-0.0995$$

With this $$\cos\theta=\sqrt{1-x^{2}}\approx0.995$$ and

$$d_{\text{max}}=a+b\sin\theta =8\times10^{-4}+4\times10^{-5}(-0.0995) \approx7.960\times10^{-4}\text{ m}$$

Finally, the maximum fringe speed is

$$v_{\max} =\frac{8(6\times10^{-7})(3.2\times10^{-6})(0.995)} {(7.960\times10^{-4})^{2}} \approx2.4\times10^{-5}\text{ m s}^{-1}$$

Expressing in micrometres per second:

$$v_{\max}\approx2.4\times10^{-5}\text{ m s}^{-1} =24\text{ }\mu\text{m s}^{-1}$$

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