Answer:
The shortest distance covered by the wave between the given points is [tex]\bf{4.2 \times 10^{-8}~m}[/tex].
Explanation:
The Phase velocity ([tex]v_{p}[/tex]) of any wave can be written as
[tex]v_{p} = \dfrac{\omega}{k}[/tex]
where '[tex]\omega[/tex]' is the angular velocity and '[tex]k[/tex]' is the wave number.
Also, if '[tex]f[/tex]' is the frequency of the wave, then
[tex]\omega = 2 \pi f[/tex]
Given, [tex]v_{p} = 3 \times 10^{8}~m~s^{-1}[/tex], [tex]f = 6 \times 10^{14}~Hz[/tex] and the phase difference between two points is [tex]\Delta \phi = 30^{0} = \dfrac{\pi}{6}[/tex].
If the wave travels by the shortest distance of [tex]\Delta x[/tex] between these points, then we can write
[tex]&& \Delta x \times k = \Delta \phi\\&or,& \Delta x = \dfrac{\Delta \phi}{k} = \dfrac{\Delta \phi v_{p}}{2 \pi f} = \dfrac{(\pi/6) v_{p}}{2 \pi f} = \dfrac{\pi v_{p}}{12 \pi f} = \dfrac{v_{p}}{12 f}\\&or,& \Delta x = \dfrac{3 \times 10^{8}~m~s^{-1}}{12 \times 6 \times 10^{14}~Hz} = 4.2 \times 10^{-8}~m[/tex]
[tex]\v_{p} = \dfrac{\omega}{k}[/tex][tex]\v_{p} = \dfrac{\omega}{k}[/tex]
