Question

) (Problem 2.35 from text) Consider a light wave having a phase velocity of 3x10-8 m/s and a frequency of 6x1014 Hz. What is the shortest distance along the wave between any two points that have a phase difference of 300

Answer 1

The shortest distance covered by the wave between the given points is
`\bf{4.2 * 10^(-8)~m}`.

Step-by-step explanation:

The Phase velocity (
`v_(p)`) of any wave can be written as

`v_(p) = (\omega)/(k)`

where '
`\omega`' is the angular velocity and '
`k`' is the wave number.

Also, if '
`f`' is the frequency of the wave, then

`\omega = 2 \pi f`

Given,
`v_(p) = 3 * 10^(8)~m~s^(-1)`,
`f = 6 * 10^(14)~Hz` and the phase difference between two points is
`\Delta \phi = 30^(0) = (\pi)/(6)`.

If the wave travels by the shortest distance of
`\Delta x` between these points, then we can write

`&& \Delta x * k = \Delta \phi\\&or,& \Delta x = (\Delta \phi)/(k) = (\Delta \phi v_(p))/(2 \pi f) = ((\pi/6) v_(p))/(2 \pi f) = (\pi v_(p))/(12 \pi f) = (v_(p))/(12 f)\\&or,& \Delta x = (3 * 10^(8)~m~s^(-1))/(12 * 6 * 10^(14)~Hz) = 4.2 * 10^(-8)~m`

`\v_(p) = (\omega)/(k)`
`\v_(p) = (\omega)/(k)`