# Question #0f123

Feb 21, 2017

Phase angle of the light radiation.

#### Explanation:

“k” is often used for a particular constant, usually with a clarifying subscript. In this case we can use dimensional analysis to at least see what form it takes. Lambda “λ” is usually wavelength in this context, and ‘r’ would be a radius to define an circumference (2πr).

So if we use SI units with meters for length, we have a circumference divided by a wavelength will equal our ‘k’.
$k = \frac{2 \cdot \pi \cdot r}{\lambda}$.
This is a dimensionless number that is the “phase angle” of the light. (See https://www.quora.com/If-the-wave-is-completely-in-phase-the-circumference-of-the-orbit-must-be-equal-to-an-integral-multiple-of-the-wavelength-λ )
And: http://www.insula.com.au/physics/1111/L3.html

(distance from start)/wavelength = phase angle/$\left(2 \cdot \pi\right)$
$\frac{x}{\lambda} = \frac{\phi}{2} \cdot \pi$
rearranged to $\lambda = \frac{2 \cdot \pi \cdot r}{k}$. Where I have substituted ‘k’ for the ‘phi’ and ‘r’ for the ‘x’.

$k = \frac{2 \cdot \pi \cdot r}{\lambda}$.