Question

In a phasor diagram for any point on the viewing screen for the two slit experiment in Fig 35-10, the resultant wave phasor rotates$\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{°}$in $\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{16}}\mathbf{ }\mathbf{\text{s}}$. What is the wavelength?

Short answer

Thus, the wavelength of light is $450nm$.

Step-by-step solution

According to the question.

It is given that

$$\begin{gathered} \Delta \varphi =60° \\ =\frac{\pi }{3}\text{rad} \end{gathered}$$

The figure according to the question is:

The phasors rotate with constant angular velocity

$$\begin{gathered} \mathbf{\omega }\mathbf{=}\frac{\mathbf{\Delta \varphi }}{\mathbf{\Delta t}} \\ =\frac{\pi }{3\times 2.5\times {10}^{-16} \text{s}} \\ =4.19\times {10}^{15} \text{rad/s} \end{gathered}$$

The wavelength of light in medium. 

Since, light waves travelling in a medium (air) where the wave speed is approximately $\mathit{c}$, then $\mathit{k}\mathit{c}\mathbf{=}\mathit{\omega }\mathbf{,}\mathbf{\text{where}}\mathbf{}\mathit{k}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{\lambda }}$, which leads to

$$\begin{gathered} \lambda =\frac{2\pi c}{\omega } \\ =\frac{2\mathrm{\pi }\times 3\times {10}^8\mathrm{m}/\mathrm{s}}{4.19\times {10}^{15}rad/s} \\ =4.498\times {10}^{-7}m \\ =450nm \end{gathered}$$

Hence, the wavelength of light is, $450nm$.